id stringlengths 32 32 | problem stringlengths 28 2.84k | output stringlengths 800 95.3k | source stringclasses 5
values | license stringclasses 1
value | dataset stringclasses 1
value | split stringclasses 1
value | difficulty stringclasses 30
values | Answer stringlengths 8 12.3k | wrong_cot stringlengths 109 615k ⌀ | wrong_plan stringlengths 525 65.3k | wrong_answer stringlengths 0 84.7k ⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|
8be667fb42682957580ebc058aeceb7a | Find the sum of the weights of edges of the Minimum-Cost Arborescence with the root r for a given weighted directed graph G = (V, E).
Constraints
* 1 ≤ |V| ≤ 100
* 0 ≤ |E| ≤ 1,000
* 0 ≤ wi ≤ 10,000
* G has arborescence(s) with the root r
Input
|V| |E| r
s0 t0 w0
s1 t1 w1
:
s|E|-1 t|E|-1 w|E|-1
, where |V| is the ... | <think>
Okay, I need to find the minimum-cost arborescence for a given directed graph with a specified root. Hmm, arborescence is like a directed tree where all edges point away from the root, and there's exactly one path from the root to every other node. Right, so the problem is to compute the sum of the weights of t... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | def find_min_cost_arborescence(V, E, root):
if len(V) == 1:
return 0
min_in_edges = {}
for node in V:
if node == root:
continue
min_edge = None
min_weight = float('inf')
for e in E:
if e['to'] == node and e['weight'] < min_weight:
... | I need to find the minimum-cost arborescence rooted at `r`. An arborescence is a directed tree where all edges point away from the root, and every node except the root has exactly one incoming edge.
My first idea: For each node (except the root), pick the incoming edge with the smallest weight. That should give me the... | The correct approach to finding the minimum-cost arborescence (directed minimum spanning tree rooted at a given node) involves Edmonds' algorithm, which iteratively contracts cycles formed by selecting the minimum-weight incoming edge for each non-root vertex, then recursively solving on a contracted graph. The key ins... | def solve():
n, m, r = map(int, input().split())
edges = []
for _ in range(m):
s, t, w = map(int, input().split())
edges.append((s, t, w))
# For each node except root, find the minimum incoming edge
min_in_weight = {}
min_in_edge = {}
for node in range(n):
i... |
0cb6b310e716e5df4a6b6d4eb076a6e3 | Today, as usual, Jerry comes up with a new puzzle for Tom. Jerry knows very well that Tom is well aware of the concepts of fibonacci series. So he intends to present the problem with a twist. By choosing any random integer Jerry asks Tom to tell whether the given number is a fibonacci number.
Input
The first line of t... | <think>
Okay, let's see. The problem is to determine if a given number is a Fibonacci number. Hmm. So, how do I check that efficiently, especially considering the constraints where N can be up to 1e9?
First, I remember that a number is part of the Fibonacci sequence if and only if 5*N^2 +4 or 5*N^2 -4 is a perfect squ... | codechef | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import math
def is_perfect_square(x):
s = math.isqrt(x)
return s * s == x
t = int(input())
for _ in range(t):
n = int(input())
x = 5 * n * n
if is_perfect_square(x + 4) or is_perfect_square(x - 4):
print("is fibo")
else:
print("not fibo") | I need to check if a given number is a Fibonacci number. I know the Fibonacci sequence goes: 1, 1, 2, 3, 5, 8, 13, 21, ... and so on. So, one way is to generate Fibonacci numbers one by one and see if the input number appears in the sequence.
But wait — the input can be as large as $ 10^9 $. How many Fibonacci numbers... | The correct solution leverages a mathematical property: a positive integer $ n $ is a Fibonacci number if and only if one of $ 5n^2 + 4 $ or $ 5n^2 - 4 $ is a perfect square. This is a well-known identity derived from Binet’s formula and properties of Fibonacci numbers. The key insight is to avoid generating Fibonacci ... | t = int(input())
for _ in range(t):
n = int(input())
a, b = 1, 1
found = False
while b <= n:
if b == n:
found = True
break
c = a + b
a = b
b = c
if found:
print("is fibo")
else:
print("not fibo") |
b1c1c912ac90e98b2b001e4fd8e658ef | This is yet another problem on regular bracket sequences.
A bracket sequence is called regular, if by inserting "+" and "1" into it we get a correct mathematical expression. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not. You have a pattern of a bracket sequenc... | <think>
Okay, I need to solve this problem where I have to replace the question marks in a bracket sequence with '(' or ')' such that the resulting sequence is regular, and the total cost is minimized. Let's think about how to approach this.
First, a regular bracket sequence must be balanced. That means the number of ... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import heapq
def main():
import sys
input = sys.stdin.read().split()
ptr = 0
s = input[ptr]
ptr +=1
m = s.count('?')
q_list = []
for _ in range(m):
a = int(input[ptr])
b = int(input[ptr+1])
ptr +=2
q_list.append((a, b))
open_count = s.count('(')
... | I need to make a valid bracket sequence from a pattern with `(`, `)`, and `?`. Each `?` can become `(` or `)` with given costs. I want the cheapest valid sequence.
First, I think: I can try to simulate the sequence from left to right, keeping track of the current balance. Whenever I see a `?`, I should choose the chea... | The correct solution hinges on a greedy strategy with a priority queue: we initially assume all `?` are replaced with `)` (to minimize early balance), then whenever the running balance goes negative, we flip the most cost-effective `?` (i.e., the one with the smallest `a - b`) to `(` to restore balance. After processin... | import sys
def main():
data = sys.stdin.read().split()
if not data:
return
s = data[0]
m = s.count('?')
q_list = []
ptr = 1
for _ in range(m):
a = int(data[ptr])
b = int(data[ptr+1])
ptr += 2
q_list.append((a, b))
# Greedy assignment: as... |
e21b0e5fc03c44fb2dec67e730d79e14 | There are two cameras which observe the up line and the down line respectively on the double lane (please see the following figure). These cameras are located on a line perpendicular to the lane, and we call the line 'monitoring line.' (the red line in the figure)
<image>
Monitoring systems are connected to the cam... | <think>
Okay, let's see. The problem is to find the maximum time interval where no cars passed through either of the two monitoring lines. Hmm. So the cars are passing on two different lines, up and down. The monitoring systems for each line record the times when cars pass. We need to find the longest period where neit... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | def main():
import sys
input = sys.stdin.read().split()
ptr = 0
while True:
if ptr >= len(input):
break
n = int(input[ptr])
m = int(input[ptr+1])
ptr +=2
if n ==0 and m ==0:
break
# Read tl
tl = []
if n >0:
... | I need to find the maximum time interval where no car passed through the monitoring line. Since there are two cameras, one for each lane, I think the longest gap should be the largest time between two consecutive cars on either lane. That makes sense — if a car passes at time 10 and the next at time 50, that's a 40-sec... | The correct solution hinges on recognizing that the maximum time interval with **no cars passing on either lane** corresponds to the largest **overlap** between idle periods on the up and down lanes — not the maximum gap on a single lane. The key insight is that both lanes must be idle simultaneously for a time interva... | import sys
def main():
input = sys.stdin.read().split()
ptr = 0
while True:
n = int(input[ptr])
m = int(input[ptr+1])
ptr += 2
if n == 0 and m == 0:
break
# Read up lane times
tl = []
if n > 0:
for i in range(n):
... |
880ac66e82396460e68d79e9ccf4bf84 | Adilbek's house is located on a street which can be represented as the OX axis. This street is really dark, so Adilbek wants to install some post lamps to illuminate it. Street has n positions to install lamps, they correspond to the integer numbers from 0 to n - 1 on the OX axis. However, some positions are blocked an... | <think>
Okay, let's see. The problem is about Adilbek wanting to install lamps along a street to illuminate the entire segment from 0 to n. The lamps can be of different types (power levels from 1 to k), but he has to choose exactly one type. The goal is to find the minimal total cost for this. If it's impossible, retu... | codeforces | cc-by-4.0 | code_contests | train | 11 | def main():
import sys
input = sys.stdin.read().split()
idx = 0
n = int(input[idx]); idx +=1
m = int(input[idx]); idx +=1
k = int(input[idx]); idx +=1
blocked = []
if m >0:
blocked = list(map(int, input[idx:idx+m]))
idx +=m
a = list(map(int, input[idx:idx+k]))
... | Okay, so we have a street from 0 to n, and some positions are blocked. We need to cover the whole segment [0, n] with lamps of the same power type, each costing `a_l`, and we can only use one type.
First, I need to check if position 0 is blocked. If it is, we can't start anywhere, so it's impossible. That makes sense.... | The correct solution hinges on two key insights:
1. **The minimum lamp power required** is determined by the largest gap between consecutive unblocked positions (or from the last unblocked position to `n`), because a lamp must cover the entire gap.
2. **For each valid lamp power `l` (from the minimum required to `k... | def main():
import sys
input = sys.stdin.read().split()
idx = 0
n = int(input[idx]); idx += 1
m = int(input[idx]); idx += 1
k = int(input[idx]); idx += 1
blocked = set()
if m > 0:
blocked = set(map(int, input[idx:idx+m]))
idx += m
a = list(map(int, input[idx... |
2854c700caeb36aead0d056e349ee840 | Simple Factorial
You are asked to calculate factorials of some small positive integers.
Input
Input
Tips:
An integer T, denoting the number of testcases, followed by T lines, each containing a single integer N.
Output
For each integer N given at input, output a single line the value of N!
If you have multiple... | <think>
Okay, I need to write a Python program that calculates the factorial of several numbers given as input. Let's see. The input starts with an integer T, which is the number of test cases. Then there are T lines each with a single integer N. For each N, I have to compute N! and output it.
First, I remember that t... | codechef | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | t = int(input())
for _ in range(t):
n = int(input())
fact = 1
for i in range(1, n+1):
fact *= i
print(fact) | I need to compute factorials for multiple test cases. Since N can be up to 100, and T up to 100, I should avoid recalculating the same factorial multiple times. So I’ll use memoization to store previously computed factorials. I’ll define a recursive function `factorial(n)` that returns n! and caches results in a dictio... | The correct approach requires computing the factorial of each input number using a simple loop that multiplies integers from 1 to N. The key insight is that factorials grow quickly, but since N ≤ 100, we must handle large integers (Python handles this naturally), and we must process T test cases efficiently in O(N) per... | def factorial(n, memo={0: 1, 1: 1}):
if n in memo:
return memo[n]
memo[n] = n * factorial(n - 1, memo)
return memo[n]
t = int(input())
# Precompute all factorials from 1 to 100
precomputed = [0] * 101
for i in range(1, 101):
precomputed[i] = factorial(i)
for _ in range(t):
n = int(input())... |
01ad6a2182b351828053ede3c800073c | In an attempt to make peace with the Mischievious Mess Makers, Bessie and Farmer John are planning to plant some flower gardens to complement the lush, grassy fields of Bovinia. As any good horticulturist knows, each garden they plant must have the exact same arrangement of flowers. Initially, Farmer John has n differe... | <think>
Okay, so I need to solve this programming problem. Let me read the problem statement carefully.
The problem is about Farmer John and Bessie planting flower gardens. They have some species of flowers, and each day they get a new species. After each day, we need to compute the sum over all possible choices of k ... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
MOD = 10**9 + 7
max_d_phi = 10**6
max_fact = 2 * 10**5 + 10
# Precompute smallest prime factors (spf)
spf = list(range(max_d_phi + 1))
for i in range(2, int(max_d_phi**0.5) + 1):
if spf[i] == i:
for j in range(i*i, max_d_phi + 1, i):
if spf[j] == j:
spf[j] = i
# Pre... | I need to compute, after each day, the sum over all possible choices of `k` species of the maximum number of gardens possible.
First, I understand that for any `k` species, the maximum number of gardens is the GCD of their flower counts. For example, if I have species with 4, 6, and 8 flowers, GCD(4,6,8) = 2, so I can... | The correct solution hinges on a deep insight: for a fixed number of gardens `g`, the maximum number of gardens we can make from a set of `k` flower species is the greatest common divisor (GCD) of their counts. But instead of computing GCDs for every combination of `k` species (which would be too slow), we use number t... | import sys
MOD = 10**9 + 7
def mod_pow(base, exp, mod):
result = 1
base %= mod
while exp > 0:
if exp & 1:
result = (result * base) % mod
base = (base * base) % mod
exp >>= 1
return result
def mod_inverse(a, mod):
return mod_pow(a, mod - 2, mod)
def precompute_f... |
a72176885daf756ec29facc18b89c1f7 | The famous joke programming language HQ9+ has only 4 commands. In this problem we will explore its subset — a language called HQ...
Input
The only line of the input is a string between 1 and 106 characters long.
Output
Output "Yes" or "No".
Examples
Input
HHHH
Output
Yes
Input
HQHQH
Output
No
Input
HH... | <think>
Okay, let's see. I need to solve this problem where I have to determine if a given input string outputs "Yes" or "No" based on some rules related to the HQ9+ language subset. But since the problem statement was partially destroyed, I have to infer the correct condition from the examples.
First, let's look at ... | codeforces | cc-by-4.0 | code_contests | train | interview | s = input().strip()
h_count = s.count('H')
print("Yes" if h_count % 2 == 0 else "No") | Okay, so I need to figure out whether a given string in the HQ... language is valid. The examples show that "HHHH" is "Yes" — that’s 4 H’s, which is even. "HQHQH" is "No" — that’s 3 H’s, which is odd. "HHQHHQH" is also "No" — that’s 4 H’s? Wait, let me count: H, H, Q, H, H, Q, H — that’s five H’s? No, wait: positions 0... | The correct solution hinges on the observation that the language HQ... only accepts programs where the number of 'H' commands is even. The 'Q' commands are irrelevant to the acceptance condition — they just trigger a side effect (printing "Hello, World!") but don't affect the validity of the program. The key insight is... | s = input().strip()
segments = s.split('Q')
for segment in segments:
h_count = 0
for char in segment:
if char == 'H':
h_count += 1
if h_count % 2 == 1:
print("No")
exit()
print("Yes") |
21c0161e37c8619015e0b515ce1311f1 | Valera loves to participate in competitions. Especially in programming contests. Today he has participated in the contest with his team, consisting of n students (including Valera). This contest was an individual competition, so each student in the team solved problems individually.
After the contest was over, Valera ... | <think>
Okay, let's tackle this problem. Hmm. So the task is to find a distribution of points for each student in a team that meets certain conditions. Let's see the requirements again.
The input gives us n (number of students), k (the top k students whose sum is sk), l and r (the min and max each student can have), s... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM | n, k, l, r, sall, sk = map(int, input().split())
# Compute the top k scores
x = sk // k
remainder_top = sk % k
top = [x + 1] * remainder_top + [x] * (k - remainder_top)
remaining = []
if n > k:
s_rest = sall - sk
remaining_students = n - k
base_remaining = l
sum_remaining_base = remaining_students * b... | I need to assign scores to *n* students such that:
- Each student scores between *l* and *r*.
- The sum of all scores is *sall*.
- The sum of the top *k* scores is *sk*.
My idea is: the top *k* students should get as many points as possible to "save" points for the others — wait, no, actually, if I give them too many,... | The correct approach involves constructing a valid score distribution by first assigning scores to the top *k* students such that their total is exactly *sk*, while respecting the bounds *l* and *r*, and then assigning the remaining *n−k* students with scores between *l* and *r* such that their total is *sall − sk*. Th... | n, k, l, r, sall, sk = map(int, input().split())
# Assign top k scores as evenly as possible
x = sk // k
remainder = sk % k
top = [x + 1] * remainder + [x] * (k - remainder)
# Assign remaining scores as high as possible
remaining_students = n - k
s_rest = sall - sk
# Start with minimum l for each
base = l
sum_base = ... |
820e1759338c48d99aeb0d14a3948729 | Raju has a very pretty girlfriend. He chats with her all day and night through his phone. His phone lacks a QWERTY keypad, instead it has a T9 keypad.
This takes a lot of time to type a message in his phone and he often fears that his girlfriend might get angry if he doesn't reply on time. So he sets a time limit to t... | <think>
Okay, I need to solve this programming problem. Let's read the problem carefully.
The problem is about Raju typing a message on a T9 keypad. The keypad layout is specific. The goal is to calculate how much of the message he can type within a given time Z, considering the time taken to move his thumb between ke... | hackerearth | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | # Precompute the character to (key, presses) mapping
mapping = {}
# Key 0
key0 = ['_', '0']
for i, c in enumerate(key0, 1):
mapping[c] = (0, i)
# Key 1
key1 = ['.', ',', '?', '!', '1']
for i, c in enumerate(key1, 1):
mapping[c] = (1, i)
# Keys 2-9
key_letters = {
2: 'abc',
3: 'def',
4: 'ghi',
... | I see that Raju is typing on a T9 keypad with limited keys, and each key has multiple letters. I remember that in T9, to type a letter, you press the key multiple times — for example, pressing key 2 once gives 'a', twice gives 'b', etc. So I need to map each character to its key and how many times it needs to be presse... | The correct solution requires mapping each character to its corresponding key and number of presses on a T9 keypad, then simulating the typing process by tracking the thumb’s current key position and accumulating time for movement (based on key distance) and key presses. The key insight is that movement time depends on... | # Precompute the character to (key, presses) mapping
mapping = {}
# Key 0
key0 = ['_', '0']
for i, c in enumerate(key0, 1):
mapping[c] = (0, i)
# Key 1
key1 = ['.', ',', '?', '!', '1']
for i, c in enumerate(key1, 1):
mapping[c] = (1, i)
# Keys 2-9
key_letters = {
2: 'abc',
3: 'def',
4: 'ghi',
... |
f57d1a119b3ee41dac1938362c55f6a9 | This problem has unusual memory constraint.
At evening, Igor and Zhenya the financiers became boring, so they decided to play a game. They prepared n papers with the income of some company for some time periods. Note that the income can be positive, zero or negative.
Igor and Zhenya placed the papers in a row and dec... | <think>
Okay, let's try to understand this problem. So, the game is between Igor and Zhenya. They take turns picking papers from the left and right ends. Igor starts by taking 1 or 2 papers from the left. Then, each next player must take k or k+1 papers, where k is the number the previous player took. The goal is to fi... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from functools import lru_cache
def main():
sys.setrecursionlimit(1 << 25)
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
prefix = [0] * (n + 1)
for i in range(n):
prefix[i+1] = prefix[i] + a[i]
@lru_cache(maxsize=None)
def dfs(l, r, m):
... | I see this is a two-player game with alternating moves, and both players want to optimize the final difference: Igor wants to maximize (his sum - opponent's sum), Zhenya wants to minimize it. Since both play optimally, I think this is a classic minimax problem.
The game starts with Igor taking 1 or 2 papers from the l... | The correct solution uses dynamic programming with memoization, tracking the current left and right boundaries of the remaining papers, the number of papers taken in the previous move (which determines the allowed moves for the current player), and alternating turns based on whose turn it is. The key insight is that th... | import sys
from functools import lru_cache
def main():
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
# Precompute prefix sums for O(1) range sum
prefix = [0] * (n + 1)
for i in range(n):
prefix[i+1] = prefix[i] + a[i]
@lru_cache(maxsize=None)
... |
e5c4cd3563e1018999c84483ca7498d9 | Wabbit is trying to move a box containing food for the rest of the zoo in the coordinate plane from the point (x_1,y_1) to the point (x_2,y_2).
He has a rope, which he can use to pull the box. He can only pull the box if he stands exactly 1 unit away from the box in the direction of one of two coordinate axes. He will... | <think>
Okay, I need to find the minimum time Wabbit needs to move the box from (x1, y1) to (x2, y2). Let's think about how the movement works.
First, the key is understanding how pulling the box affects both the box's position and Wabbit's movement. When Wabbit pulls the box, he moves in a direction, pulling the box ... | codeforces | cc-by-4.0 | code_contests | train | 7 | t = int(input())
for _ in range(t):
x1, y1, x2, y2 = map(int, input().split())
dx = abs(x2 - x1)
dy = abs(y2 - y1)
if dx == 0 or dy == 0:
print(dx + dy)
else:
print(dx + dy + 2) | I start by analyzing the problem. The box moves only when Wabbit pulls it, and he can only pull if he's exactly 1 unit away along a coordinate axis. When he pulls, the box moves 1 unit toward him, and he moves 1 unit further in the same direction. So each pull takes 1 second and moves the box 1 unit.
I realize that Wa... | The correct solution hinges on recognizing that when the box needs to move both horizontally and vertically (i.e., neither `dx` nor `dy` is zero), Wabbit must perform an extra "detour" to reposition himself to pull the box in the second direction. The key insight is that pulling the box in one direction requires Wabbit... | t = int(input())
for _ in range(t):
x1, y1, x2, y2 = map(int, input().split())
dx = abs(x2 - x1)
dy = abs(y2 - y1)
print(dx + dy) |
f92a87ca98fccfd4abc33838c6a50c9b | Draw a frame which has a height of H cm and a width of W cm. For example, the following figure shows a frame which has a height of 6 cm and a width of 10 cm.
........#
........#
........#
........#
Constraints
* 3 ≤ H ≤ 300
* 3 ≤ W ≤ 300
Input
The input consists of multiple datasets. Each dataset consists of tw... | <think>
Okay, I need to solve this programming problem. Let's read the problem carefully. The task is to draw a frame with given height H and width W. The example shows that for H=6 and W=10, the frame is made up of '#' and '.' characters.
First, I should understand the structure of the frame. Let's look at the examp... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | while True:
line = input().strip()
H, W = map(int, line.split())
if H == 0 and W == 0:
break
for i in range(H):
if i == 0 or i == H - 1:
print('#' * W)
else:
print('#' + '.' * (W - 2) + '#')
print() | I need to draw a frame with height H and width W. The example shows that the outer border is made of `#` and the inside is `.`. So, the first row should be all `#`, the last row should be all `#`, and the middle rows should have `#` only at the first and last positions, with `.` in between.
But I'm not sure how to do ... | The correct solution hinges on recognizing that the frame consists of only two types of rows: the top and bottom rows (all `#`), and the middle rows (with `#` at the start and end, and `.` in between). The key insight is that the frame is hollow — only the border is made of `#`, and the interior is filled with `.`. A c... | while True:
line = input().strip()
H, W = map(int, line.split())
if H == 0 and W == 0:
break
for i in range(H):
for j in range(W):
if i == 0 or i == H - 1 or j == 0 or j == W - 1:
print('#', end='')
else:
print('.', end='')
... |
55d3b18f842a7b31694102620a5bb1fd | After Misha's birthday he had many large numbers left, scattered across the room. Now it's time to clean up and Misha needs to put them in a basket. He ordered this task to his pet robot that agreed to complete the task at certain conditions. Before the robot puts a number x to the basket, Misha should answer the quest... | <think>
Okay, let's tackle this problem. Hmm. So the problem is about processing a sequence of numbers and, for each new number, determining if it can be formed by XORing some subset of the numbers already in the basket. And if yes, output the indexes of those numbers.
Wait, right. The basket starts empty. For each n... | codeforces | cc-by-4.0 | code_contests | train | competition | m = int(input())
basis = []
for i in range(m):
x = int(input().strip())
# Check phase
check_val = x
check_mask = set()
sorted_basis = sorted(basis, key=lambda elem: (-elem[0].bit_length(), elem[0]))
for elem_val, elem_mask in sorted_basis:
if check_val == 0:
break
hi... | I need to check if the current number can be written as XOR of some previous numbers. Since XOR is associative and commutative, and we’re in GF(2), I should think in terms of linear algebra.
I remember that any set of numbers can be reduced to a basis using Gaussian elimination. But I don’t want to do full elimination... | The correct approach involves maintaining a basis of linearly independent numbers in the vector space over GF(2), where each number is represented as a binary vector. The key insight is that we can use Gaussian elimination over GF(2) to maintain a basis such that any new number can be checked for linear dependence effi... | m = int(input())
basis = [] # list of (value, index_set)
for i in range(m):
x = int(input().strip())
temp_x = x
temp_set = set()
# Process basis in the order it was added (no sorting)
for val, idx_set in basis:
if temp_x == 0:
break
# Check if the highest bit of va... |
c62f2f0502787a756e934847f7cb91ae | There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies.
Snuke can perform the following operation any number of times:
* Choose a box containing at least one candy, and eat one of the candies in the chosen box.
His objective is as follows:
* Any two neighboring boxes conta... | <think>
Okay, let's see. The problem is to find the minimum number of operations needed so that any two neighboring boxes have a total of at most x candies. The operation is eating one candy from a box.
Hmm. So the key here is to adjust the pairs of adjacent boxes. For each pair (i, i+1), the sum a_i + a_{i+1} must b... | atcoder | cc-by-4.0 | code_contests | train | 0 | n, x = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in range(n - 1):
current_sum = a[i] + a[i+1]
if current_sum > x:
need = current_sum - x
subtract = min(need, a[i+1])
a[i+1] -= subtract
ans += subtract
need -= subtract
if need > 0... | I see that the goal is to make sure that every two neighboring boxes have at most `x` candies in total. So for each adjacent pair `(i, i+1)`, if their sum exceeds `x`, I need to remove some candies.
My first idea: just go through each pair and calculate how much is "over" the limit. For each pair where `a[i] + a[i+1] ... | The correct approach involves greedily fixing neighboring pairs from left to right, ensuring that each adjacent pair sums to at most `x`. The key insight is that when `a[i] + a[i+1] > x`, we should reduce the right box (`a[i+1]`) first (since it will affect the next pair), and only if necessary reduce the left box (`a[... | n, x = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in range(n - 1):
if a[i] + a[i+1] > x:
ans += a[i] + a[i+1] - x
print(ans) |
23ee3de2eeb960311f02e7e95ee83e54 | There are n parrots standing in a circle. Each parrot has a certain level of respect among other parrots, namely r_i. When a parrot with respect level x starts chattering, x neighbours to the right and to the left of it start repeating the same words in 1 second. Their neighbours then start repeating as well, and so on... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
So, the problem is about parrots arranged in a circle. Each parrot has a respect level r_i. When a parrot starts chattering, its x left and right neighbors (x being its respect level) start repeating after 1 second. The... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
def main():
n = int(sys.stdin.readline())
r = list(map(int, sys.stdin.readline().split()))
if n == 1:
print(0)
return
# Precompute log_table for sparse table
log_table = [0] * (n + 1)
for i in range(2, n + 1):
log_table[i] = log_table[i // 2] + 1
k_ma... | I need to simulate the chattering process for each starting parrot. The idea is that when a parrot starts, it spreads its chatter to its neighbors based on its respect level. But I'm not sure exactly how the spreading works.
Let me re-read the problem: "When a parrot with respect level x starts chattering, x neighbour... | The correct solution uses a sparse table to efficiently query the maximum respect level in any interval of the circular array, enabling a binary search-like propagation of chattering across the circle. The key insight is that at each second, the chattering spreads outward from the starting parrot by a distance determin... | import sys
def main():
n = int(sys.stdin.readline())
r = list(map(int, sys.stdin.readline().split()))
if n == 1:
print(0)
return
result = []
for start in range(n):
covered = [False] * n
covered[start] = True
queue = [start]
time = 0
... |
c5b4107ac8cace546d99deeb218b49a9 | You are given three positive (greater than zero) integers c, d and x.
You have to find the number of pairs of positive integers (a, b) such that equality c ⋅ lcm(a, b) - d ⋅ gcd(a, b) = x holds. Where lcm(a, b) is the least common multiple of a and b and gcd(a, b) is the greatest common divisor of a and b.
Input
The... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The problem is about finding the number of pairs (a, b) of positive integers that satisfy the equation c * lcm(a, b) - d * gcd(a, b) = x. The input gives t test cases, each with c, d, x. For each case, I have to output ... | codeforces | cc-by-4.0 | code_contests | train | HARD | import sys
def main():
input = sys.stdin.read().split()
idx = 0
t = int(input[idx])
idx += 1
max_spf = 2 * 10**7
spf = list(range(max_spf + 1))
for i in range(2, int(max_spf**0.5) + 1):
if spf[i] == i:
for j in range(i*i, max_spf + 1, i):
if spf[j] == j:... | I need to find the number of pairs $(a, b)$ such that $c \cdot \text{lcm}(a, b) - d \cdot \gcd(a, b) = x$.
First, I recall that $\text{lcm}(a, b) \cdot \gcd(a, b) = a \cdot b$, but I'm not sure if that helps directly. Maybe I can fix $\gcd(a, b) = g$, then write $a = g \cdot m$, $b = g \cdot n$, with $\gcd(m, n) = 1$.... | The correct solution hinges on a key mathematical insight: for any pair $(a, b)$, we can write $a = g \cdot m$, $b = g \cdot n$, where $g = \gcd(a, b)$, and $\gcd(m, n) = 1$. Then $\text{lcm}(a, b) = g \cdot m \cdot n$. Substituting into the equation:
$$
c \cdot \text{lcm}(a, b) - d \cdot \gcd(a, b) = x \Rightarrow c ... | import sys
def main():
input = sys.stdin.read().split()
idx = 0
t = int(input[idx])
idx += 1
# Precompute SPF up to 10^7, but we need up to 2e7, so this is wrong
max_spf = 10**7
spf = list(range(max_spf + 1))
for i in range(2, int(max_spf**0.5) + 1):
if spf[i] == i:
... |
19d436a316a23363e3cf769af595676f | You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
You can perform the followin... | <think>
Okay, let's see. The problem is to find the maximum possible length of a substring with all distinct characters after reversing at most one substring. Hmm. So I need to think about how reversing a substring can help extend the maximum length.
First, let's understand the problem. Without any reversal, the maxim... | codeforces | cc-by-4.0 | code_contests | train | HARD | def main():
import sys
from collections import defaultdict
s = sys.stdin.readline().strip()
n = len(s)
if n == 0:
print(0)
return
# Compute left and left_mask arrays
left = [0] * n
left_mask = [0] * n
current_mask = 0
start = 0
for i in range(n):
c =... | I need to find the longest substring with all distinct characters after reversing at most one substring. Since the string only uses letters from 'a' to 't', there are at most 20 distinct characters. That means the maximum possible distinct substring length is 20 — but we can do better than brute force.
First, I think:... | The correct solution hinges on efficiently computing the longest substring with distinct characters after at most one reversal, using bitmask DP and precomputed left/right maximums. The key insight is that reversing a substring can merge two disjoint segments of distinct characters if they don’t share any common letter... | def main():
s = input().strip()
n = len(s)
if n == 0:
print(0)
return
# Compute left distinct lengths and masks
left_len = [0] * n
left_mask = [0] * n
seen = set()
start = 0
for i in range(n):
c = s[i]
while c in seen:
removed = s[start]
... |
5b11d9760a5322782729439efbd53cd7 | For a given weighted graph G(V, E) and a source r, find the source shortest path to each vertex from the source (SSSP: Single Source Shortest Path).
Constraints
* 1 ≤ |V| ≤ 100000
* 0 ≤ di ≤ 10000
* 0 ≤ |E| ≤ 500000
* There are no parallel edges
* There are no self-loops
Input
An edge-weighted graph G (V, E) and the... | <think>
Okay, I need to solve this problem where I have to find the shortest paths from a source node to all other nodes in a weighted graph. The constraints are pretty tight—up to 100,000 vertices and 500,000 edges. Hmm, so the standard Dijkstra's algorithm is O((V + E) log V), which should handle that, right? Because... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import heapq
import sys
def main():
n, m, r = map(int, sys.stdin.readline().split())
adj = [[] for _ in range(n)]
for _ in range(m):
s, t, d = map(int, sys.stdin.readline().split())
adj[s].append((t, d))
INF = 1 << 60
dist = [INF] * n
dist[r] = 0
heap = []
heapq.hea... | I need to solve the shortest path from a source vertex to all other vertices in a weighted graph. Since the graph can have up to 100,000 vertices and 500,000 edges, I need an efficient algorithm. I remember Dijkstra’s algorithm uses a heap to always pick the next closest node, but I’m not sure if I need to use it. Mayb... | The correct approach to the Single Source Shortest Path (SSSP) problem in a weighted directed graph with non-negative edge weights is Dijkstra’s algorithm, which uses a priority queue (min-heap) to efficiently extract the vertex with the smallest tentative distance at each step. The key insights are:
- Use a min-heap t... | import sys
def main():
n, m, r = map(int, sys.stdin.readline().split())
edges = []
for _ in range(m):
s, t, d = map(int, sys.stdin.readline().split())
edges.append((s, t, d))
INF = float('inf')
dist = [INF] * n
dist[r] = 0
# Bellman-Ford with early termination
... |
86cc8e4dabca6efbce70aec6f8762f2e | You are given an integer number n. The following algorithm is applied to it:
1. if n = 0, then end algorithm;
2. find the smallest prime divisor d of n;
3. subtract d from n and go to step 1.
Determine the number of subtrations the algorithm will make.
Input
The only line contains a single integer n (2 ≤ ... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully. So the task is to determine how many subtraction steps are performed based on the given algorithm when starting with a number n.
The algorithm steps are:
1. If n is 0, stop.
2. Find the smallest prime divisor d of n.
3... | codeforces | cc-by-4.0 | code_contests | train | 8 | n = int(input())
if n % 2 == 0:
print(n // 2)
else:
max_div = int(n**0.5) + 1
d = None
for i in range(3, max_div, 2):
if n % i == 0:
d = i
break
if d is None:
d = n
if d == n:
print(1)
else:
print(1 + (n - d) // 2) | I need to simulate the algorithm step by step. The algorithm says: while `n > 0`, find the smallest prime divisor `d` of `n`, subtract it, and repeat. So I can just loop until `n` becomes 0, and count how many times I subtract.
But checking the smallest prime divisor every time might be slow. Maybe I can optimize it.
... | The correct solution hinges on recognizing two key insights:
1. If `n` is even, its smallest prime divisor is 2, so we subtract 2 repeatedly until reaching 0 — this takes `n // 2` steps.
2. If `n` is odd, we must find its smallest prime divisor `d`. If no such divisor exists (i.e., `n` is prime), then we subtract `... | n = int(input())
if n % 2 == 0:
print(n // 2)
else:
max_div = int(n**0.5)
d = None
for i in range(3, max_div, 2):
if n % i == 0:
d = i
break
if d is None:
d = n
if d == n:
print(1)
else:
print(1 + (n - d) // 2) |
fa257be9cfb542899c22f4ad82cc3d99 | You have initially a string of N characters, denoted by A1,A2...AN. You have to print the size of the largest subsequence of string A such that all the characters in that subsequence are distinct ie. no two characters in that subsequence should be same.
A subsequence of string A is a sequence that can be derived from A... | <think>
Okay, let's see. The problem is to find the size of the largest subsequence of a string where all characters are distinct. Hmm. So a subsequence can be formed by deleting some characters without changing the order, right?
Wait, the largest possible such subsequence would just be the maximum number of unique ch... | codechef | cc-by-4.0 | code_contests | train | 2 | t = int(input())
for _ in range(t):
s = input().strip()
print(len(set(s))) | I need to find the largest subsequence where all characters are distinct. Since a subsequence maintains order, I can't just rearrange characters — but I can skip some. So I should try to build such a subsequence greedily.
Let me think: I can go through the string from left to right, and for each character, decide whet... | The correct solution hinges on the observation that the largest subsequence with all distinct characters is simply the number of unique characters in the string, since we can always pick one occurrence of each distinct character while preserving order (by selecting the first occurrence of each character, for example). ... | t = int(input())
for _ in range(t):
s = input().strip()
seen = set()
count = 0
for char in s:
if char not in seen:
seen.add(char)
count += 1
print(count) |
f5949fb179ed5853dcfa8436121dbc31 | There is BMI (Body Mass Index) as an index showing the degree of obesity. The BMI value is calculated using the following formula.
BMI = weight (kg) / (height (m)) 2
The closer the BMI value is to the standard value, the more "ideal body shape" is considered. Therefore, if the standard value of BMI is 22, create a pr... | <think>
Okay, I need to solve this programming problem. Let's read the problem carefully.
The task is to find, for each dataset, the person whose BMI is closest to 22. If there's a tie, pick the one with the smallest reception number (p). The input has multiple datasets, each starting with n, followed by n lines of da... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
def main():
lines = [line.strip() for line in sys.stdin if line.strip()]
idx = 0
while idx < len(lines):
n = int(lines[idx])
idx += 1
if n == 0:
break
people = []
for _ in range(n):
if idx >= len(lines):
break
... | I need to find the person whose BMI is closest to 22. First, I recall that BMI = weight / (height in meters)^2. But height is given in centimeters, so I need to convert it to meters by dividing by 100. That makes sense.
Wait — but maybe I can avoid floating-point arithmetic? Let me think: if I keep height in centimete... | The correct approach requires calculating the BMI for each person using their height in meters (converted from centimeters by dividing by 100), then computing the absolute difference between their BMI and the ideal value of 22. The person with the smallest difference is the closest to the ideal body shape. In case of a... | import sys
def main():
lines = []
for line in sys.stdin:
line = line.strip()
if line:
lines.append(line)
idx = 0
while idx < len(lines):
n = int(lines[idx])
idx += 1
if n == 0:
break
best_p = None
min_diff = f... |
04da4df7694953c2f72688d297d099df | Petr is a detective in Braginsk. Somebody stole a huge amount of money from a bank and Petr is to catch him. Somebody told Petr that some luxurious car moves along the roads without stopping.
Petr knows that it is the robbers who drive the car. The roads in Braginsk are one-directional and each of them connects two in... | <think>
Okay, let's see. The problem is to find an intersection that lies on every possible cycle in the graph. Because the robbers can be driving around any cycle indefinitely, so the intersection we choose must be part of all possible cycles. Because no matter where the robbers start, their cycle must include this in... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from sys import stdin
from collections import deque
def main():
sys.setrecursionlimit(1 << 25)
n, m = map(int, stdin.readline().split())
edges = [[] for _ in range(n+1)]
reverse_edges = [[] for _ in range(n+1)]
for _ in range(m):
u, v = map(int, stdin.readline().split())
... | I need to find an intersection that all possible infinite paths will eventually reach. Since the robber moves forever and the graph is finite, they must eventually enter a cycle. So I should look for nodes that are in cycles that can't be escaped — that is, **strongly connected components (SCCs)** that have no outgoing... | The correct solution hinges on identifying strongly connected components (SCCs) in the directed graph using Kosaraju’s algorithm, then finding the unique sink SCC (one with no outgoing edges to other components). Within that sink SCC, the only valid intersection is one that lies in a single cycle — specifically, a comp... | import sys
from collections import deque
def main():
n, m = map(int, sys.stdin.readline().split())
edges = [[] for _ in range(n+1)]
reverse_edges = [[] for _ in range(n+1)]
for _ in range(m):
u, v = map(int, sys.stdin.readline().split())
edges[u].append(v)
reverse_edges[v].appen... |
e6f00c28b7fb38575286e0934cfe0cdf | Petya and Vasya arranged a game. The game runs by the following rules. Players have a directed graph consisting of n vertices and m edges. One of the vertices contains a chip. Initially the chip is located at vertex s. Players take turns moving the chip along some edge of the graph. Petya goes first. Player who can't m... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
So, the problem is about a game played on a directed graph. Petya and Vasya take turns moving a chip along the edges. Petya starts first. If a player can't move, they lose. If the game goes on for 1e6 moves, it's a draw... | codeforces | cc-by-4.0 | code_contests | train | HARD | from collections import deque
n, m = map(int, input().split())
adj = [[] for _ in range(n + 1)]
for i in range(n):
parts = list(map(int, input().split()))
ci = parts[0]
adj[i + 1] = parts[1:] if ci > 0 else []
s = int(input())
# Check for Win condition
visited = [[False] * 2 for _ in range(n + 1)]
prev_no... | I need to solve this game theory problem where Petya controls both players' moves. The goal is to determine if he can force a win or at least tie.
First, I think: if I can find a path from the starting vertex `s` to any dead-end (a vertex with no outgoing edges), and the number of moves is odd (since Petya goes first)... | The correct solution hinges on a two-phase analysis: first, checking if Petya can force a win by reaching a terminal state on his turn (i.e., after an odd number of moves), using a BFS that tracks both node and turn parity. Second, if no such winning path exists, we analyze whether a draw is possible — which occurs if ... | from collections import deque
n, m = map(int, input().split())
adj = [[] for _ in range(n + 1)]
for i in range(n):
parts = list(map(int, input().split()))
ci = parts[0]
adj[i + 1] = parts[1:] if ci > 0 else []
s = int(input())
# Step 1: Find all reachable nodes from s
reachable = set()
q = deque([s])
reac... |
ab17f1acb5b477301c1efdb52d17e12f | Reversed number is a number written in arabic numerals but the order of digits is reversed.
The first digit becomes last and vice versa. For example, if the number is 1245,it will become
5421 .Note that all the leading zeros are omitted. That means if the number ends with a zero,
the zero is lost by reversing (e.g.... | <think>
Okay, let's see. I need to solve this problem where I have to add two reversed numbers and then reverse their sum again, making sure to remove any leading zeros in the output. Hmm. So the steps are: reverse each number, add them, reverse the sum, and then output that. Wait, but the original numbers are already ... | codechef | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n = int(input())
for _ in range(n):
a, b = input().split()
rev_a = int(a[::-1])
rev_b = int(b[::-1])
total = rev_a + rev_b
rev_total = int(str(total)[::-1])
print(rev_total) | Okay, so I need to reverse two numbers, add them, and then reverse the sum. The example shows 24 and 1 → 42 + 1 = 43 → reversed is 34. That matches.
But wait — the problem says the input numbers are already reversed. So 24 means the original number was 42, and 1 means the original was 1. So I should reverse each input... | The correct approach requires reversing each input number (to recover the original numbers), adding them, then reversing the sum to produce the final output. The key insight is that we must treat the input numbers as reversed forms, so we reverse them back to get the actual values to add, then reverse the sum to output... | n = int(input())
for _ in range(n):
a, b = input().split()
# Reverse the input strings to get the original numbers
rev_a = a[::-1]
rev_b = b[::-1]
# Convert to lists of digits for manual addition
digits_a = list(rev_a)
digits_b = list(rev_b)
# Pad the shorter list with zer... |
12bd2aefdc24d44c99c1f60111792002 | In the Kingdom of AtCoder, people use a language called Taknese, which uses lowercase English letters.
In Taknese, the plural form of a noun is spelled based on the following rules:
* If a noun's singular form does not end with `s`, append `s` to the end of the singular form.
* If a noun's singular form ends with `s`... | <think>
Okay, let's see. I need to write a Python program that takes a word and forms its plural according to the given rules. Hmm. So the rules are: if the singular doesn't end with 's', add 's'. If it does end with 's', add 'es'.
Wait, the examples: like 'apple' becomes 'apples', which makes sense. 'bus' becomes 'b... | atcoder | cc-by-4.0 | code_contests | train | 0 | s = input().strip()
if s.endswith('s'):
print(s + 'es')
else:
print(s + 's') | I need to figure out how to form the plural in Taknese. The rules say:
- If the word doesn't end with `'s'`, add `'s'`.
- If it does end with `'s'`, add `'es'`.
But I'm not sure if "ends with" means just the last character or if it's about the letter `'s'` appearing somewhere in the word. Let me think about the examp... | The correct approach is straightforward: check if the input string ends with the letter `'s'`. If it does, append `'es'`; otherwise, append `'s'`. The key insight is that this rule depends solely on the last character of the string, and the solution should be efficient with O(1) time complexity for the check and O(n) o... | s = input().strip()
has_s = False
for char in s:
if char == 's':
has_s = True
break
if has_s:
print(s + 'es')
else:
print(s + 's') |
24d883511bc38cd2fc71e2c589ff54fc | Problem
There are n vertices that are not connected to any of the vertices.
An undirected side is stretched between each vertex.
Find out how many sides can be stretched when the diameter is set to d.
The diameter represents the largest of the shortest distances between two vertices.
Here, the shortest distance is the... | <think>
Okay, let's see. I need to find the maximum number of edges that can be added to a graph with n vertices such that the diameter of the resulting graph is exactly d. Hmm. Wait, the problem says when the diameter is set to d. Oh right, so the graph's diameter (the longest shortest path between any two nodes) must... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n, d = map(int, input().split())
if d == 1:
print(n * (n - 1) // 2)
elif d == 2:
print(n * (n - 1) // 2 - 1)
else:
m = n - d - 1
path_edges = d
clique_edges = m * (m - 1) // 2
connect_edges = m * (d - 1)
total = path_edges + clique_edges + connect_edges
print(total) | I need to find the maximum number of edges in a graph with `n` vertices and diameter `d`. The diameter is the longest shortest path between any two vertices.
First, I know that a complete graph has diameter 1 and the most edges: `n*(n-1)/2`. So if `d = 1`, the answer is just the complete graph.
But if `d > 1`, I can'... | The correct solution hinges on constructing a graph with maximum edges while ensuring the diameter is exactly `d`. The key insights are:
- When `d = 1`, the graph must be complete (all vertices connected), giving `n(n-1)/2` edges.
- When `d = 2`, the graph can be almost complete, but one edge must be missing to prevent... | n, d = map(int, input().split())
if d == 1:
print(n * (n - 1) // 2)
else:
m = n - d - 1
if m < 0:
m = 0
path_edges = d
clique_edges = m * (m - 1) // 2
connect_edges = m * (d - 1)
total = path_edges + clique_edges + connect_edges
print(total) |
ee3cff357eeb2c984d3bb8e002065377 | Problem G Rendezvous on a Tetrahedron
One day, you found two worms $P$ and $Q$ crawling on the surface of a regular tetrahedron with four vertices $A$, $B$, $C$ and $D$. Both worms started from the vertex $A$, went straight ahead, and stopped crawling after a while.
When a worm reached one of the edges of the tetrahe... | <think>
Okay, so I need to solve this problem where two worms crawl on a tetrahedron and determine if they end up on the same face. Let me think through how to approach this.
First, the problem says that each worm starts at vertex A and moves straight until they hit an edge, then transitions to the adjacent face, main... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import math
def get_final_face(first_edge, angle_deg, length):
angle = math.radians(angle_deg)
cos = math.cos(angle)
sin = math.sin(angle)
# Calculate the distance to the first edge
denominator = cos + (sin / math.sqrt(3))
distance_to_edge = 1.0 / denominator if denominator != 0 else float... | Okay, so I have two worms starting at vertex A, each going in a direction defined by which edge they hit first and the angle from edge AX to their path. They move along the surface, and when they hit an edge, they continue on the adjacent face — but the angle to the edge stays the same. I need to figure out which face ... | The correct solution hinges on modeling the worm’s path as a straight line on the surface of the tetrahedron, with reflections across edges preserving the angle relative to the edge. The key insight is that the path can be unfolded into a plane by reflecting the tetrahedron’s faces, and the worm’s trajectory becomes a ... | import math
def get_final_face(first_edge, angle_deg, length):
# Convert angle to radians
angle_rad = math.radians(angle_deg)
cos_val = math.cos(angle_rad)
sin_val = math.sin(angle_rad)
# Distance to first edge
denominator = cos_val + (sin_val / math.sqrt(3))
if denominator == 0:
... |
0db3cfa4c383a10579a81871e44e4d78 | In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any ... | <think>
Okay, let's try to figure out how to solve this problem. Hmm. The problem is about finding the maximum number of roads we can remove such that two pairs of cities can reach each other within certain time constraints. The roads form a connected graph, so initially all cities are reachable. But after removing som... | codeforces | cc-by-4.0 | code_contests | train | HARD | import sys
from collections import deque
def main():
n, m = map(int, sys.stdin.readline().split())
adj = [[] for _ in range(n+1)]
for _ in range(m):
a, b = map(int, sys.stdin.readline().split())
adj[a].append(b)
adj[b].append(a)
s1, t1, l1 = map(int, sys.stdin.readline().sp... | I start by reading the problem: we have a connected undirected graph with n cities and m roads. We want to destroy as many roads as possible, but still be able to go from s1 to t1 in at most l1 hours, and from s2 to t2 in at most l2 hours.
My first thought: if the shortest path from s1 to t1 is longer than l1, it's im... | The correct solution hinges on the insight that we want to **maximize the number of roads destroyed**, which is equivalent to **minimizing the number of roads we must keep**. The key idea is that we need to preserve a subgraph where both required paths (s1→t1 in ≤l1 hours, s2→t2 in ≤l2 hours) exist. The optimal strateg... | import sys
from collections import deque
def bfs_path(start, end, adj, n):
dist = [-1] * (n+1)
parent = [-1] * (n+1)
q = deque([start])
dist[start] = 0
while q:
u = q.popleft()
if u == end:
break
for v in adj[u]:
if dist[v] == -1:
dist... |
0db3cfa4c383a10579a81871e44e4d78 | In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any ... | <think>
Okay, let's tackle this problem. Hmm, the task is to find the maximum number of roads we can destroy such that two paths exist: one from s1 to t1 in l1 hours or less, and another from s2 to t2 in l2 hours or less. The roads form a connected graph, so we need to find a subgraph that satisfies these constraints a... | codeforces | cc-by-4.0 | code_contests | train | HARD | import sys
from collections import deque
def main():
n, m = map(int, sys.stdin.readline().split())
adj = [[] for _ in range(n+1)]
for _ in range(m):
a, b = map(int, sys.stdin.readline().split())
adj[a].append(b)
adj[b].append(a)
s1, t1, l1 = map(int, sys.stdin.readline().split()... | I see this problem as two separate shortest path problems: one from s1 to t1 within l1 hours, and another from s2 to t2 within l2 hours. I need to preserve enough roads so that both paths exist, and then destroy as many as possible. So the maximum number of roads I can destroy is m minus the minimum number of roads nee... | The correct solution hinges on the idea of finding the shortest paths from key nodes (s1, t1, s2, t2) and then exploring all possible "shared" paths between pairs of nodes (u, v) to minimize the total number of edges used in both required journeys. The key insight is that we can model the problem as finding a path from... | import sys
from collections import deque
def bfs(start, n, adj):
dist = [-1] * (n + 1)
q = deque([start])
dist[start] = 0
while q:
u = q.popleft()
for v in adj[u]:
if dist[v] == -1:
dist[v] = dist[u] + 1
q.append(v)
return dist
def get_pa... |
286bc26c2f8ab87cc869d6904d77a34f | This issue is a reactive issue. It is necessary to create a program that derives the correct answer by interactively responding to the program prepared on the server side.
Example
Input
Output | <think>
Okay, I need to solve this programming problem where the server sends some input and my program has to respond correctly. Let me think about the example given. The input and output weren't specified, but maybe it's something that requires interactive input handling. Oh right, the problem mentions it's a reactiv... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
for line in sys.stdin:
line = line.strip()
# Process the input line and compute the response
response = line # Replace with actual processing logic
print(response)
sys.stdout.flush() | I see that the problem involves interacting with a server, and the reference solution shows reading from `sys.stdin` line by line. But I'm not sure how the interaction works exactly. Maybe the server sends multiple lines at once, and I need to process them all before responding? That seems safer than responding immedia... | The correct approach requires reading input line by line from standard input, processing each line in real time, and immediately outputting a response—this is a classic interactive problem where the server expects a response after each input. The key insight is that the program must not wait for all input to be read be... | import sys
input_lines = []
for line in sys.stdin:
line = line.strip()
input_lines.append(line)
# Process all lines at once
responses = []
for line in input_lines:
response = line # Replace with actual processing logic
responses.append(response)
# Output all responses at once
for response in respons... |
c08e068f4d4396e169d6e06e19572f21 | Rahul has to buy a few wooden planks ,the shop he visited has 50 different types of wooden planks available.Each type of plank is marked from 1-50.The rate(per square feet) of each plank is calculated as the number marked on the
plank multiplied by sum of the digits of the number marked on the plank.He told the shopkee... | <think>
Okay, let's see. I need to solve this problem where Rahul buys some wooden planks and we have to calculate the total amount he has to pay. Hmm, let's break down the problem step by step.
First, the input includes the number of planks, the specific plank numbers, and the dimensions. Then, for each plank number,... | hackerearth | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n = int(input())
planks = list(map(int, input().split()))
length, breadth = map(int, input().split())
area = length * breadth
total = 0
for num in planks:
s = sum(int(d) for d in str(num))
rate = num * s
if 1 <= num <= 10:
discounted = rate * 0.75
elif 11 <= num <= 20:
discounted = rat... | Okay, so Rahul wants to buy `n` planks, each with a specific number. Each plank has a rate per square foot based on its number and the sum of its digits. I need to compute that rate.
First, I'll read the number of planks, then the list of plank numbers, then the dimensions. The area is length × breadth. That part is c... | The correct solution requires calculating the cost per plank based on its number and the sum of its digits, applying tiered discounts depending on the plank’s number range, rounding the discounted rate to the nearest integer, and then multiplying by the area of each plank (length × breadth). The total cost is the sum o... | n = int(input())
planks = list(map(int, input().split()))
length, breadth = map(int, input().split())
area = length * breadth
total = 0
for num in planks:
s = sum(int(d) for d in str(num))
rate = num * s
if num <= 10:
discounted = rate * 0.75
elif num <= 20:
discounted = rate * 0.7
... |
e2133ed23d70b3ffbeacbeef77277752 | You're given a tree with n vertices rooted at 1.
We say that there's a k-ary heap of depth m located at u if the following holds:
* For m = 1 u itself is a k-ary heap of depth 1.
* For m > 1 vertex u is a k-ary heap of depth m if at least k of its children are k-ary heaps of depth at least m - 1.
Denote dpk(... | <think>
Okay, so I need to solve this programming problem. Let's read the problem carefully.
The task is to compute the sum of dp_k(u) for all possible k and u. Each dp_k(u) is the maximum depth of a k-ary heap rooted at u. The definition of a k-ary heap here is a bit specific. For a node to form a k-ary heap of depth... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from sys import stdin
from collections import defaultdict, deque
def main():
sys.setrecursionlimit(1 << 25)
n = int(stdin.readline())
edges = [[] for _ in range(n+1)]
for _ in range(n-1):
u, v = map(int, stdin.readline().split())
edges[u].append(v)
edges[v].append(u)
... | I see this is a tree problem where we need to compute `dpk(u)` — the maximum depth of a k-ary heap rooted at `u`. The definition says that for depth 1, just the node itself works. For depth >1, we need at least `k` children that each have a k-ary heap of depth at least `m-1`.
So for each node, I need to figure out, fo... | The correct solution hinges on a post-order traversal to compute `dpk(u)` — the maximum depth of a k-ary heap rooted at `u` — for each node and each possible `k`. The key insight is that for a node `u` to have a k-ary heap of depth `m > 1`, at least `k` of its children must have k-ary heaps of depth at least `m-1`. Thi... | import sys
from collections import deque
def main():
n = int(sys.stdin.readline())
graph = [[] for _ in range(n+1)]
for _ in range(n-1):
u, v = map(int, sys.stdin.readline().split())
graph[u].append(v)
graph[v].append(u)
parent = [0] * (n+1)
children = [[] for _ in rang... |
41a4d13e6590bab8747785efccc2f1d6 | You have a set of n weights. You know that their masses are a_1, a_2, ..., a_n grams, but you don't know which of them has which mass. You can't distinguish the weights.
However, your friend does know the mass of each weight. You can ask your friend to give you exactly k weights with the total mass m (both parameters ... | <think>
Okay, so I need to solve this programming problem where the goal is to determine the maximum number of weights we can uniquely identify after making a single query. Let's see. The problem statement says that we can ask for exactly k weights with total mass m, and if possible, the friend will point to a valid su... | codeforces | cc-by-4.0 | code_contests | train | HARD | n = int(input())
a = list(map(int, input().split()))
from itertools import combinations
from collections import Counter
max_reveal = 0
for k in range(1, n + 1):
sum_map = {}
for subset in combinations(a, k):
s = sum(subset)
cnt = Counter(subset)
if s not in sum_map:
sum_map... | I need to find the maximum number of weights whose masses I can determine with one query. The query lets me pick `k` weights and ask for total mass `m`. My friend gives me any valid subset if it exists.
I think: if I can find a value `m` and a size `k` such that **only one combination of `k` weights sums to `m`**, the... | The correct approach hinges on the insight that a single query can only reveal the masses of a subset of weights if the total mass `m` and size `k` uniquely determine the multiset of masses in the selected subset. If multiple different multisets of size `k` can sum to the same `m`, then the query is ambiguous and canno... | from itertools import combinations
n = int(input())
a = list(map(int, input().split()))
max_reveal = 0
for k in range(1, n + 1):
sum_count = {}
for combo in combinations(a, k):
s = sum(combo)
sum_count[s] = sum_count.get(s, 0) + 1
for s in sum_count:
if sum_count[s] == 1:
... |
af43308eec9ef1811dfcd4b9f37c5cf1 | Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>.
Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 a... | <think>
Okay, let's tackle this problem. Hmm. The goal is to find the minimal number of operations to make the sequence beautiful, where the GCD of all elements is greater than 1. If it's not possible, output NO.
First, I need to understand what the operation does. The operation allows us to choose an index i (1-base... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM_HARD | import math
n = int(input())
a = list(map(int, input().split()))
def compute_gcd(arr):
current_gcd = arr[0]
for num in arr[1:]:
current_gcd = math.gcd(current_gcd, num)
if current_gcd == 1:
break
return current_gcd
gcd = compute_gcd(a)
if gcd > 1:
print("YES")
print(0... | I start by reading the problem. We can replace two adjacent elements `ai, ai+1` with `ai - ai+1` and `ai + ai+1`. The goal is to make the GCD of the whole sequence > 1, preferably using as few moves as possible.
First, I check if the GCD of the original array is already > 1. If yes, answer is 0. That seems straightfor... | The correct solution hinges on two key insights:
1. If the GCD of the original array is already greater than 1, no moves are needed.
2. Otherwise, we must consider parity (even/odd) of the numbers. The operation `ai, ai+1 → ai - ai+1, ai + ai+1` preserves the sum of the array modulo 2, and importantly, it allows us... | import math
n = int(input())
a = list(map(int, input().split()))
def compute_gcd(arr):
current_gcd = arr[0]
for num in arr[1:]:
current_gcd = math.gcd(current_gcd, num)
if current_gcd == 1:
break
return current_gcd
gcd = compute_gcd(a)
if gcd > 1:
print("YES")
print(0... |
8f93923a41942f7599875c7aac1c087e |
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of points on a plane.
Each of the next n lines contains two real coordinates xi and yi of the <image> point, specified with exactly 2 fractional digits. All coordinates are between - 1000 and 1000, inclusive.
Output
Output a single real ... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
So, the task is to find the angle θ such that when all points are rotated clockwise around the origin by θ degrees, the maximum y-coordinate among all points minus the minimum y-coordinate is minimized. And we need to o... | codeforces | cc-by-4.0 | code_contests | train | 13 | import math
n = int(input())
points = [tuple(map(float, input().split())) for _ in range(n)]
min_spread = float('inf')
best_theta = 0.0
for i in range(36000):
theta = i * 0.01
rad = math.radians(theta)
sin_t = math.sin(rad)
cos_t = math.cos(rad)
current_ys = []
for x, y in points:
... | I see that we need to find the angle θ that minimizes the vertical spread of the points after rotating the coordinate system. The idea is to rotate all points by θ and then compute the difference between the highest and lowest y-coordinates — that’s the spread. We want the smallest possible spread.
I recall that the o... | The correct approach involves rotating the coordinate system by an angle θ and measuring the vertical spread (difference between max and min y-coordinates after rotation). The goal is to find the angle θ that minimizes this spread, which corresponds to the orientation where the points are most "compact" vertically — es... | import math
n = int(input())
points = [tuple(map(float, input().split())) for _ in range(n)]
min_spread = float('inf')
best_theta = 0.0
for i in range(36000):
theta = i * 0.01 # in degrees
sin_t = math.sin(theta) # WRONG: theta is in degrees, but math.sin expects radians
cos_t = math.cos(theta) # WRON... |
555516a5097ca87b001951033f937f41 | Positive integer x is called divisor of positive integer y, if y is divisible by x without remainder. For example, 1 is a divisor of 7 and 3 is not divisor of 8.
We gave you an integer d and asked you to find the smallest positive integer a, such that
* a has at least 4 divisors;
* difference between any two di... | <think>
Okay, let's tackle this problem. So, the task is to find the smallest positive integer 'a' for each test case such that two conditions are met: 'a' has at least 4 divisors, and the difference between any two divisors is at least 'd'. Hmm.
Wait, the example helps. For d=1, the answer is 6. Its divisors are 1,2,... | codeforces | cc-by-4.0 | code_contests | train | 8 | import sys
import math
def sieve(max_limit):
sieve = [True] * (max_limit + 1)
sieve[0] = sieve[1] = False
for i in range(2, int(math.sqrt(max_limit)) + 1):
if sieve[i]:
sieve[i*i : max_limit + 1 : i] = [False] * len(sieve[i*i : max_limit + 1 : i])
return sieve
# Precompute sieve up... | I need to find the smallest positive integer `a` such that it has at least 4 divisors, and the difference between any two divisors is at least `d`.
First, I recall that the divisors of a number come in pairs. For example, if `a = 12`, divisors are `[1, 2, 3, 4, 6, 12]`. The condition requires that the difference betwe... | The correct approach hinges on constructing a number `a` with at least four divisors such that the difference between any two divisors is at least `d`. The optimal strategy is to use two distinct primes `p` and `q`, where `p ≥ 1 + d` and `q ≥ p + d`, and set `a = p * q`. This gives exactly four divisors: `1, p, q, pq`,... | import sys
import math
def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
for i in range(3, int(math.isqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
t = int(sys.stdin.readline())
for _ in range(t):
d... |
d3017fa464658a2fead5afe037523d54 | Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a va... | <think>
Okay, let's see. The problem is to find the minimum number of heaters needed to warm up the entire house array. Each heater, when turned on, can cover a range of positions. The key here is to select the heaters in such a way that their ranges cover all positions, using as few heaters as possible.
First, I need... | codeforces | cc-by-4.0 | code_contests | train | 8 | n, r = map(int, input().split())
a = list(map(int, input().split()))
heaters = [i for i, val in enumerate(a) if val == 1]
if not heaters:
print(-1)
else:
intervals = []
for pos in heaters:
start = max(0, pos - r + 1)
end = min(n-1, pos + r - 1)
intervals.append((start, end))
in... | I need to warm up the whole house with as few heaters as possible. The house is an array of size n, and each heater at position i can warm up positions from `i - r + 1` to `i + r - 1`. So each heater covers a range of length `2*r - 1`, centered at its position.
My first idea: go from left to right. For each position, ... | The correct approach involves modeling each heater as an interval it can cover, then using a greedy interval covering strategy: sort the intervals by their starting position, and at each step, extend coverage as far as possible by choosing the heater whose interval starts before or at the current uncovered position and... | n, r = map(int, input().split())
a = list(map(int, input().split()))
# Find all heater positions
heaters = []
for i in range(n):
if a[i] == 1:
heaters.append(i)
# If no heaters, impossible
if not heaters:
print(-1)
else:
current_end = -1
count = 0
i = 0
# Go through the house from left... |
05152813232aa743166aa5d67e13d9bc | The state space of the output of this problem (and as a matter of fact, all the problems in this Cook-Off) is - 2 to the power T -
where T is the number of test cases (so be extra careful!). Each test case consists of T lines consisting of "YES" or "NO".
If a test case accurately represents the output that you would pr... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The task is to determine for each of the T test cases whether the candidate output (the input's test cases) matches the correct output that my code would produce. For each test case in the input, I have to output "YES" ... | codechef | cc-by-4.0 | code_contests | train | MEDIUM | T = int(input())
test_cases = []
for _ in range(T):
lines = [input().strip() for _ in range(T)]
test_cases.append(lines)
correct = None
for candidate in test_cases:
valid = True
for j in range(T):
expected = "YES" if test_cases[j] == candidate else "NO"
if candidate[j] != expected:
... | Okay, so I need to figure out which of the T candidate outputs is the actual correct one. The problem says that if my output matches the input for a test case, I should print "YES", otherwise "NO". But the output I print depends on which candidate is correct.
Wait — the input gives me T test cases, each with T lines o... | The correct approach hinges on finding a *single consistent output* across all test cases such that for each test case `j`, the output for that case is `"YES"` if and only if the candidate output for case `j` matches the actual correct output. This is a self-referential consistency check: the correct output must be one... | T = int(input())
test_cases = []
for _ in range(T):
lines = [input().strip() for _ in range(T)]
test_cases.append(lines)
# Pick the first test case as the correct output
correct = test_cases[0]
# For each test case, print "YES" if it matches the correct output, else "NO"
for i in range(T):
if test_cases[i... |
591d569b69fd5363f43a78cfed8f9816 | You are given a positive integer n. Output its prime factorization.
If n = a1b1 a2b2 ... akbk (bi > 0), where ak are prime numbers, the output of your program should look as follows: a1*...*a1*a2*...*a2*...*ak*...*ak, where the factors are ordered in non-decreasing order, and each factor ai is printed bi times.
Input... | <think>
Okay, I need to find the prime factors of a given integer n and output them in the specified format. Let's think about how to approach this.
First, the problem says to break down n into its prime factors and list each prime as many times as its exponent, all multiplied together with asterisks. For example, 245... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM_HARD | n = int(input())
factors = []
i = 2
while i * i <= n:
while n % i == 0:
factors.append(i)
n //= i
i += 1
if n > 1:
factors.append(n)
print('*'.join(map(str, factors))) | I need to find the prime factorization of n. The output should list each prime factor repeated according to its exponent, in non-decreasing order, separated by asterisks.
First, I know that prime factorization means breaking n down into primes. So I need to find all prime numbers that divide n and count how many times... | The correct approach involves trial division by integers starting from 2, dividing out each prime factor completely before moving to the next, and handling the remaining number greater than 1 as a final prime factor. The key insight is that we only need to check divisors up to √n, since if n has a factor greater than √... | n = int(input())
factors = []
i = 2
while i <= n:
if n % i == 0:
factors.append(i)
n //= i
else:
i += 1
print('*'.join(map(str, factors))) |
78b6828f712006ffaf2bc5c5efc4bb79 | After hard work Igor decided to have some rest.
He decided to have a snail. He bought an aquarium with a slippery tree trunk in the center, and put a snail named Julia into the aquarium.
Igor noticed that sometimes Julia wants to climb onto the trunk, but can't do it because the trunk is too slippery. To help the sna... | <think>
Okay, let's see. So the problem is about a snail trying to climb up a tree trunk using ropes. Each rope has a lower end at li and a higher end at ri. The snail can move down the trunk at any time, but can only move up via the ropes. The key is that the snail can't go below x or above y. For each query (x, y), w... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import bisect
class SegmentTree:
def __init__(self, data):
self.n = len(data)
self.size = 1
while self.size < self.n:
self.size <<= 1
self.tree = [[] for _ in range(2 * self.size)]
for i in range(self.n):
self.tree[self.size + i] = [data[i]]
f... | I start by reading the problem. The snail can start at height `x`, and can move up only via ropes — but only from the lower end to the higher end. It can’t jump down or climb down freely unless it's on a rope. But the problem says it can move down anywhere on the trunk, so the snail can descend freely to any height ≥ x... | The correct solution hinges on efficiently answering queries about the maximum reachable height given constraints on minimum (x) and maximum (y) bounds, using ropes that allow upward movement from `li` to `ri`. The key insight is that ropes can only be used to climb *up* from their lower end, and once used, the snail c... | import sys
def main():
input = sys.stdin.read().split()
ptr = 0
n = int(input[ptr])
ptr += 1
m = int(input[ptr])
ptr += 1
ropes = []
for _ in range(m):
l = int(input[ptr])
r = int(input[ptr+1])
ptr += 2
ropes.append((l, r))
q = int(input[ptr])
... |
b84d42f407ed4455970fb0686cc4b0c0 | Volodya is an odd boy and his taste is strange as well. It seems to him that a positive integer number is beautiful if and only if it is divisible by each of its nonzero digits. We will not argue with this and just count the quantity of beautiful numbers in given ranges.
Input
The first line of the input contains the ... | <think>
Okay, I need to solve this problem where I have to count the number of beautiful numbers between li and ri for multiple test cases. A beautiful number is divisible by each of its non-zero digits. Hmm, the constraints are up to 1e18, which is really big, so brute-forcing each number in that range isn't feasible.... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from math import gcd
def compute_divisors():
divisors = set()
for a in range(4):
for b in range(3):
for c in range(2):
for d in range(2):
num = (2**a) * (3**b) * (5**c) * (7**d)
divisors.add(num)
return sorted(divisors)
... | I need to count how many numbers in [L, R] are divisible by each of their nonzero digits. Let's think about this.
First, I know that checking every number from L to R is impossible because the range can be up to 9×10¹⁸. So I can't iterate through them.
I recall digit DP from previous problems. The idea is to build th... | The correct solution hinges on a few key insights:
1. A number is "beautiful" if it's divisible by each of its nonzero digits.
2. The LCM of all digits 1–9 is 2520, so any number divisible by 2520 is divisible by all digits 1–9. But more importantly, the LCM of the nonzero digits in a number must divide the number ... | null |
7da7d846b4cadb56abab62f4c42c601c | The evil mastermind, Moriarty, strikes again! He sent Sherlock an image of a wall, with N integers painted on it. Sherlock is asked to solve the puzzle and find the next clue.
Sherlock knows that Moriarty is a deceitful person. All the numbers painted on the wall are not valid. Only the numbers that are Coprime with t... | <think>
Okay, I need to solve this problem where I have to find all the numbers in an array that are coprime with a given X. Let's think about how to approach this.
First, what does coprime mean? Oh right, two numbers are coprime if their greatest common divisor (GCD) is 1. So for each number in the array A, I need to... | hackerearth | cc-by-4.0 | code_contests | train | EASY | import math
t = int(input())
for _ in range(t):
n, x = map(int, input().split())
a = list(map(int, input().split()))
valid = [num for num in a if math.gcd(num, x) == 1]
print(' '.join(map(str, valid)) if valid else -1) | I need to find numbers in the array that are coprime with X. Two numbers are coprime if they share no common prime factors. So, I should first find the prime factors of X. But I don't need to factorize X completely—maybe I can just check divisibility by small primes like 2, 3, and 5, since those are the most common. If... | The correct approach requires checking for each number in the array whether it is coprime with X, i.e., their greatest common divisor (GCD) is 1. The most efficient way to do this is using the built-in `math.gcd` function. However, a flawed reasoning path might incorrectly assume that coprimality can be determined by c... | def get_small_primes(limit):
sieve = [True] * (limit + 1)
sieve[0] = sieve[1] = False
for i in range(2, int(limit**0.5) + 1):
if sieve[i]:
for j in range(i*i, limit + 1, i):
sieve[j] = False
return [i for i in range(2, limit + 1) if sieve[i]]
small_primes = get_small... |
5bf0b31ecc4fdf84586bfbe5132422f7 | The Great Mushroom King descended to the dwarves, but not everyone managed to see him. Only the few chosen ones could see the King.
We know that only LCM(k2l + 1, k2l + 1 + 1, ..., k2r + 1) dwarves can see the Great Mushroom King. Numbers k, l, r are chosen by the Great Mushroom King himself in some complicated manner... | <think>
Okay, let's see. The problem is to compute the LCM of a sequence of numbers generated by ki^(2^x) +1 for x from li to ri, modulo a prime pi. Then, output that LCM mod pi. Hmm. So first, I need to understand how to compute this efficiently, especially given the constraints where t can be up to 1e5 and li and ri ... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
import math
def multiplicative_order(a, p):
if math.gcd(a, p) != 1:
return -1
order = p - 1
temp = order
factors = {}
for i in [2] + list(range(3, int(math.isqrt(temp)) + 1, 2)):
while temp % i == 0:
factors[i] = factors.get(i, 0) + 1
temp //= i
... | I see that we need to compute the LCM of numbers of the form $ k^{2^i} + 1 $ for $ i $ from $ l $ to $ r $, modulo a prime $ p $. Since $ l $ and $ r $ can be as large as $ 10^{18} $, I can't iterate over all $ i $ from $ l $ to $ r $ — that would be impossible. So I need a smarter way.
I recall that in modular arithm... | The correct solution hinges on understanding that the number of dwarves who can see the King is the LCM of the sequence $ k^{2^l} + 1, k^{2^{l+1}} + 1, \dots, k^{2^r} + 1 $, taken modulo a prime $ p $. The key insight is that these numbers are of the form $ k^{2^i} + 1 $, and when considered modulo a prime $ p $, their... | import sys
def solve():
input = sys.stdin.read().split()
t = int(input[0])
idx = 1
for _ in range(t):
ki = int(input[idx])
li = int(input[idx+1])
ri = int(input[idx+2])
p = int(input[idx+3])
idx += 4
if p == 2:
if ki % 2 == 1:
... |
0b2342e2e3c7a81ecda6f5ceae35c582 | As you all know, the plum harvesting season is on! Little Milutin had his plums planted in an orchard that can be represented as an n by m matrix. While he was harvesting, he wrote the heights of all trees in a matrix of dimensions n by m.
At night, when he has spare time, he likes to perform various statistics on his... | <think>
Okay, let's tackle this problem. Hmm, the goal is to find the minimal height in the entire matrix, but there's this special property about the leftmost minima in each row. The problem states that for each row i, the leftmost minimum's column L(i) is <= L(i+1). And even in any submatrix formed by selecting some ... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
n, m = map(int, sys.stdin.readline().split())
current_min = float('inf')
prev_col = 0 # Initially, for the first row, start from column 1
for i in range(1, n + 1):
start_col = prev_col if prev_col != 0 else 1
min_val = None
min_col = start_col
# Iterate from start_col to m to find the left... | I see that the orchard is an $ n \times m $ matrix, and I need to find the minimum value. The problem says there's a special property: for each row, the leftmost column with the minimum height in that row, say $ L(i) $, satisfies $ L(i) \leq L(i+1) $. That means the leftmost minimum column never moves left as we go dow... | The correct approach hinges on leveraging the key monotonicity property: the leftmost minimum column $ L(i) $ in each row is non-decreasing across rows, i.e., $ L(i) \leq L(i+1) $. This allows us to efficiently search for the global minimum by scanning each row starting from the column of the previous row’s minimum, av... | import sys
n, m = map(int, sys.stdin.readline().split())
current_min = float('inf')
for i in range(1, n + 1):
# For each row, scan from column 1 to m to find the leftmost minimum
min_val = None
min_col = 1
for j in range(1, m + 1):
print(f'? {i} {j}')
sys.stdout.flush()
val = i... |
0d66a3cece19162d35e0ce61bc991a65 | For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print the value with the specified $... | <think>
Okay, I need to solve this programming problem. Let's read through the problem statement carefully to understand the requirements.
So the problem is about handling a dictionary that maps string keys to integer values. We need to perform several operations: insert, get, delete, and dump. Each operation has spec... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
from bisect import bisect_left, bisect_right
def main():
q = int(sys.stdin.readline())
M = {}
sorted_keys = []
for _ in range(q):
query = sys.stdin.readline().split()
if not query:
continue
if query[0] == '0':
key = query[1]
x ... | I need to store key-value pairs with unique keys and support four operations: insert, get, delete, and dump. I'll use a dictionary since it's efficient for insert, get, and delete — O(1) average case.
For `get`, I can just use `M.get(key, 0)` — that's straightforward.
For `delete`, I can just remove the key from the ... | The correct solution efficiently maintains a dictionary `M` for O(1) average-case insert, get, and delete operations, while keeping a separate sorted list `sorted_keys` to support fast range queries via binary search. The key insight is that `dump(L, R)` requires all keys in lexicographic range `[L, R]`, which can be f... | import sys
def main():
q = int(sys.stdin.readline())
M = {}
for _ in range(q):
query = sys.stdin.readline().split()
if not query:
continue
if query[0] == '0':
key = query[1]
x = int(query[2])
M[key] = x
elif query[0] == '1... |
de468d3f44e40873ce02103d5e55088c | Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of tree... | <think>
Okay, let's see. The problem is to determine if there's more than one rooted tree that matches a given sequence a0, a1, ..., ah. If there's only one, output "perfect". Otherwise, output two different trees in the parent array format.
Hmm. The key is to find two different structures that have the same a_i seque... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM | h = int(input())
a = list(map(int, input().split()))
n = sum(a)
parents = [0] # Parent array for the first tree
split_level = -1
# Check for split_level
for i in range(h):
if a[i] >= 2 and a[i + 1] >= 2:
split_level = i
break
if split_level == -1:
print("perfect")
else:
# Build first tree... | I need to build two different rooted trees that have the same level counts `a0, a1, ..., ah`. The idea is: if I can assign children to parents in two different ways while keeping the number of nodes per level the same, then the trees are non-isomorphic and I can output them.
At first, I think: “If any level has more t... | The correct approach hinges on recognizing that when a level `i` has at least 2 nodes and the next level `i+1` also has at least 2 nodes, we can construct two non-isomorphic trees by rearranging how children are assigned to parents — specifically, by changing one child’s parent from the first node of the current level ... | null |
f57d1a119b3ee41dac1938362c55f6a9 | This problem has unusual memory constraint.
At evening, Igor and Zhenya the financiers became boring, so they decided to play a game. They prepared n papers with the income of some company for some time periods. Note that the income can be positive, zero or negative.
Igor and Zhenya placed the papers in a row and dec... | <think>
Okay, let's see. I need to solve this problem where two players are taking papers from either end, and the goal is to compute the maximum possible difference between Igor's sum and Zhenya's sum when both play optimally. Hmm.
First, let's understand the game mechanics. Igor starts by taking 1 or 2 papers from ... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from functools import lru_cache
def main():
sys.setrecursionlimit(1 << 25)
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + a[i]
suffix = [0] * (n + 2)
for i in range(n, 0,... | I see this is a two-player game with alternating moves, and both players play optimally. Igor goes first and can take 1 or 2 papers from the left. Then, the number of papers a player can take is based on what the opponent just took — they must take `k` or `k+1` papers, where `k` is the number the previous player took.
... | The correct solution uses dynamic programming with memoization, tracking the current segment of papers `[i, j]`, the number of papers `k` the previous player took (which constrains the current player’s move), and whose turn it is (Igor on the left, Zhenya on the right). The key insight is that the move size is determin... | import sys
from functools import lru_cache
def main():
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + a[i]
@lru_cache(maxsize=None)
def dp(i, j, is_igor_turn):
if i > j:
... |
4886cb31eb88d09f7fd3624927c82195 | This problem is interactive.
You should guess hidden number x which is between 1 and M = 10004205361450474, inclusive.
You could use up to 5 queries.
In each query, you can output an increasing sequence of k ≤ x integers, each between 1 and M, inclusive, and you will obtain one of the following as an answer:
* ei... | <think>
Okay, I need to solve this problem where I have to guess a hidden number x between 1 and M=10004205361450474 using up to 5 queries. The key is to structure each query such that each response narrows down the possible range as much as possible.
Hmm, the problem statement says that each query is a list of numbe... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | def compute_max(q):
if q == 0:
return 0
return (10**4 + 1) * compute_max(q - 1) + 10**4
def generate_query(low, remaining_queries):
if remaining_queries == 0:
return []
k = min(low, 10**4)
step = compute_max(remaining_queries - 1)
res = []
current = low
for _ in range(k)... | I see this is an interactive problem where I need to guess a number between 1 and M = 10^16. I have only 5 queries, and each query can give me a lot of information — I can send up to 10,000 numbers at once. That’s a lot! So I should use that.
But how? I can’t just send all numbers — that’s impossible. I need to narrow... | The correct approach hinges on a recursive binary-search-like strategy where each query partitions the search space using a carefully chosen sequence of numbers, leveraging the fact that we can make up to 5 queries and each query can contain up to 10⁴ numbers. The key insight is that with `q` remaining queries, the max... | import sys
def main():
low = 1
high = 10004205361450474
remaining_queries = 5
while remaining_queries > 0:
# Calculate step size for 10000 evenly spaced numbers
k = 10000
if high - low + 1 < k:
k = high - low + 1
# Generate k evenly spaced numbe... |
97ac2d24138ba12314f87b0bde46c92d | Bimal decided to join a non Hill'ffair club because he is not interested in cultural activities at
all. He thought that joining a technical club would help him in gaining technical prowess. However, on joining
the most technical club that he could find, he realised that most of the work in all technical clubs is
not te... | <think>
Okay, let's see. The problem is to compute a very large number modulo 1e9+7. The number can have up to 1e5 digits, so we can't convert it to a normal integer because that would be way too big and cause overflow or take too much time. Oh right, in Python, integers can be big, but converting a string with 1e5 dig... | hackerearth | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | MOD = 10**9 + 7
T = int(input())
for _ in range(T):
n = int(input()) # Read and ignore the number of digits
num = input().strip()
result = 0
for c in num:
result = (result * 10 + int(c)) % MOD
print(result) | I see that the number can have up to $10^5$ digits, which is huge — way more than what fits in a normal int in most languages. But wait, Python handles big integers automatically! So maybe I don’t need to worry about overflow or modular arithmetic during processing. I can just read the number as a string, convert it to... | The correct approach involves processing the large number digit by digit while maintaining the result modulo $10^9 + 7$. This works because of the modular arithmetic property:
$$
(ab + c) \mod m = ((a \mod m) \cdot b + c) \mod m
$$
This allows us to avoid storing the entire number (which could be up to $10^5$ digits ... | MOD = 10**9 + 7
T = int(input())
for _ in range(T):
n = int(input()) # Read and ignore the number of digits
num_str = input().strip()
num = int(num_str)
print(num % MOD) |
a88c80a7ffe80f7f6abaadd6e9e7351b | Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is ... | <think>
Okay, let's tackle this problem. So the task is to determine for multiple queries whether a substring of S can be transformed into a substring of T using the given operations. The operations are a bit tricky. Let's see.
First, the operations allowed are:
1. Replace a single character with its double (A become... | atcoder | cc-by-4.0 | code_contests | train | 0 | def compute_prefix(s):
n = len(s)
a = [0] * (n + 1)
b = [0] * (n + 1)
for i in range(n):
a[i+1] = a[i] + (s[i] == 'A')
b[i+1] = b[i] + (s[i] == 'B')
return a, b
s = input().strip()
t = input().strip()
q = int(input())
a_s, b_s = compute_prefix(s)
a_t, b_t = compute_prefix(t)
for _... | I see that we can replace `A` with `BB` and `B` with `AA`, and also remove `AAA` or `BBB`. These operations seem like they could be used to transform one string into another. My first idea is to simulate the process: for each query, I’ll try to reduce both substrings to their "simplest" form and see if they match.
I t... | The correct solution hinges on recognizing that the two operations preserve a certain invariant modulo 3: specifically, the value `(count_A + 2 * count_B) mod 3` remains unchanged under both operations. This invariant allows us to compare substrings efficiently by computing this value for both substrings and checking e... | def reduce_string(s):
seen = set()
stack = [s]
while stack:
curr = stack.pop()
if curr in seen:
continue
seen.add(curr)
# Try to remove any AAA or BBB
found = False
for i in range(len(curr) - 2):
if curr[i] == curr[i+1] == curr[i+2]:
... |
36b83185981de122e14e1f644066d56d | Ivan is a programming teacher. During the academic year, he plans to give n lectures on n different topics. Each topic should be used in exactly one lecture. Ivan wants to choose which topic will he explain during the 1-st, 2-nd, ..., n-th lecture — formally, he wants to choose some permutation of integers from 1 to n ... | <think>
Okay, let's tackle this programming problem. Hmm, the problem statement is a bit complex, so I need to parse it carefully.
So, Ivan needs to arrange n topics into a permutation q. There are two main constraints here. First, the prerequisite structure: each topic (except one) has a prerequisite p_i, which must ... | codeforces | cc-by-4.0 | code_contests | train | 11 | n, k = map(int, input().split())
p = list(map(int, input().split())) # p[i] is the prerequisite of topic (i+1)
root = None
for i in range(n):
if p[i] == 0:
root = i + 1
break
children = [[] for _ in range(n + 1)]
for i in range(n):
topic = i + 1
parent = p[i]
if parent != 0:
ch... | I understand that we have a tree of prerequisites: each topic (except one) has a parent, and the root has no parent. So I need to do a topological sort on this tree. I remember that topological sort can be done using DFS or Kahn’s algorithm (in-degree based). Since it's a tree, there's only one path from root to each n... | The correct solution hinges on two key insights:
1. The prerequisite constraints form a rooted tree (with the topic having `p_i = 0` as the root), and any valid topological order must respect parent-before-child relationships.
2. The special pairs `(x, y)` must appear **consecutively** in the lecture order — that i... | n, k = map(int, input().split())
p = list(map(int, input().split()))
root = None
for i in range(n):
if p[i] == 0:
root = i + 1
break
children = [[] for _ in range(n + 1)]
for i in range(n):
topic = i + 1
parent = p[i]
if parent != 0:
children[parent].append(topic)
# Check speci... |
088d197cc8f688b14935b9ef36665d9a | Rebel spy Heidi has just obtained the plans for the Death Star from the Empire and, now on her way to safety, she is trying to break the encryption of the plans (of course they are encrypted – the Empire may be evil, but it is not stupid!). The encryption has several levels of security, and here is how the first one lo... | <think>
Okay, let's see. I need to solve this problem where Heidi has to split an array into two non-overlapping contiguous parts, and maximize the sum of their modulo p scores. Hmm.
So the problem says that each part must have at least one element, and they are contiguous and non-overlapping. So the split is like div... | codeforces | cc-by-4.0 | code_contests | train | 9 | import bisect
n, p = map(int, input().split())
A = list(map(int, input().split()))
# Compute left_max
prefix_sum = [0]
for num in A:
prefix_sum.append((prefix_sum[-1] + num) % p)
sorted_left = [0]
left_max = []
for i in range(n):
x = prefix_sum[i+1]
# Compute candidates
min_y = sorted_left[0]
cand... | I need to split the array into two contiguous parts, each with at least one element. The score of each part is the sum of its elements modulo `p`, and I want to maximize the total score.
First, I think: "Okay, I can try every possible split point — from index 0 to index N-2 (so the first part has at least one element,... | The correct approach involves splitting the array into two contiguous non-overlapping parts and maximizing the sum of their scores, where the score of a part is the sum of its elements modulo `p`. A key insight is that for any split at index `i` (where the first part is `A[0:i+1]` and the second is `A[i+1:n]`), we want... | import bisect
def max_subarray_score(arr, p):
n = len(arr)
prefix = [0]
for num in arr:
prefix.append((prefix[-1] + num) % p)
sorted_prefix = [0]
max_score = 0
for i in range(n):
x = prefix[i+1]
min_y = sorted_prefix[0]
candidate1 = (x - min_y) % p
pos = ... |
373d3d7364dd3b2c4aade4030d85559d | Thomson is very weak in set theory. Recently, he came across the following problem:
Given a set X with N distinct elements, in how many ways can you select two sets A and B such that both A and B are subsets of X and A is also a subset of B.
Help Thomson solve the above problem. As the answer can be very large, print i... | <think>
Okay, let's see. The problem is to find the number of ways to select two sets A and B such that A is a subset of B and both are subsets of X, where X has N elements. And we need to compute this for multiple test cases efficiently, especially since N can be as big as 1e9. Hmm.
First, I need to understand the ma... | hackerearth | cc-by-4.0 | code_contests | train | MEDIUM | MOD = 10**9 + 7
T = int(input())
for _ in range(T):
n = int(input())
print(pow(3, n, MOD)) | Let me think about this step by step.
We are given a set $ X $ with $ N $ distinct elements, and we need to count the number of pairs $ (A, B) $ such that $ A \subseteq B \subseteq X $.
Hmm, okay. So both $ A $ and $ B $ are subsets of $ X $, and $ A $ must be a subset of $ B $. That means every element in $ A $ mu... | The correct solution hinges on recognizing that for each element in the set $ X $ of size $ N $, there are exactly three independent choices when forming pairs of subsets $ (A, B) $ such that $ A \subseteq B \subseteq X $:
- The element is in neither $ A $ nor $ B $,
- The element is in $ B $ but not in $ A $,
- ... | MOD = 10**9 + 7
T = int(input())
for _ in range(T):
n = int(input())
result = 1
for i in range(n):
result = (result * 3) % MOD
print(result) |
7101aa5cda89d1f1df05a0a97e5c6875 | Witua is a little student from the University of Lviv. He enjoys studying math. Witua knows a lot of famous mathematicians like Eratosthenes, Pythagoras, Fermat, Diophantus, Furko, Gauss and so on. However, his favorite one is Euler. The only thing Witua likes more than Euler is Euler’s totient function φ. He is explor... | <think>
Okay, let's see. The problem is to find, for a given N, the integer i in the range [2, N] that maximizes φ(i)/i. Hmm. Euler's totient function φ(i) counts the number of integers up to i that are coprime with i. The ratio φ(i)/i would then be the probability that a randomly chosen number from 1 to i is coprime w... | codechef | cc-by-4.0 | code_contests | train | interview | def is_prime(n):
if n <= 1:
return False
elif n <= 3:
return True
elif n % 2 == 0:
return False
d = n - 1
s = 0
while d % 2 == 0:
d //= 2
s += 1
bases = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
for a in bases:
if a >= n:
con... | I need to find the value of $ i $ between 2 and $ N $ that maximizes $ \phi(i)/i $.
I know that $ \phi(i) $ counts the number of integers less than $ i $ that are coprime to $ i $.
So $ \phi(i)/i $ is like the "density" of coprime numbers below $ i $.
I recall that $ \phi(i)/i = \prod_{p|i} (1 - 1/p) $.
So the... | The correct solution hinges on a deep insight about Euler's totient function:
For any integer $ i $,
$$
\frac{\phi(i)}{i} = \prod_{p \mid i} \left(1 - \frac{1}{p}\right)
$$
where the product is over the distinct prime factors of $ i $.
This means that to **maximize** $ \phi(i)/i $, we want to **minimize** the pro... | def is_prime(n):
if n < 2:
return False
if n == 2:
return True
if n % 2 == 0:
return False
if n == 3:
return True
if n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
... |
1747fa4ba26a6fd040d8da1637541336 | Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are thr... | <think>
Okay, let's see. I need to solve this problem where I have to split a word into parts such that each part doesn't have any typos. The definition of a typo is a substring of at least three consonants in a row with at least two different letters. So, the goal is to insert the minimal number of spaces to break the... | codeforces | cc-by-4.0 | code_contests | train | interview | vowels = {'a', 'e', 'i', 'o', 'u'}
def split_consonant_group(group):
parts = []
current_part = []
for c in group:
temp = current_part + [c]
if len(temp) >= 3:
last3 = temp[-3:]
if len(set(last3)) >= 2:
parts.append(''.join(current_part))
... | I need to find typos in the word. A typo happens when there are 3 or more consonants in a row with at least two different letters. So I should scan the word and look for sequences of consonants.
First, I'll define vowels: 'a', 'e', 'i', 'o', 'u'. All others are consonants.
Now, I'll go through the word character by c... | The correct approach requires scanning the word and identifying contiguous blocks of consonants. For each such block, we must determine if it contains a typo—i.e., three or more consonants in a row with at least two different letters. If so, we must split the block into smaller segments using the minimum number of spac... | vowels = {'a', 'e', 'i', 'o', 'u'}
s = input().strip()
words = []
current = []
consonants = []
for char in s:
if char in vowels:
# Process the current consonant group
if consonants:
if len(consonants) < 3:
current.extend(consonants)
elif all(c == consonants[... |
623366d93011804c4e7fe74e430d9ad7 | Finally, a basketball court has been opened in SIS, so Demid has decided to hold a basketball exercise session. 2 ⋅ n students have come to Demid's exercise session, and he lined up them into two rows of the same size (there are exactly n people in each row). Students are numbered from 1 to n in each row in order from ... | <think>
Okay, let's see. So the problem is about selecting students from two rows such that the indices are strictly increasing and no two consecutive are from the same row. And we need to maximize the sum of their heights. Hmm.
First, I need to model this problem. Since the selection has to be in increasing order of ... | codeforces | cc-by-4.0 | code_contests | train | 9 | n = int(input())
h1 = list(map(int, input().split()))
h2 = list(map(int, input().split()))
prev0 = 0 # max sum ending with row1, up to current index
prev1 = 0 # max sum ending with row2, up to current index
for i in range(n):
current0 = h1[i] + prev1
current1 = h2[i] + prev0
prev0, prev1 = max(prev0, cu... | I need to pick students from left to right, with strictly increasing indices, and no two consecutive players from the same row. The goal is to maximize total height.
First, I think: "I can start with any student, and then keep picking the next tallest one that’s in a different row and has a higher index." So maybe a g... | The correct solution uses dynamic programming where we track the maximum sum achievable ending at each row (top or bottom) up to the current column, ensuring that no two consecutive players are from the same row. The key insight is that at each position `i`, we can either take the current student from row 1 (adding the... | n = int(input())
h1 = list(map(int, input().split()))
h2 = list(map(int, input().split()))
prev0 = 0
prev1 = 0
for i in range(n):
current0 = h1[i] + prev1
current1 = h2[i] + prev0
prev0 = current0
prev1 = current1
print(max(prev0, prev1)) |
a0ef6ad9a617ce36513245dae469e685 | You are given some Tetris field consisting of n columns. The initial height of the i-th column of the field is a_i blocks. On top of these columns you can place only figures of size 2 × 1 (i.e. the height of this figure is 2 blocks and the width of this figure is 1 block). Note that you cannot rotate these figures.
Yo... | <think>
Okay, let's try to solve this problem. Hmm, the problem says that we have a Tetris field with n columns, each having some initial height. We can place 2x1 blocks (vertically, since they can't be rotated) on top of these columns. Then, as long as all columns have at least one block, we subtract one from each. We... | codeforces | cc-by-4.0 | code_contests | train | 7 | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
all_even = all(x % 2 == 0 for x in a)
all_odd = all(x % 2 == 1 for x in a)
if not (all_even or all_odd):
print("NO")
else:
if n % 2 == 0:
total = sum(a)
if total % 2 !... | Okay, so I need to figure out if I can clear the whole Tetris field by placing 2×1 blocks. Each time I place a block, I add 2 to one column, and then every column loses 1. The process repeats until all columns are zero.
Let me think about what happens in one full cycle:
- I pick a column, say column i, and add 2 to ... | The correct solution hinges on two key insights:
1. All columns must have the same parity (all even or all odd) — otherwise, it's impossible to balance the field through 2×1 placements.
2. If all values are even or all are odd, then the total number of blocks (after accounting for the 2-block additions) must allow ... | import heapq
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
# Simulate the process
# Use a max heap
heap = [-x for x in a]
heapq.heapify(heap)
# Keep track of moves
moves = 0
max_moves = 100000 # safety limit
while any(x ... |
5a42f6627646d58415e10bd6c23db85c | You've got a table of size n × m. On the intersection of the i-th row (1 ≤ i ≤ n) and the j-th column (1 ≤ j ≤ m) there is a non-negative integer ai, j. Besides, you've got a non-negative integer k.
Your task is to find such pair of integers (a, b) that meets these conditions:
* k ≤ a ≤ n - k + 1;
* k ≤ b ≤ m -... | <think>
Okay, let's see. I need to solve this programming problem. Let's read the problem statement carefully.
So, the problem is about finding a position (a, b) in a grid such that when you take a k x k square centered at (a, b), the sum of all elements in that square is the maximum possible. The constraints are that... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | n, m, k = map(int, input().split())
# Initialize grid with 1-based indexing
grid = [[0] * (m + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
row = list(map(int, input().split()))
for j in range(1, m + 1):
grid[i][j] = row[j - 1]
# Compute prefix sums
prefix = [[0] * (m + 1) for _ in range(n + 1)... | I need to find a pair (a, b) such that the k×k submatrix starting at (a, b) has the maximum sum, and (a, b) must be within bounds so that the submatrix doesn't go out of the grid.
First, I think: "a" should start from 1, and go up to n - k + 1, because if I start at row a, I need k rows, so a ≤ n - k + 1. Same for b: ... | The correct approach involves identifying a k×k submatrix centered at position (a, b) such that the sum of elements in that submatrix is maximized, with the constraint that the center (a, b) must be at least k units away from the borders (i.e., the submatrix fits entirely within the grid). The key insight is to use a 2... | n, m, k = map(int, input().split())
# Read the grid
grid = []
for _ in range(n):
row = list(map(int, input().split()))
grid.append(row)
max_sum = -1
best_a, best_b = 0, 0
# Iterate through all possible top-left corners of k×k submatrix
# But the student doesn't know the constraint k <= a <= n-k+1
# So they u... |
2f01106bbb22b8f6f584c5fade47b7b8 | You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The ... | <think>
Okay, so I need to solve this problem where I have to determine if a certain amount S can be formed using a coins of value n and b coins of 1. Let's think about how to approach this.
First, the constraints are pretty big—up to 1e9, so any solution that's O(a) is definitely out of the question. So I need a math... | codeforces | cc-by-4.0 | code_contests | train | 7 | import sys
q = int(sys.stdin.readline())
for _ in range(q):
a, b, n, s = map(int, sys.stdin.readline().split())
max_x = min(a, s // n)
remainder = s - max_x * n
print("YES" if remainder <= b else "NO") | Okay, so we have `a` coins of value `n`, and `b` coins of value `1`, and we want to make exactly `S`. We need to pick `x` coins of value `n` and `y` coins of value `1`, such that `x*n + y = S`, with `0 ≤ x ≤ a` and `0 ≤ y ≤ b`.
Let me think: for a given `x`, the required `y` would be `S - x*n`. That must be between 0 ... | The correct approach hinges on the observation that we can use at most `a` coins of value `n`, so the maximum number of such coins we can use is `min(a, S // n)`. After using that many `n`-value coins, the remaining amount must be made up using `1`-value coins, and we can only use up to `b` of them. Thus, the remainder... | import sys
q = int(sys.stdin.readline())
for _ in range(q):
a, b, n, s = map(int, sys.stdin.readline().split())
max_x = min(a, s // n)
found = False
# Try from max_x down to 0
# But this could be up to 10^9 iterations per test case
for x in range(max_x, -1, -1):
remainder = s - x * n
... |
6feb418950046d69da53da0d379db90c | You are given an integer sequence 1, 2, ..., n. You have to divide it into two sets A and B in such a way that each element belongs to exactly one set and |sum(A) - sum(B)| is minimum possible.
The value |x| is the absolute value of x and sum(S) is the sum of elements of the set S.
Input
The first line of the input c... | <think>
Okay, I need to find the minimum possible difference between the sums of two sets A and B formed by dividing the numbers from 1 to n. Let's think about how to approach this problem.
First, the sum of all numbers from 1 to n is S = n*(n+1)/2. The goal is to split this sum into two parts such that their absolute... | codeforces | cc-by-4.0 | code_contests | train | 7 | n = int(input())
print((n * (n + 1) // 2) % 2) | Let me think about this step by step.
We need to divide the numbers from 1 to n into two sets so that the absolute difference of their sums is minimized.
First, I know the total sum is $ S = \frac{n(n+1)}{2} $. If I can split this sum into two equal parts, the difference would be 0. That happens only if $ S $ is even... | The correct solution hinges on the observation that the total sum of the sequence $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ is either even or odd. If the total sum is even, we can split the set into two subsets with equal sum (difference 0). If it's odd, the best we can do is a difference of 1, since we can't split an odd... | n = int(input())
sumA = 0
sumB = 0
for i in range(n, 0, -1):
if sumA <= sumB:
sumA += i
else:
sumB += i
print(abs(sumA - sumB)) |
5af99262dadd445aaf6311359102f366 | C： AA グラフ (AA Graph)
Problem
Given a graph as an ASCII Art (AA), please print the length of shortest paths from the vertex s to the vertex t. The AA of the graph satisfies the following constraints.
A vertex is represented by an uppercase alphabet and symbols `o` in 8 neighbors as follows.
ooo
oAo
ooo
Horizontal... | <think>
Okay, I need to solve this problem where I have to find the shortest path from vertex s to t in an ASCII art graph. Hmm, let's read the problem statement carefully.
So the vertices are uppercase letters surrounded by o's in their 8 neighbors. The edges are either '-' (horizontal) or '|' (vertical), and each ed... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
from collections import deque, defaultdict
def main():
h, w, s, t = sys.stdin.readline().split()
h = int(h)
w = int(w)
grid = [sys.stdin.readline().strip() for _ in range(h)]
# Collect vertices
vertices = {}
for i in range(h):
for j in range(w):
c = grid[... | I need to find the shortest path from vertex `s` to `t` in this ASCII graph. First, I should locate where `s` and `t` are. Each uppercase letter is a vertex, so I can scan the grid and record the position of each letter.
Now, how do edges work? From the example, I see that vertices are surrounded by `o` in a 3x3 patte... | The correct solution hinges on accurately parsing the ASCII graph to extract vertex positions and then building a proper adjacency list by tracing edges from each vertex through the `o`, `-`, or `|` symbols to the next vertex. The key insight is that edges are only valid if they connect two vertices via a path of `o`, ... | import sys
from collections import deque
def main():
line = sys.stdin.readline().split()
h = int(line[0])
w = int(line[1])
s = line[2]
t = line[3]
grid = []
for _ in range(h):
grid.append(sys.stdin.readline().strip())
# Find start and end positions
start = None
end ... |
d834baf52801e4cead15d295f7a874fe | Let's denote as <image> the number of bits set ('1' bits) in the binary representation of the non-negative integer x.
You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and <image> is maximum possible. If there are multiple such numbers find the sma... | <think>
Okay, let's see. The problem is to find, for each query [l, r], the number x in that range that has the maximum number of set bits (1s in binary). And if there are multiple such numbers, pick the smallest one. Hmm.
So the key is to maximize the count of 1s. The maximum possible number of 1s is when the number ... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM_HARD | def solve():
import sys
input = sys.stdin.read().split()
n = int(input[0])
idx = 1
for _ in range(n):
l = int(input[idx])
r = int(input[idx + 1])
idx += 2
def find_x(l, r):
if l == r:
return l
d = l ^ r
high... | I need to find the number in `[l, r]` with the most set bits, and if there are ties, pick the smallest one.
First, I know that numbers with more 1s in binary are better. So ideally, I want a number like `111...1` (all 1s) if it's in the range. But such a number might be too big or too small.
I think I can try to buil... | The correct approach hinges on the insight that to maximize the number of set bits (popcount) in a range `[l, r]`, we want a number as close to `r` as possible that has as many 1s as possible in its binary representation. The key idea in the reference solution is to find the highest bit where `l` and `r` differ, then t... | def solve():
import sys
input = sys.stdin.read().split()
n = int(input[0])
idx = 1
for _ in range(n):
l = int(input[idx])
r = int(input[idx + 1])
idx += 2
ans = 0
# Go from high bit to low bit (60 down to 0)
for i in range(60, -1, -1):
... |
b0daaae2a34fb4025a46970280738c9d | Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, both Memory and Lexa get some integer in the range [ - k;k] (i.e. one integer among - k, - k + 1, - k + 2, ..., - 2, - 1, 0, 1, 2, ..., k - 1, k) and add... | <think>
Okay, I need to solve this problem where I have to calculate the number of possible games where Memory ends up with a strictly higher score than Lexa after t turns. Let's see the input constraints and the examples to understand better.
So, the problem says that in each turn, both players add a random integer b... | codeforces | cc-by-4.0 | code_contests | train | interview | MOD = 10**9 + 7
a, b, k, t = map(int, input().split())
T = b - a
max_sum = k * t
# Initialize DP for 0 steps
dp_prev = [0] * (2 * max_sum + 1)
dp_prev[max_sum] = 1 # sum 0
for step in range(1, t + 1):
dp_current = [0] * (2 * max_sum + 1)
prefix_prev = [0] * (2 * max_sum + 2)
for i in range(2 * max_sum +... | I need to count the number of games where Memory ends with a strictly higher score than Lexa. Memory starts with `a`, Lexa with `b`. After `t` turns, each turn both pick a number from `[-k, k]`, so each has a total gain between `-t*k` and `t*k`.
Let me denote:
- `M_total = a + sum_M`, where `sum_M` is Memory’s total g... | The correct solution uses dynamic programming to compute the distribution of the total score difference after `t` turns, leveraging prefix sums for efficiency. The key insight is to model the net change in score (Memory's score minus Lexa's score) over `t` turns, where each turn contributes a value in `[-2k, 2k]` (sinc... | MOD = 10**9 + 7
a, b, k, t = map(int, input().split())
# Net gain range: from -2*k*t to 2*k*t
# We'll use offset = 2*k*t
max_net = 2 * k * t
size = 2 * max_net + 1
offset = max_net
# dp[i] = number of ways to have net gain i - offset
dp = [0] * size
dp[offset] = 1 # net gain 0
# For each turn, net change can be fr... |
329c1b452a93342e919486e9b5c47532 | Link to Russian translation of the problem.
Kevin has an array A of N integers, but he doesn't like it - he only likes an array if all its neighboring elements are different (e.g. he likes the array [2,4,2,5,3,2,4], but not [1,2,3,3,4]).
Kevin can reverse any contiguous subarray of his array any number of times. For ... | <think>
Okay, let's try to figure out how to solve this problem. So the task is to find the minimum number of reversals needed to make sure that no two neighboring elements in the array are the same. If it's impossible, we return -1.
First, I need to understand what a reversal can do. Reversing a subarray can change ... | hackerearth | cc-by-4.0 | code_contests | train | MEDIUM | n = int(input())
a = list(map(int, input().split()))
# Check if the array is already valid
valid = True
for i in range(n - 1):
if a[i] == a[i + 1]:
valid = False
break
if valid:
print(0)
else:
# Check frequency condition
from collections import Counter
freq = Counter(a)
max_freq... | Okay, so Kevin wants an array where no two neighboring elements are the same. He can reverse any contiguous subarray any number of times. I need to find the minimum number of reversals.
First, I’ll check if the array is already valid. That’s easy: just loop through and see if any adjacent elements are equal. If not, a... | The correct solution hinges on two key insights: first, that a valid array (with no adjacent duplicates) can only exist if no element appears more than ⌈n/2⌉ times (a classic constraint from rearrangement problems like "rearrange string so no two adjacent are same"); second, that if this condition is satisfied, the min... | n = int(input())
a = list(map(int, input().split()))
# Check if already valid
valid = True
for i in range(n - 1):
if a[i] == a[i + 1]:
valid = False
break
if valid:
print(0)
else:
from collections import Counter
freq = Counter(a)
max_freq = max(freq.values())
if max_freq > (n + ... |
ccfd2750cfb86cf7660b0646a913164c | Johnny drives a truck and must deliver a package from his hometown to the district center. His hometown is located at point 0 on a number line, and the district center is located at the point d.
Johnny's truck has a gas tank that holds exactly n liters, and his tank is initially full. As he drives, the truck consumes ... | <think>
Okay, let's see. So the problem is about Johnny driving a truck and needing to find the minimum cost for fuel to reach the destination. Hmm.
First, I need to understand the input. The distance is d, the tank capacity is n liters, and there are m gas stations. The truck starts at 0 with a full tank, which is n ... | codeforces | cc-by-4.0 | code_contests | train | interview | d, n, m = map(int, input().split())
stations = [tuple(map(int, input().split())) for _ in range(m)]
stations.sort()
stations.append((d, 0)) # Add virtual station at destination
# Check if any segment exceeds tank capacity
prev = 0
for x, p in stations:
if x - prev > n:
print(-1)
exit()
prev = ... | I need to find the minimum cost to deliver the package. Johnny starts with a full tank of `n` liters and consumes 1 liter per unit distance. There are `m` gas stations with different prices. The goal is to minimize the total cost.
First, I think about the constraints: the tank can only hold `n` liters, so Johnny can o... | The correct solution hinges on two key insights: first, ensuring that no gap between consecutive stations (or from start to first station, or last station to destination) exceeds the tank capacity `n`, otherwise the trip is impossible. Second, using a greedy strategy with a "next cheaper station" optimization via a mon... | d, n, m = map(int, input().split())
stations = []
for _ in range(m):
x, p = map(int, input().split())
stations.append((x, p))
# Sort stations by position
stations.sort()
# Add destination as a station with price 0
stations.append((d, 0))
# Check if any gap exceeds tank capacity
prev = 0
for x, p in stations:... |
ab33d865c9c0d2c6dba9d8d50d48230d | There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.
Adjacent slimes with the same color will fuse into one larger slime withou... | <think>
Okay, I need to solve this problem where I have to find out how many slimes are left after adjacent ones of the same color merge. Let's see. The problem says that adjacent slimes with the same color fuse into one. So, for example, if the string is "aabba", it becomes "aba" because the two a's merge, the two b's... | atcoder | cc-by-4.0 | code_contests | train | 0 | n = int(input())
s = input().strip()
if n == 0:
print(0)
else:
count = 1
prev = s[0]
for c in s[1:]:
if c != prev:
count += 1
prev = c
print(count) | Okay, so we have a string of slimes, and adjacent slimes with the same color fuse into one. I need to figure out how many slimes remain after all possible fusions.
Hmm, the key idea is that whenever two adjacent slimes have the same color, they merge. So I should simulate this process step by step. But how?
Maybe I c... | The correct approach is to count the number of color changes in the string — each time the current slime's color differs from the previous one, a new group (and thus a new slime after fusion) begins. The total number of slimes after fusion is simply the number of such transitions plus one (for the first slime). This ca... | n = int(input())
s = input().strip()
slimes = list(s)
changed = True
while changed:
changed = False
i = 0
while i < len(slimes) - 1:
if slimes[i] == slimes[i + 1]:
slimes.pop(i + 1)
changed = True
else:
i += 1
print(len(slimes)) |
6ae0931fd34c656697a927b4092b0e87 | In the last mission, MDCS has successfully shipped N AI robots to Mars. Before they start exploring, system initialization is required so they are arranged in a line. Every robot can be described with three numbers: position (x_i), radius of sight (r_i) and IQ (q_i).
Since they are intelligent robots, some of them wil... | <think>
Okay, let's try to figure out how to solve this problem. So, the problem is about counting the number of pairs of robots that can see each other and have IQ differences within K. Hmm. The main challenge here is to efficiently find all such pairs without checking every possible pair, which would be O(n^2) and no... | codeforces | cc-by-4.0 | code_contests | train | HARD | import bisect
n, K = map(int, input().split())
robots = []
for _ in range(n):
x, r, q = map(int, input().split())
robots.append((x, r, q))
# Sort robots by their x-coordinate
robots.sort(key=lambda x: x[0])
iq_map = {}
total = 0
for j in range(n):
x_j, r_j, q_j = robots[j]
# Iterate over all possibl... | I need to count how many robot pairs talk. First, I understand that two robots talk if:
1. They can see each other (mutual visibility).
2. Their IQs differ by at most K.
So I start by reading the input and storing each robot’s x, r, q.
I think: "I can loop through every pair of robots and check both conditions." That... | The correct solution hinges on efficiently counting pairs of robots that can see each other *and* have IQs within K of each other. The key insights are:
- Robots must be sorted by position to enable binary search for visibility.
- For each robot, only robots with IQs in `[q_j - K, q_j + K]` can potentially talk.
- We n... | n, K = map(int, input().split())
robots = []
for _ in range(n):
x, r, q = map(int, input().split())
robots.append((x, r, q))
total = 0
for i in range(n):
x_i, r_i, q_i = robots[i]
for j in range(n):
if i == j:
continue
x_j, r_j, q_j = robots[j]
# Check mutual visibil... |
dfc913dc88440a7fb3b43f045a2d80a0 | Ted: Robin, get me my legal pad. It's Pros and Cons Time!
There is a long list of n girls in front of Barney, and he is to calculate the optimal "happiness" he can find by selecting exactly 2 girls. (Why 2? No one knows!)
Ted, as a fan of pros and cons, suggests to make a list, a method for estimating the maximum hap... | <think>
Okay, let's try to figure out how to solve this problem. So, the problem is about choosing exactly two girls from a list, and calculating the maximum happiness Barney can get. Each girl contributes to the happiness in two ways: if selected, their favour is added, and if not selected, their anger is subtracted. ... | hackerearth | cc-by-4.0 | code_contests | train | EASY | import sys
def main():
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
sum_anger = 0
max1 = -float('inf')
max2 = -float('inf')
for _ in range(n):
f, a = map(int, sys.stdin.readline().split())
sum_anger += a
... | Okay, so Barney has to pick exactly two girls, and happiness is calculated as the sum of their favour values minus the anger values of the girls he *doesn’t* pick. That means if I pick girl A and girl B, the happiness is:
`favour_A + favour_B - (anger of all girls except A and B)`
Hmm, that’s the same as:
`favour_A ... | The correct solution hinges on a key insight: the total happiness when selecting two girls is `f1 + f2 - (sum of anger of all girls not selected)`. Since exactly two girls are chosen, the unselected girls contribute their anger values to the total happiness as a negative term. This means the happiness becomes `f1 + f2 ... | import sys
def main():
t = int(sys.stdin.readline())
for _ in range(t):
n = int(sys.stdin.readline())
girls = []
total_anger = 0
for _ in range(n):
f, a = map(int, sys.stdin.readline().split())
girls.append((f, a))
total_anger += a
... |
d422b9067922998538c412618974384a | A function <image> is called Lipschitz continuous if there is a real constant K such that the inequality |f(x) - f(y)| ≤ K·|x - y| holds for all <image>. We'll deal with a more... discrete version of this term.
For an array <image>, we define it's Lipschitz constant <image> as follows:
* if n < 2, <image>
* if n ... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The problem is about the Lipschitz constant of arrays. The Lipschitz constant L(h) for an array h is defined as the minimal non-negative integer such that for all i < j, |h[i] - h[j]| ≤ L * (j - i). If the array has les... | codeforces | cc-by-4.0 | code_contests | train | HARD | import sys
def main():
n, q = map(int, sys.stdin.readline().split())
h = list(map(int, sys.stdin.readline().split()))
d = [abs(h[i+1] - h[i]) for i in range(n-1)]
for _ in range(q):
l, r = map(int, sys.stdin.readline().split())
L = l - 1
R = r - 2
if L > R:
... | I need to find the sum of Lipschitz constants of all subarrays within each query [l, r]. First, I recall that the Lipschitz constant L of a function (or array) is the smallest K such that |f(x) - f(y)| ≤ K·|x - y| for all x, y. For an array, this means that for any two indices i and j, the difference in values must be ... | The correct solution hinges on a key insight: the Lipschitz constant of a subarray is equal to the maximum absolute difference between adjacent elements in that subarray. This is because the Lipschitz condition is most tightly constrained by adjacent elements — any non-adjacent pair (i, j) will have a smaller ratio |h[... | import sys
def main():
n, q = map(int, sys.stdin.readline().split())
h = list(map(int, sys.stdin.readline().split()))
for _ in range(q):
l, r = map(int, sys.stdin.readline().split())
L = l - 1
R = r - 1
total = 0
# Iterate over all subarrays of h[L:R+1]... |
937fd7b346d4dcbf3e13f31d1be2ee1f | A Ministry for Defense sent a general to inspect the Super Secret Military Squad under the command of the Colonel SuperDuper. Having learned the news, the colonel ordered to all n squad soldiers to line up on the parade ground.
By the military charter the soldiers should stand in the order of non-increasing of their h... | <think>
Okay, let's see. The problem is to find the minimum number of swaps needed to make the first soldier the maximum height and the last soldier the minimum height. Oh right, but the swaps are between neighboring soldiers, so each swap takes one second, and moving a soldier from position i to position j takes |i-j|... | codeforces | cc-by-4.0 | code_contests | train | EASY | n = int(input())
arr = list(map(int, input().split()))
max_val = max(arr)
max_pos = arr.index(max_val)
min_val = min(arr)
min_pos = len(arr) - 1 - arr[::-1].index(min_val)
if min_pos < max_pos:
total = max_pos + (n - 1 - (min_pos + 1))
else:
total = max_pos + (n - 1 - min_pos)
print(total) | I need to make the first soldier the tallest and the last soldier the shortest. So, I should find the tallest soldier and move him to the front, and find the shortest soldier and move him to the end.
First, I'll find the index of the tallest soldier. Then, I'll count how many adjacent swaps it takes to move him from h... | The correct solution hinges on two key insights:
1. The general only cares about the first soldier being the tallest and the last soldier being the shortest.
2. We can achieve this by moving the tallest soldier to the front and the shortest soldier to the end, using adjacent swaps. The cost is the number of swaps n... | n = int(input())
arr = list(map(int, input().split()))
max_val = max(arr)
min_val = min(arr)
# Make a copy to simulate
current = arr[:]
cost = 0
# Move the leftmost max to the front
max_idx = current.index(max_val)
while max_idx > 0:
current[max_idx], current[max_idx - 1] = current[max_idx - 1], current[max_idx]... |
7b6e86408b4d63cc210763821ec969b6 | You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and... | <think>
Okay, let's see. I need to solve this programming problem. The task is to create a string res of length 3n, with exactly n a's, b's, and c's. The catch is that this string should not contain the substrings s or t. If it's impossible, output "NO", else output "YES" and the string.
Hmm. So the first thing I need... | codeforces | cc-by-4.0 | code_contests | train | 11 | import sys
from itertools import permutations
n = int(sys.stdin.readline())
s = sys.stdin.readline().strip()
t = sys.stdin.readline().strip()
# Check all permutations of 'abc' for validity
for perm in permutations(['a', 'b', 'c']):
pair1 = perm[0] + perm[1]
pair2 = perm[1] + perm[2]
pair3 = perm[2] + perm... | I need to build a string of length `3n` with exactly `n` of each character `'a'`, `'b'`, `'c'`, and make sure that neither `s` nor `t` appears as a substring.
First, I'll try to build the string greedily: start with an empty string, and at each step, try to append one of `'a'`, `'b'`, `'c'` that:
- Doesn't exceed the ... | The correct solution hinges on two key insights:
1. Since we need exactly `n` of each character and the forbidden substrings are only length 2, we can try constructing the result using repeating patterns of 2 or 3 characters.
2. The most efficient and correct approach is to check all cyclic permutations of `'abc'` ... | import sys
def solve():
n = int(sys.stdin.readline().strip())
s = sys.stdin.readline().strip()
t = sys.stdin.readline().strip()
# Greedy construction: try to build string character by character
# Use counts and last character to avoid forbidden substrings
count = {'a': 0, 'b': 0, 'c': 0}
... |
8edcdcae6d793106af4c6277c3817cb0 | Vasya has an array a_1, a_2, ..., a_n.
You don't know this array, but he told you m facts about this array. The i-th fact is a triple of numbers t_i, l_i and r_i (0 ≤ t_i ≤ 1, 1 ≤ l_i < r_i ≤ n) and it means:
* if t_i=1 then subbarray a_{l_i}, a_{l_i + 1}, ..., a_{r_i} is sorted in non-decreasing order;
* if t_i... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The problem is about constructing an array that satisfies certain conditions based on given facts. Each fact says whether a particular subarray is sorted in non-decreasing order or not. The challenge is to find such an ... | codeforces | cc-by-4.0 | code_contests | train | MEDIUM_HARD | n, m = map(int, input().split())
type0 = []
type1 = []
for _ in range(m):
t, l, r = map(int, input().split())
if t == 1:
type1.append((l, r))
else:
type0.append((l, r))
# Check if any type0 interval is fully contained in a type1 interval
for l, r in type0:
contained = False
for L, R... | I need to build an array that satisfies all the given facts. The idea is to start with a simple array, like [1, 2, 3, ..., n], which is sorted everywhere. Then, I’ll go through each t=0 constraint and make sure the subarray is *not* sorted.
For each t=0 fact (l, r), I can just make sure that somewhere in the subarray,... | The correct solution hinges on two key insights:
1. If a subarray is required to be *not sorted* (t=0), but it is entirely contained within a subarray that must be *sorted* (t=1), then the facts are contradictory — such a case is impossible.
2. For each t=0 interval, there must exist at least one adjacent pair (j, ... | n, m = map(int, input().split())
a = list(range(1, n + 1))
facts = []
for _ in range(m):
t, l, r = map(int, input().split())
facts.append((t, l, r))
# For each t=0 fact, try to break the order at the first position
for t, l, r in facts:
if t == 0:
# Try to make a[l] > a[l+1] by decreasing a[l+1]
... |
1e1acf2afcf98e025f8ca91c1783bed1 | Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plas... | <think>
Okay, let's try to figure out how to solve this problem. So, the problem is about Kolya wanting to buy as much kefir as possible with his rubles. The options are buying plastic bottles or glass ones, which can be returned for some money back. Hmm.
The input gives us n rubles, then a, b, c. The plastic bottle c... | codeforces | cc-by-4.0 | code_contests | train | interview | n = int(input())
a = int(input())
b = int(input())
c = int(input())
d = b - c
if d < a:
if n < b:
max_liters = n // a
else:
k = (n - b) // d + 1
remaining = n - k * d
option1 = k + remaining // a
option2 = n // a
max_liters = max(option1, option2)
else:
optio... | Okay, so Kolya has `n` rubles. He can buy plastic bottles for `a` rubles each, or glass bottles for `b` rubles each, and return glass bottles to get `c` rubles back. He wants to maximize the number of liters.
First, I know that plastic bottles are straightforward: he can just buy `n // a` of them. That’s easy.
Glass ... | The correct solution hinges on recognizing two distinct strategies:
1. Buying only plastic bottles (no return, simple division).
2. Using glass bottles strategically—buying one, returning it to get money back, and repeating this cycle to effectively reduce the cost per liter to `b - c` (net cost after return).
T... | n = int(input())
a = int(input())
b = int(input())
c = int(input())
# Try to simulate buying glass bottles and returning them
# But this will be too slow for large n
max_liters = n // a # at least buy plastic bottles
# Try to buy glass bottles in a cycle
money = n
glass_bottles = 0
# But if we can't buy a glass bot... |
b577befd63301ec6895f08a4b30d972f | A patient has been infected with an unknown disease. His body can be seen as an infinite grid of triangular cells which looks as follows:
<image>
Two cells are neighboring if they share a side. Therefore, each cell (x, y) has exactly three neighbors:
* (x+1, y)
* (x-1, y)
* (x+1, y-1) if x is even and (x-1,... | <think>
Okay, so I need to solve this programming problem. Let's read the problem statement carefully.
The problem is about a grid of triangular cells, and the spread/recovery of a disease. The key points are:
- The grid is infinite, but the initial infected cells are all in a 500x500 area.
- Each cell has three neig... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import sys
from collections import deque
MOD = 998244353
def main():
n = int(sys.stdin.readline())
coords = [tuple(map(int, sys.stdin.readline().split())) for _ in range(n)]
color_count = [0, 0]
edges = 0
cell_map = {}
adj = [[] for _ in range(n)]
dirs_even = [(1, 0), (-1, 0), (1, -1)... | I see that each cell has three neighbors, and the grid has a strange neighbor rule depending on whether `x` is even or odd. But I notice that the parity of `x + y` seems to alternate in a checkerboard pattern. So I decide to color each cell based on `(x + y) % 2`. That gives me two colors: 0 and 1.
Now, I think about ... | The correct solution hinges on two key insights:
1. The triangular grid has a checkerboard-like coloring based on `(x + y) % 2`, and the infection dynamics depend on this bipartite structure.
2. If infected cells are not all in the same color class (i.e., both colors are present), then the system can enter a cycle ... | import sys
MOD = 998244353
def main():
n = int(sys.stdin.readline())
coords = []
for _ in range(n):
x, y = map(int, sys.stdin.readline().split())
coords.append((x, y))
# Count infected cells in each color class (x+y) % 2
color_count = [0, 0]
for x, y in coords:
col... |
9948cc615d9c024300a02a5cb7d5891e | Mike takes part in olympiads in informatics. You think he is a rookie? Wrong! He is an experienced and well-prepared competitor! He participated in many important contests and won some of them. Now his level is rather high.
In order to keep fit, Mike decided to improve his training sessions. He downloaded N task pack... | <think>
Okay, let's see. The problem is about Mike solving tasks from different packages, and we need to find the optimal way for him to maximize successful packages while minimizing failed ones. The constraints are a bit tricky. So, the goal is to first minimize the number of failed packages, and then maximize the suc... | codechef | cc-by-4.0 | code_contests | train | 3 | import bisect
def main():
import sys
input = sys.stdin.read().split()
N = int(input[0])
X = int(input[1])
A = list(map(int, input[2:2+N]))
min_required = [(a + 1) // 2 for a in A]
sum_S = sum(min_required)
if sum_S <= X:
X_remaining = X - sum_S
costs = [a - m f... | I see that Mike wants to minimize failed packages and, among those, maximize successful ones. So I need to decide which packages to fully solve and which to partially solve.
Let me think: a package is failed if he solves less than half the tasks. So to avoid failure, he must solve at least ⌈Ai/2⌉ tasks in package i. T... | The correct solution hinges on two key insights:
1. To minimize failed packages, we should prioritize completing entire packages (i.e., solving all tasks in them) when possible, because a package is failed if fewer than half are solved.
2. A package is *successful* if all tasks are solved, and *failed* if fewer tha... | import sys
def main():
input = sys.stdin.read().split()
N = int(input[0])
X = int(input[1])
A = list(map(int, input[2:2+N]))
# Calculate conversion cost: extra tasks to make a package successful
conversion_costs = []
for a in A:
min_req = (a + 1) // 2
conversion_cost = ... |
5439aa4f20841de38e070f53fee878a9 | Recent improvements in information and communication technology have made it possible to provide municipal service to a wider area more quickly and with less costs. Stimulated by this, and probably for saving their not sufficient funds, mayors of many cities started to discuss on mergers of their cities.
There are, of... | <think>
Okay, let's tackle this programming problem. Hmm. The goal is to find the shortest possible name for a merged city that includes all the original city names as consecutive substrings. Oh right, and the cities can be overlapped as long as their entire names are present. But overlapping is allowed only when parts... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
from itertools import combinations
def main():
while True:
line = sys.stdin.readline()
if not line:
break
line = line.strip()
if not line:
continue
n = int(line)
if n == 0:
break
cities = [sys.stdin.readline().st... | I see this problem as one where I need to merge city names by overlapping common parts. The goal is to make the shortest possible name that contains all original city names as substrings. Since the names are in uppercase and can be overlapped, I think I can just try to merge them in some order and keep overlapping as m... | The correct solution hinges on several key insights:
1. **Redundant city names** (those fully contained in others) can be safely removed, since including the longer name automatically includes the shorter one.
2. The problem reduces to finding the **shortest superstring** that contains all remaining city names as c... | import sys
from itertools import permutations
def main():
while True:
line = sys.stdin.readline().strip()
if not line:
continue
n = int(line)
if n == 0:
break
cities = [sys.stdin.readline().strip() for _ in range(n)]
# Precompute over... |
80ce32e3ef5aa560d3a88900dbd083ee | You are given a string s, consisting of lowercase English letters, and the integer m.
One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
So, the problem is about selecting a subset of characters from a string such that every contiguous subsegment of length m has at least one selected character. Then, we can rearrange those selected characters to form the... | codeforces | cc-by-4.0 | code_contests | train | competition | import bisect
from collections import defaultdict
m = int(input())
s = input().strip()
n = len(s)
if m > n:
print("")
exit()
# Preprocess positions (1-based)
pos = defaultdict(list)
for i, c in enumerate(s):
pos[c].append(i + 1)
res = []
current_start = 1
max_start = n - m + 1
while current_start <= m... | I need to find the lexicographically smallest string we can form by selecting some positions such that every contiguous segment of length `m` contains at least one selected position.
First, I think: to get the lexicographically smallest result, I should prefer selecting 'a' over 'b', 'b' over 'c', etc. So I should try... | The correct approach involves greedily selecting the lexicographically smallest character that can "cover" the current segment starting at `current_start`, using the rightmost occurrence of that character within the current window `[current_start, current_start + m - 1]` to maximize the coverage of future segments. Thi... | m = int(input())
s = input().strip()
n = len(s)
# Preprocess positions (1-based)
pos = {}
for i, c in enumerate(s):
if c not in pos:
pos[c] = []
pos[c].append(i + 1)
res = []
current_start = 1
max_start = n - m + 1
while current_start <= max_start:
# Find the smallest character that appears in [... |
e65bf506387c47bfb952dff3f7b89208 | Leha and Noora decided to go on a trip in the Baltic States. As you know from the previous problem, Leha has lost his car on the parking of the restaurant. Unfortunately, requests to the watchman didn't helped hacker find the car, so friends decided to go hitchhiking.
In total, they intended to visit n towns. However ... | <think>
Okay, let's see. We have this problem where Leha and Noora want to visit as many towns as possible by choosing the days they visit each town and then selecting a subsequence where both the indices and the days are strictly increasing. The goal is to maximize the length of such a subsequence. Hmm.
So the proble... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import bisect
n = int(input())
last = [-10**18]
for _ in range(n):
l, r = map(int, input().split())
pos = bisect.bisect_left(last, r)
k_max = pos - 1
x = max(last[k_max] + 1, l)
if x <= r:
if k_max + 1 >= len(last):
last.append(x)
else:
if x < last[k_max + 1... | I need to find the maximum number of towns I can visit such that both the town indices and the visit days are strictly increasing. So, I can’t visit town 3 before town 2, and I can’t visit town 2 on day 5 and town 3 on day 4.
My idea: since we want to maximize the number of towns, I should try to visit towns as early ... | The correct solution uses a greedy strategy with a dynamic programming-like structure, maintaining a list `last` where `last[i]` represents the earliest possible day on which a valid sequence of `i` towns can end. The key insight is that for each town, we should try to extend the longest possible sequence by choosing t... | n = int(input())
last = [0] # last[k] = minimum day for a sequence of length k
for _ in range(n):
l, r = map(int, input().split())
# Find the largest k such that last[k] < r using linear scan
k = 0
for i in range(len(last)):
if last[i] < r:
k = i
else:
break
... |
c58135bbd63d5cd3d5453ddadaf8a27b | There are league games and tournament games in sports competitions. In soccer league games, points are given to each of the wins, losses, and draws, and the rankings are competed based on the points. The points are win (3 points), negative (0 points), and draw (1 point), respectively.
Enter the number of teams and the... | <think>
Okay, let's see. I need to solve this programming problem where I have to read multiple datasets of team results and output their rankings based on points. The points are calculated from wins, losses, and draws. The tricky part is handling the sorting correctly, especially when points are tied. Oh right, if poi... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
def main():
datasets = []
while True:
line = sys.stdin.readline()
if not line:
break
n = int(line.strip())
if n == 0:
break
teams = []
for _ in range(n):
parts = sys.stdin.readline().split()
name = parts[... | I need to read the number of teams and then process each team’s data. For each team, I calculate points as 3 times wins plus draws — that part seems straightforward.
I’ll store each team’s name and points in a list. Then I need to sort them by points in descending order. I can use `sorted()` with `reverse=True` for th... | The correct approach requires reading multiple datasets until `n = 0`, computing points as `3 * wins + draws`, and sorting teams by descending points. In case of ties, the original input order must be preserved. The key insight is using a stable sort with a custom key that prioritizes higher points first and, for equal... | import sys
def main():
first_dataset = True
while True:
line = sys.stdin.readline().strip()
if not line:
break
n = int(line)
if n == 0:
break
teams = []
for _ in range(n):
parts = sys.stdin.readline().split()
... |
b1bbaed48a70a5d0baea4921e501b184 | Mr. Frugal bought a new house. He feels deeply in love with his new house because it has a comfortable living room in which he can put himself completely at ease. He thinks his new house is a really good buy.
But, to his disappointment, the floor of its living room has some scratches on it.
The floor has a rectangle ... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The task is to find the minimum number of square carpets needed to cover all the scratched panels (marked as 1) without covering any of the flawless ones (0). The carpets can't be folded, can overlap, and their sizes ar... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
def main():
sys.setrecursionlimit(1000000)
while True:
W, H = map(int, sys.stdin.readline().split())
if W == 0 and H == 0:
break
grid = []
for _ in range(H):
row = list(map(int, sys.stdin.readline().split()))
grid.append(row)
... | I need to cover all scratched panels with the fewest square carpets. Since bigger carpets cover more area, I should use them when possible. The carpets must be square, can overlap, and can't cover flawless panels. So my idea is: at each step, find the largest square that can be placed such that it covers only scratched... | The correct solution hinges on modeling the problem as a **minimum set cover** problem, where each possible square carpet (that covers only scratched panels and no flawless ones) is a "set" of scratched positions it can cover. The goal is to find the smallest collection of such sets that covers all scratched panels. To... | import sys
def main():
while True:
line = sys.stdin.readline().strip()
if not line:
break
W, H = map(int, line.split())
if W == 0 and H == 0:
break
grid = []
for _ in range(H):
row = list(map(int, sys.stdin.readline().split()))
... |
61daea303e18209f1a663fd48566e35b | In a far away land, there are two cities near a river. One day, the cities decide that they have too little space and would like to reclaim some of the river area into land.
The river area can be represented by a grid with r rows and exactly two columns — each cell represents a rectangular area. The rows are numbered ... | <think>
Okay, let's see. The problem is about determining whether the current player can win the game given the current state of the grid. The game is between two cities taking turns to reclaim cells under certain constraints, and the winner is the one who makes the last move.
First, I need to model the game state an... | codeforces | cc-by-4.0 | code_contests | train | 9 | r, n = map(int, input().split())
reclaimed = set()
blocked = set()
for _ in range(n):
ri, ci = map(int, input().split())
reclaimed.add((ri, ci))
# Mark the blocked cells from this reclaimed cell
for delta in (-1, 0, 1):
blocked_row = ri + delta
if 1 <= blocked_row <= r:
bloc... | Okay, so we have a river with r rows and 2 columns. The cities take turns reclaiming cells, but they can't reclaim a cell if it's adjacent (vertically or horizontally) to a reclaimed cell in the opposite column. Wait — actually, the problem says: if a cell (r,c) is reclaimed, then you can't reclaim (r-1, 3-c), (r, 3-c)... | The correct approach involves modeling the game as a combinatorial impartial game where each move removes a cell and blocks its symmetric neighbors in the opposite column. The key insight is that the game state can be reduced to a set of available moves, and a player wins if they can make a move that leaves no valid mo... | r, n = map(int, input().split())
reclaimed = set()
blocked = set()
for _ in range(n):
ri, ci = map(int, input().split())
reclaimed.add((ri, ci))
# Only block the same row, opposite column — incorrect!
blocked.add((ri, 3 - ci))
# Compute available cells: not reclaimed and not blocked
available = set()
... |
3150cd2fb3343f91bcf86e7fdfdfa148 | Smart Beaver recently got interested in a new word game. The point is as follows: count the number of distinct good substrings of some string s. To determine if a string is good or not the game uses rules. Overall there are n rules. Each rule is described by a group of three (p, l, r), where p is a string and l and r (... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The problem is about counting the number of distinct good substrings of a given string s. A substring is good if it complies with all the given rules. Each rule has a string p and two integers l and r. To comply with a ... | codeforces | cc-by-4.0 | code_contests | train | interview | class State:
def __init__(self):
self.transitions = {}
self.link = None
self.length = 0
self.count = 0
def build_sam(s):
sam = []
initial = State()
sam.append(initial)
last = initial
for c in s:
curr = State()
curr.length = last.length + 1
... | I need to find all distinct substrings of `s` and check each one against all rules. First, I’ll generate all substrings using two nested loops. Since the problem asks for *distinct* substrings, I’ll store them in a set to avoid duplicates. That’s straightforward.
Now, for each substring `t`, I need to check if it comp... | The correct approach involves generating all distinct substrings of the input string `s`, then checking each one against all `n` rules to see if it complies with each rule’s frequency constraints. The key insight is that for each substring `t`, we need to count how many times it appears in each rule’s string `p`. A nai... | s = input().strip()
n = int(input())
rules = []
for _ in range(n):
parts = input().split()
p, l, r = parts[0], int(parts[1]), int(parts[2])
rules.append((p, l, r))
# Generate all distinct substrings of s
substrings = set()
for i in range(len(s)):
for j in range(i + 1, len(s) + 1):
substrings.ad... |
28415922c4236680e3a67f85de94888f | For $n$ lists $L_i$ $(i = 0, 1, ..., n-1)$, perform a sequence of the following operations.
* insert($t$, $x$): Insert an integer $x$ at the end of $L_t$.
* dump($t$): Print all elements in $L_t$.
* splice($s$, $t$): Transfer elements of $L_s$ to the end of $L_t$. $L_s$ becomes empty.
In the initial state, $L_i$ $(... | <think>
Okay, let's see. I need to solve this programming problem. Let's read the problem statement carefully.
So, the problem is about handling a sequence of operations on n different lists. The operations are insert, dump, and splice. The challenge is to process these operations efficiently given the constraints, es... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | class Node:
__slots__ = ('value', 'next')
def __init__(self, value):
self.value = value
self.next = None
n, q = map(int, input().split())
lists = [{'head': None, 'tail': None} for _ in range(n)]
for _ in range(q):
parts = input().split()
if parts[0] == '0':
t = int(parts[1])
... | I see that we have up to 500,000 operations and 1,000 lists. Each list starts empty, and we need to support insert, dump, and splice.
The dump operation just prints the list, so I can use a regular Python list for each `L_i`. Inserting at the end is `append`, which is O(1). That seems fine.
For splicing, I need to mo... | The correct solution uses a linked list with head and tail pointers for each list to efficiently support appending and splicing operations in O(1) time. The key insight is that splicing two lists can be done by linking the tail of one to the head of another, then updating the destination's tail — all without copying el... | n, q = map(int, input().split())
lists = [[] for _ in range(n)]
for _ in range(q):
parts = input().split()
if parts[0] == '0':
t = int(parts[1])
x = int(parts[2])
lists[t].append(x)
elif parts[0] == '1':
t = int(parts[1])
print(' '.join(map(str, lists[t])))
elif ... |
bef5030efb1b82e572d26ade420e112a | Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 /... | <think>
Okay, I need to write a Python program that evaluates Reverse Polish Notation (RPN) expressions. Let's think about how to approach this.
First, RPN uses a stack. For each token in the input, if it's a number, push it onto the stack. If it's an operator, pop the top two elements, apply the operator, and push th... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import sys
for line in sys.stdin:
line = line.strip()
if not line:
continue
tokens = line.split()
stack = []
for token in tokens:
if token in '+-*/':
b = stack.pop()
a = stack.pop()
if token == '+':
res = a + b
elif tok... | I see that the input is in Reverse Polish Notation, which means operators come after their operands. The example shows that `10 2 12 - /` evaluates to `-1`, which is `10 / (2 - 12)`. So, I need to parse the tokens and compute step by step.
First, I’ll split the input line into tokens. That part seems straightforward. ... | The correct approach involves using a stack to evaluate Reverse Polish Notation (RPN) expressions: iterate through each token, push numbers onto the stack, and when encountering an operator, pop two operands, apply the operation, and push the result back. The key insight is that the order of popping matters—since the f... | import sys
for line in sys.stdin:
line = line.strip()
if not line:
continue
tokens = line.split()
stack = []
for token in tokens:
if token in '+-*/':
b = stack.pop()
a = stack.pop()
if token == '+':
res = a + b
elif tok... |
a90bacf94cc55a2d42cefe0082f70191 | Game is played on the field of the size 1 x N cells. The cells are numbered from 1 to N. In the i-th cell there are 2 positive integers - Ai and Bi.
Initially, the player stands at the fictive cell with the index 0 that is located right before all the N cells of the board. Then, he makes moves. Each move consists in m... | <think>
Okay, so I need to solve this programming problem. Let me read the problem statement carefully again to make sure I understand it.
The problem is about a game played on a 1xN grid. Each cell has two values, Ai and Bi. The player starts at cell 0 (before the first cell) and moves forward up to K cells each step... | hackerearth | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n, k = map(int, input().split())
cells = [tuple(map(int, input().split())) for _ in range(n)]
INF = float('inf')
min_penalty = INF
# Initialize DP: dp[j] contains list of (maxA, maxB) pairs
dp = [[] for _ in range(n + 1)]
dp[0] = [(0, 0)]
for j in range(n + 1):
if not dp[j]:
continue
for a, b in dp[j]... | I need to find the minimal penalty, which is `max(A) * max(B)` over visited cells. I start at position 0 and can jump up to K steps forward, ending at position N+1. The goal is to minimize the product of the maximum A and maximum B encountered along the way.
I think I can use dynamic programming. Let `dp[i]` be the mi... | The correct solution uses dynamic programming where `dp[i]` stores all non-dominated (maxA, maxB) pairs achievable when reaching cell `i`. The key insight is that we only need to keep Pareto-optimal pairs — if one pair has both values ≤ another, it can be safely discarded. This keeps the state space manageable and avoi... | n, k = map(int, input().split())
cells = []
for _ in range(n):
a, b = map(int, input().split())
cells.append((a, b))
# dp[i] = (maxA, maxB) for the best path to cell i (minimal product)
# We'll use a list to store the best state for each position
# But we'll store only one state per position: the one with mini... |
f2c7182f715cf8a5760847282d80e76b | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows:
* Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `888888... | <think>
Okay, let's see. The problem is about a string that transforms each day by expanding certain digits. After a huge number of days (5e15), we need to find the K-th character. Hmm, directly simulating this is impossible because the string would be way too large. So there's definitely a pattern or mathematical appr... | atcoder | cc-by-4.0 | code_contests | train | 0 | S = input().strip()
K = int(input())
count = 0
for c in S:
if c == '1':
count += 1
if count >= K:
print('1')
exit()
else:
if count < K:
print(c)
exit()
else:
continue | I see that the string grows every day by expanding each digit: `2` → `22`, `3` → `333`, etc. The problem asks for the K-th character after $5 \times 10^{15}$ days. That’s a huge number of days — I can’t simulate this directly. But maybe I can simulate a few days and see a pattern.
Let me try with the first example: `1... | The correct approach hinges on the observation that after a large number of days (like $5 \times 10^{15}$), the string has grown so vast that we cannot simulate it directly. Instead, we must realize that the transformation is deterministic and each digit expands exponentially in length over time. The key insight is tha... | S = input().strip()
K = int(input())
total_length = 0
for c in S:
# Assume that after 5e15 days, every digit's block is so large that its length exceeds 10^18
# So we treat the block length as 10**18 + 1 for any digit
block_length = 10**18 + 1
# If adding this block would make the total length at ... |
e34f5e8853cf129cc6c932656caac8fc | Yaroslav has n points that lie on the Ox axis. The coordinate of the first point is x1, the coordinate of the second point is x2, ..., the coordinate of the n-th point is — xn. Now Yaroslav wants to execute m queries, each of them is of one of the two following types:
1. Move the pj-th point from position xpj to pos... | <think>
Okay, let's see. We need to solve this programming problem. So, the problem is about handling a bunch of points on a line and processing queries that either move a point or ask for the sum of distances between all pairs in a certain interval. Hmm, that's the gist of it.
First, let's parse the problem statement... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | from bisect import bisect_left, bisect_right
class FenwickTree:
def __init__(self, size):
self.n = size
self.tree = [0] * (self.n + 1)
def update(self, idx, delta):
while idx <= self.n:
self.tree[idx] += delta
idx += idx & -idx
def query(self, idx):... | I see that we have up to 10⁵ points and 10⁵ queries, so I need an efficient way to handle updates and range queries. The points are on a line, and I need to answer queries about the sum of distances between all pairs in a given interval [l, r].
First, I think: "I can keep the points sorted in a list. When a point move... | The correct solution hinges on efficiently maintaining a dynamic set of points and answering range queries about the sum of pairwise distances within a segment. The key insight is that for a sorted list of coordinates in a range, the sum of all pairwise distances can be computed using a formula involving prefix sums: ... | import bisect
n = int(input())
points = list(map(int, input().split()))
m = int(input())
# Keep a sorted list of current points
sorted_points = sorted(points)
for _ in range(m):
parts = list(map(int, input().split()))
if parts[0] == 1:
p, d = parts[1], parts[2]
p -= 1 # 0-based index
... |
765ef144b6e007f22c881f0ce03a2ad2 | All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!
It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that th... | <think>
Okay, let's tackle this problem. Hmm, the wizard needs to make as many episodes as possible, each consisting of a path of two roads (so three cities visited in sequence, x-y-z), and each road can be used at most once. So the main constraint is that each road is used in at most one episode, either in one directi... | codeforces | cc-by-4.0 | code_contests | train | HARD | n, m = map(int, input().split())
edges = []
adj = [[] for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
if a > b:
a, b = b, a
edges.append((a, b))
adj[a].append((a, b))
adj[b].append((a, b))
used = set()
episodes = []
degree = [0] * (n + 1)
for a, b in edges:
... | I need to maximize the number of episodes, where each episode is a sequence of three cities connected by two roads: `x-y` and `y-z`. The wizard can't reuse any road, so each road can be used at most once. So, each episode consumes two roads that share a common city (the middle one).
At first, I think: "I can just find... | The correct solution hinges on modeling the problem as finding a maximum set of edge-disjoint paths of length 2 (i.e., three-node sequences `x-y-z`) such that no road is used more than once. The key insight is that each episode uses two edges incident to the middle city `y`, so the maximum number of episodes is limited... | n, m = map(int, input().split())
edges = []
adj = [[] for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
edges.append((a, b))
adj[a].append((a, b))
adj[b].append((a, b))
episodes = []
# Try to form episodes by iterating over all possible middle cities
# But don't sort by degre... |
bb799fd7429af01726c43740ad9341c1 | In the fest of Paradigm, every organiser is given a special kind of pager. The pager consists of a screen and a single big red button. To type a message in it, one has to push the red button that many time at which the alphabet occurs in the alphabetical order. For example for typing D, one needs to push button 4 times... | <think>
Okay, let's see. I need to solve this problem where I have to calculate the total number of button presses required to type a given set of strings. The problem statement says that each letter is typed by pressing a button as many times as its position in the alphabet. Like A is 1 press, B is 2, up to Z which is... | codechef | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n = int(input())
for _ in range(n):
s = input().strip()
print(sum(ord(c) - ord('A') + 1 for c in s)) | I need to figure out how many times the red button must be pressed to type each letter. The example says A needs 1 push, B needs 2, C needs 3, D needs 4, and so on. So it's just the alphabetical position.
But instead of using `ord(c) - ord('A') + 1`, I think I should create a list or dictionary to map each letter to i... | The correct approach is straightforward: for each character in the string, compute its position in the alphabet (A=1, B=2, ..., Z=26) and sum these values. This can be efficiently done using `ord(c) - ord('A') + 1`. The key insight is that the alphabetical position is directly computable via ASCII values.
However, a p... | n = int(input())
# Create a dictionary mapping each uppercase letter to its alphabetical position
letter_map = {}
for i in range(26):
letter_map[chr(ord('A') + i)] = i + 1
for _ in range(n):
s = input().strip()
total_pushes = 0
for char in s:
total_pushes += letter_map[char]
print(total_pus... |
7e361c5d300039fd743920af3a6e86b7 | Darshit is planning to celebrate the birthday of his friend, Diksha. There are two types of gifts that Diksha wants from Darshit: one is black and the other is white. To make her happy, Darshit has to buy B number of black gifts and W number of white gifts.
The cost of each black gift is X units.
The cost of every wh... | <think>
Okay, let's see. The problem is to find the minimum cost for Darshit to buy B black gifts and W white gifts. The costs are X per black, Y per white. But there's also the option to convert one color to another with a conversion cost Z per gift. So, for each gift, maybe it's cheaper to buy the other color and con... | hackerearth | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | def main():
import sys
input = sys.stdin.read().split()
idx = 0
T = int(input[idx])
idx += 1
for _ in range(T):
B = int(input[idx])
W = int(input[idx + 1])
idx += 2
X = int(input[idx])
Y = int(input[idx + 1])
Z = int(input[idx + 2])
idx += ... | I see that Darshit needs B black gifts and W white gifts. The prices are X for black, Y for white, and Z to convert between colors. At first glance, it seems like I should just compare the direct cost of buying everything: B*X + W*Y. But the problem mentions conversion, so maybe I can save money by buying one color and... | The correct approach hinges on realizing that Darshit can either buy gifts in their original color or convert them from the other color, whichever is cheaper. For black gifts, the effective cost is the minimum of buying them directly (X) or buying white gifts and converting them (Y + Z). Similarly, for white gifts, it'... | def main():
import sys
input = sys.stdin.read().split()
idx = 0
T = int(input[idx])
idx += 1
for _ in range(T):
B = int(input[idx])
W = int(input[idx + 1])
idx += 2
X = int(input[idx])
Y = int(input[idx + 1])
Z = int(input[idx + 2])
idx += ... |
eef357a9b3332c1f5f0a6d38d41b1b90 | You are given an integer k and n distinct points with integer coordinates on the Euclidean plane, the i-th point has coordinates (x_i, y_i).
Consider a list of all the (n(n - 1))/(2) pairs of points ((x_i, y_i), (x_j, y_j)) (1 ≤ i < j ≤ n). For every such pair, write out the distance from the line through these two po... | <think>
Okay, let's see. So the problem is to find the k-th smallest distance from the origin to the line formed by every pair of points given. Hmm. The input has n points, and we have to consider all possible pairs, calculate their line's distance to (0,0), then find the k-th smallest in that list.
First, I need to f... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | import math
from collections import defaultdict
def main():
import sys
input = sys.stdin.read
data = input().split()
idx = 0
n = int(data[idx])
k = int(data[idx+1])
idx += 2
points = []
for _ in range(n):
x = int(data[idx])
y = int(data[idx+1])
points.append(... | I need to find the k-th smallest distance from the origin to the line formed by each pair of points. The total number of pairs is $n(n-1)/2$, which could be huge — up to $5 \times 10^9$ for $n = 10^5$. But I don't think I need to compute all of them explicitly. Wait — actually, maybe I do? The problem asks for the k-th... | The correct solution hinges on two key insights: first, efficiently identifying pairs of points whose line passes through the origin (distance 0), and second, using binary search over the answer combined with a geometric condition to count how many lines have distance ≤ a given value. The crucial geometric formula for ... | import math
import sys
def main():
data = sys.stdin.read().split()
n = int(data[0])
k = int(data[1])
points = []
idx = 2
for i in range(n):
x = int(data[idx])
y = int(data[idx+1])
points.append((x, y))
idx += 2
distances = []
for i in range(n):
f... |
facd48e401f91a020adf5f3d83468c84 | Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.
Iahub asks Iahubina: can you build a rooted tree, such that
* each ... | <think>
Okay, I need to solve this programming problem. Let's read the problem statement carefully.
The task is to determine if there's a rooted tree that satisfies two conditions:
1. Each internal node (a node with at least one child) has at least two children.
2. For each node i, the subtree rooted at i has exactly... | codeforces | cc-by-4.0 | code_contests | train | competition | from collections import Counter
def can_partition(freq, target, min_children):
if target < 0:
return False
if min_children <= 0 and target == 0:
return True
sum_available = sum(s * count for s, count in freq.items())
if sum_available < target:
return False
if min_children > ... | I need to build a rooted tree with `n` nodes, where every internal node has at least two children, and each node `i` has `c[i]` nodes in its subtree.
First, I notice that the root must have a subtree of size `n`, so one of the `c[i]` values must be `n`. That seems necessary.
Also, if there's a node with subtree size ... | The correct solution hinges on a recursive decomposition of the tree structure: the root must have a subtree size of `n`, and its children must collectively account for `n - 1` nodes (since the root itself is one node). Each internal node must have at least two children, so any subtree of size `s` used as a child must ... | def solve():
n = int(input())
c = list(map(int, input().split()))
# Check if 2 is in the list - invalid because a node with subtree size 2 must have only one child
if 2 in c:
print("NO")
return
# Check that there is exactly one node with subtree size n (the root)
if... |
da3a26c20b92c7836b4b3fd179d71ae4 | You are given a sequence of n positive integers d1, d2, ..., dn (d1 < d2 < ... < dn). Your task is to construct an undirected graph such that:
* there are exactly dn + 1 vertices;
* there are no self-loops;
* there are no multiple edges;
* there are no more than 106 edges;
* its degree set is equal to d.... | <think>
Okay, let's see. I need to construct an undirected graph that satisfies certain conditions based on a given degree set. Hmm. The problem says the degree set is the sorted distinct values from the degree sequence. So the graph's degrees must include all the elements in d and no others.
First, the graph must hav... | codeforces | cc-by-4.0 | code_contests | train | interview | n = int(input())
d = list(map(int, input().split()))
m = d[-1] + 1
seq = d.copy()
seq += [d[-1]] * (m - n)
sum_seq = sum(seq)
if sum_seq % 2 != 0:
for i in range(n, len(seq)):
if seq[i] == d[-1]:
seq[i] = d[-2]
break
degrees = [(seq[i], i + 1) for i in range(m)]
edges = []
while True... | I need to build a graph with `dn + 1` vertices, where the distinct degrees are exactly the values in the input list `d`. The degrees must be from `d`, and no others.
I think: "I can just create a vertex for each degree in `d`, and then add extra vertices to make up the total count to `dn + 1`." But wait — I have `n` d... | The correct approach involves constructing a graph with exactly `dn + 1` vertices such that the set of distinct vertex degrees is exactly the given sequence `d`. The key insight is that we can use a greedy algorithm (like the Havel–Hakimi process) to simulate degree reduction, but we must ensure that the final degree s... | n = int(input())
d = list(map(int, input().split()))
m = d[-1] + 1
seq = d.copy()
seq += [d[-1]] * (m - n)
sum_seq = sum(seq)
if sum_seq % 2 != 0:
for i in range(n, len(seq)):
if seq[i] == d[-1]:
seq[i] = d[-2] # This will fail when n=1 because d[-2] doesn't exist
break
degrees = [(... |
6545c467232ea9225ae5587f72b78f85 | Taro is playing with a puzzle that places numbers 1-9 in 9x9 squares. In this puzzle, you have to arrange the numbers according to the following rules.
* One number appears exactly once in the same column
* A number appears exactly once on the same line
* In each of the 3x3 ranges separated by double lines, a number a... | <think>
Okay, I need to solve this problem where I have to check a Sudoku grid for rule violations and mark the incorrect numbers with an asterisk. Let me think about how to approach this.
First, the problem states that each row, column, and 3x3 subgrid must have exactly one of each number from 1-9. So, for each cell ... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | n = int(input())
for dataset in range(n):
grid = []
for _ in range(9):
row = list(map(int, input().split()))
grid.append(row)
row_dupes = [[False]*9 for _ in range(9)]
col_dupes = [[False]*9 for _ in range(9)]
subgrid_dupes = [[False]*9 for _ in range(9)]
# Check rows
... | I need to check if the puzzle is valid. The rules say each number 1–9 must appear exactly once in each row, column, and 3×3 subgrid. So, I should verify that for every row, column, and subgrid, the numbers 1 through 9 are all there — no more, no less.
First, I’ll read the input. There are multiple datasets, so I’ll lo... | The correct solution requires checking three independent constraints: each row, each column, and each 3×3 subgrid must contain the numbers 1–9 exactly once. A valid solution must ensure no number appears more than once in any of these regions. The key insight is to track duplicates per cell by checking all three constr... | n = int(input())
for dataset in range(n):
grid = []
for _ in range(9):
row = list(map(int, input().split()))
grid.append(row)
# Count frequency of each number globally
count = [0] * 10 # index 0 unused, 1-9 used
for i in range(9):
for j in range(9):
num = gr... |
c21dcba2cb63c43d1f69e4b5dc1f7a33 | Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separate... | <think>
Okay, let's tackle this problem. So, the task is to read a text and output two words: the most frequent one and the longest one. Hmm. Let's think about how to approach this step by step.
First, I need to split the input text into words. Since the input is a line with spaces separating words, splitting by space... | aizu | cc-by-4.0 | code_contests | train | UNKNOWN_DIFFICULTY | import collections
words = input().split()
counts = collections.Counter(words)
most_frequent = counts.most_common(1)[0][0]
max_len = -1
longest_word = ""
for word in words:
current_len = len(word)
if current_len > max_len:
max_len = current_len
longest_word = word
print(f"{most_frequent} {lon... | I need to find two things: the most frequent word and the longest word. First, I’ll split the input into words. Since I need frequency, I can use a dictionary to count how many times each word appears. That seems straightforward.
Now, for the most frequent word, I can just loop through the words and keep track of the ... | The correct approach requires two separate tasks: finding the most frequent word (using frequency counting) and finding the longest word (by comparing lengths). The key insight is that these two tasks are independent and must be solved separately—frequency does not correlate with length, and vice versa. A correct solut... | words = input().split()
counts = {}
for word in words:
counts[word] = counts.get(word, 0) + 1
most_frequent = max(counts, key=counts.get)
# I assume the most frequent word is also the longest word
longest_word = most_frequent
print(f"{most_frequent} {longest_word}") |
5c5acc023550745722411633a9bb1d66 | Firecrackers scare Nian the monster, but they're wayyyyy too noisy! Maybe fireworks make a nice complement.
Little Tommy is watching a firework show. As circular shapes spread across the sky, a splendid view unfolds on the night of Lunar New Year's eve.
A wonder strikes Tommy. How many regions are formed by the circl... | <think>
Okay, so I need to solve this problem where I have to find the number of regions formed by circles on a plane. The input is up to 3 circles, and I have to compute how many regions they divide the plane into. Hmm, right. Let's think about how circles can intersect and how that affects the regions.
First, I reme... | codeforces | cc-by-4.0 | code_contests | train | VERY_HARD | n = int(input())
circles = [tuple(map(int, input().split())) for _ in range(n)]
total = 0
for i in range(n):
x1, y1, r1 = circles[i]
for j in range(i + 1, n):
x2, y2, r2 = circles[j]
dx = x2 - x1
dy = y2 - y1
d_sq = dx * dx + dy * dy
sum_r = r1 + r2
sum_r_sq = s... | I need to find how many regions are formed by the circles. I know that one circle divides the plane into 2 regions: inside and outside. Two circles can make 4 regions if they intersect twice, or 3 if they touch, or 2 if one is inside the other. So the number of regions increases with intersections.
But how do I count ... | The correct solution relies on Euler’s formula for planar graphs: for a arrangement of circles, the number of regions is $ R = 2 + E - V $, where $ E $ is the number of intersection points (counted with multiplicity) and $ V $ is the number of vertices (i.e., intersection points), but actually in this case, it's simple... | n = int(input())
circles = []
for _ in range(n):
x, y, r = map(int, input().split())
circles.append((x, y, r))
# Use a grid to simulate the plane
# Coordinates from -20 to 20 in both x and y
# Use step size 0.1 to get reasonable resolution
step = 0.1
min_coord = -20
max_coord = 20
size = int((max_coord - min_c... |
b5279ed3f2e2096a37d89aca36e614a7 | You drew lottery N times. In the i-th draw, you got an item of the kind represented by a string S_i.
How many kinds of items did you get?
Constraints
* 1 \leq N \leq 2\times 10^5
* S_i consists of lowercase English letters and has a length between 1 and 10 (inclusive).
Input
Input is given from Standard Input in th... | <think>
Okay, so I need to solve this programming problem. Let's read the problem carefully.
The task is to find out how many different types of items someone got after N lottery draws. Each item is represented by a string S_i. The answer is the count of unique strings in the list of S_i's.
Hmm, right. So the problem... | atcoder | cc-by-4.0 | code_contests | train | 0 | n = int(input())
items = set()
for _ in range(n):
items.add(input().strip())
print(len(items)) | I need to count how many different kinds of items I got. So, I should go through each draw and check if the current item is already in my list of items I've seen.
I'll keep a list to store the unique items I've seen so far. For each new string, I'll check if it's already in the list. If it's not, I'll add it. Then at ... | The correct approach is to count the number of distinct strings among the N inputs. This is naturally solved by storing the strings in a set, which automatically handles duplicates, and then returning the size of the set. The key insight is that sets inherently track uniqueness, making this a perfect use case for them.... | n = int(input())
items = []
for _ in range(n):
items.append(input().strip())
items.sort()
count = 1 # at least one unique item
for i in range(1, n):
if items[i] != items[i-1]:
count += 1
print(count) |
6fd1f8eac6437dd366cfc5bc37eb03ce | Karlsson has recently discovered a huge stock of berry jam jars in the basement of the house. More specifically, there were 2n jars of strawberry and blueberry jam.
All the 2n jars are arranged in a row. The stairs to the basement are exactly in the middle of that row. So when Karlsson enters the basement, he sees exa... | <think>
Okay, let's tackle this problem. Hmm. So the problem is about Karlsson eating jars of jam such that after he's done, the remaining jars have an equal number of strawberry and blueberry. And we need to find the minimum number of jars he has to eat to achieve that.
Wait, the setup is that the jars are arranged i... | codeforces | cc-by-4.0 | code_contests | train | 9 | import sys
def main():
input = sys.stdin.read().split()
ptr = 0
t = int(input[ptr])
ptr += 1
for _ in range(t):
n = int(input[ptr])
ptr += 1
a = list(map(int, input[ptr:ptr + 2 * n]))
ptr += 2 * n
left = a[:n]
right = a[n:]
total_1 = a.count(1... | I see that Karlsson can only eat from the left or right ends, one jar at a time, and he must end up with equal numbers of strawberry and blueberry jars. So the goal is to remove some jars from the left and some from the right such that the remaining jars have equal counts.
First, I count how many strawberry (1) and bl... | The correct solution hinges on the insight that Karlsson must remove jars such that the remaining full jars have equal numbers of strawberry (1) and blueberry (2) jam. The difference between the total counts of 1s and 2s is `D = count(1) - count(2)`. To balance the remaining jars, Karlsson must remove a total of `D` mo... | import sys
def main():
input = sys.stdin.read().split()
ptr = 0
t = int(input[ptr])
ptr += 1
for _ in range(t):
n = int(input[ptr])
ptr += 1
a = list(map(int, input[ptr:ptr + 2 * n]))
ptr += 2 * n
left = a[:n]
right = a[n:]
t... |
2486198f63dd2accc39b01f01c7e2063 | Pupils decided to go to amusement park. Some of them were with parents. In total, n people came to the park and they all want to get to the most extreme attraction and roll on it exactly once.
Tickets for group of x people are sold on the attraction, there should be at least one adult in each group (it is possible tha... | <think>
Okay, let's tackle this problem. So, we need to find the minimum total price for all the visitors to the amusement park, grouping them such that each group has at least one adult. The price for a group of x people is c1 + c2*(x-1)^2. The challenge is to split everyone into groups optimally.
First, let's parse ... | codeforces | cc-by-4.0 | code_contests | train | interview | n, c1, c2 = map(int, input().split())
s = input().strip()
A = s.count('1')
P = len(s) - A
min_cost = float('inf')
for k in range(1, A + 1):
total_people = (A - k) + P
if total_people == 0:
cost = k * c1
else:
m, rem = divmod(total_people, k)
sum_sq = (k - rem) * (m ** 2) + rem * (... | I see that we have `n` people, some are adults (1s), some are pupils (0s). The total number of adults is `A`, pupils is `P`. Each group must have at least one adult, so the number of groups can't exceed `A`. But I think the best way is to use as few groups as possible — maybe just one group? That would minimize the num... | The correct solution hinges on the insight that we can form exactly `k` groups, where `k` ranges from 1 to the number of adults `A`, since each group must contain at least one adult. The remaining `P` pupils can be distributed among these `k` groups. The key idea is to minimize the total cost by trying all possible num... | n, c1, c2 = map(int, input().split())
s = input().strip()
A = s.count('1')
P = n - A
# Mistake: assume number of groups must be equal to number of adults
k = A
total_extra = n - k # extra people to distribute (non-essential adults + pupils)
if total_extra == 0:
cost = k * c1
else:
m, rem = divmod(total_extr... |
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