N = int(input())
A = list(map(int, input().split()))

total = abs(A[0])
cancel = [0 for _ in range(N)]
cancel[0] = abs(A[1]) - (abs(A[1] - A[0]) + abs(A[0]))

for i in range(1, N-1):
    total += abs(A[i]-A[i-1])
    cancel[i] = abs(A[i+1] - A[i-1]) - (abs(A[i+1] - A[i]) + abs(A[i] - A[i-1]))

total += abs(A[N-1]-A[N-2]) + abs(A[N-1])
cancel[N-1] = abs(A[N-2]) - (abs(A[N-1]) + abs(A[N-1] - A[N-2]))

for c in cancel:
    print(total+c)

Score : 300 points

Problem Statement

There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N.
Spot i is at the point with coordinate A_i.
It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis.

You planned a trip along the axis.
In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0.

However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i.
You will visit the remaining spots as planned in the order they are numbered.
You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned.

For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

Constraints

2 \leq N \leq 10^5

-5000 \leq A_i \leq 5000 (1 \leq i \leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N lines.
In the i-th line, print the total cost of travel during the trip when the visit to Spot i is canceled.

Sample Input 1

3
3 5 -1

Sample Output 1

12
8
10

Spot 1, 2 and 3 are at the points with coordinates 3, 5 and -1, respectively.
For each i, the course of the trip and the total cost of travel when the visit to Spot i is canceled, are as follows:

For i = 1, the course of the trip is 0 \rightarrow 5 \rightarrow -1 \rightarrow 0 and the total cost of travel is 5 + 6 + 1 = 12 yen.

For i = 2, the course of the trip is 0 \rightarrow 3 \rightarrow -1 \rightarrow 0 and the total cost of travel is 3 + 4 + 1 = 8 yen.

For i = 3, the course of the trip is 0 \rightarrow 3 \rightarrow 5 \rightarrow 0 and the total cost of travel is 3 + 2 + 5 = 10 yen.

Sample Input 2

5
1 1 1 2 0

Sample Output 2

4
4
4
2
4

Sample Input 3

6
-679 -2409 -3258 3095 -3291 -4462

Sample Output 3

21630
21630
19932
8924
21630
19288


a,b = map(int,input().split())
if a > b:
    big = a
    dif = a - b
    iros = '#'
else:
    big = b
    dif = b - a
    iros = '.'
print('2 {}'.format(big * 2))
if b > a:
    iro = ['#.' for Big in range(big)]
    print(''.join(iro))
    iro = ['#.' for Big in range(big - dif - 1)]
    for i in range(dif * 2 + 2):
        iro.append(iros)
    print(''.join(iro))
else:
    iro = ['.#' for Big in range(big)]
    print(''.join(iro))
    iro = ['.#' for Big in range(big - dif - 1)]
    for i in range(dif * 2 + 2):
        iro.append(iros)
    print(''.join(iro))

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 500 points

Problem Statement

You are given two integers A and B.

Print a grid where each square is painted white or black that satisfies the following conditions, in the format specified in Output section:

Let the size of the grid be h \times w (h vertical, w horizontal). Both h and w are at most 100.

The set of the squares painted white is divided into exactly A connected components.

The set of the squares painted black is divided into exactly B connected components.

It can be proved that there always exist one or more solutions under the conditions specified in Constraints section.
If there are multiple solutions, any of them may be printed.

Notes

Two squares painted white, c_1 and c_2, are called connected when the square c_2 can be reached from the square c_1 passing only white squares by repeatedly moving up, down, left or right to an adjacent square.

A set of squares painted white, S, forms a connected component when the following conditions are met:

Any two squares in S are connected.

No pair of a square painted white that is not included in S and a square included in S is connected.

A connected component of squares painted black is defined similarly.

Constraints

1 \leq A \leq 500

1 \leq B \leq 500

Input

Input is given from Standard Input in the following format:

A B

Output

Output should be in the following format:

In the first line, print integers h and w representing the size of the grid you constructed, with a space in between.

Then, print h more lines. The i-th (1 \leq i \leq h) of these lines should contain a string s_i as follows:

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted white, the j-th character in s_i should be ..

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted black, the j-th character in s_i should be #.

Sample Input 1

2 3

Sample Output 1

3 3
##.
..#
#.#

This output corresponds to the grid below:

Sample Input 2

7 8

Sample Output 2

3 5
#.#.#
.#.#.
#.#.#

Sample Input 3

1 1

Sample Output 3

4 2
..
#.
##
##

Sample Input 4

3 14

Sample Output 4

8 18
..................
..................
....##.......####.
....#.#.....#.....
...#...#....#.....
..#.###.#...#.....
.#.......#..#.....
#.........#..####.

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


import sys
input = sys.stdin.readline

MOD = 10 ** 9 + 7

N,M = map(int,input().split())
X = int(input())
UVW = [[int(x) for x in input().split()] for _ in range(M)]
UVW.sort(key = lambda x: x[2])

root = list(range(N+1))
def find_root(x):
    y = root[x]
    if x == y:
        return x
    z = find_root(y)
    root[x] = z
    return z

# min spanning tree
tree = [set() for _ in range(N+1)]
min_tree_size = 0
for u,v,w in UVW:
    ru = find_root(u)
    rv = find_root(v)
    if ru == rv:
        continue
    root[ru] = rv
    min_tree_size += w
    tree[u].add((v,w))
    tree[v].add((u,w))

# treeにおける2頂点の間の辺の最大値を求めたい
# LCA。2^n個手前およびそこまでの最大値を覚える

parent = [[0] * 12 for _ in range(N+1)]
max_wt = [[0] * 12 for _ in range(N+1)]
depth = [0] * (N+1)

def dfs(v=1,p=0,dep=0,w=0):
    parent[v][0] = p
    depth[v] = dep
    max_wt[v][0] = w
    for n in range(1,11):
        parent[v][n] = parent[parent[v][n-1]][n-1]
        max_wt[v][n] = max(max_wt[v][n-1], max_wt[parent[v][n-1]][n-1])
    for u,w in tree[v]:
        if u == p:
            continue
        dfs(u,v,dep+1,w)

dfs()

def max_wt_between(x,y):
    # LCA しながら重みの最大値を得る
    wt = 0
    dx,dy = depth[x], depth[y]
    if dx > dy:
        x,y = y,x
        dx,dy = dy,dx
    while dy > dx:
        diff = dy - dx
        step = diff & (-diff)
        n = step.bit_length() - 1
        wt = max(wt, max_wt[y][n])
        y = parent[y][n]
        dy -= step
    if x == y:
        return wt
    step = 1 << 11
    while step:
        n = step.bit_length() - 1
        rx,ry = parent[x][n], parent[y][n]
        if rx != ry:
            x,y = rx,ry
            wt = max(wt, max_wt[x][n], max_wt[y][n])
        step >>= 1
    return max(wt, max_wt[x][0], max_wt[y][0])

# 各edgeに対して、その辺を含む最小の全域木の大きさを求める
min_size = []
for u,v,w in UVW:
    if (v,w) in tree[u]:
        min_size.append(min_tree_size)
    else:
        x = max_wt_between(u,v)
        min_size.append(min_tree_size + w - x)

sm = sum(1 if s < X else 0 for s in min_size)
eq = sum(1 if s == X else 0 for s in min_size)
gr = sum(1 if s > X else 0 for s in min_size)

if eq == 0:
    answer = 0
elif sm == 0:
    # eq 内の辺が完全同色でなければよい
    answer = (pow(2,eq,MOD) - 2) * pow(2,gr,MOD) % MOD
else:
    # sm 内が完全同色でなければならない。
    # eq 内は、smの色と異なる色を持っていれば何でもよい
    answer = 2 * (pow(2,eq,MOD) - 1) * pow(2,gr,MOD) % MOD

print(answer)

Score : 900 points

Problem Statement

We have an undirected weighted graph with N vertices and M edges.
The i-th edge in the graph connects Vertex U_i and Vertex V_i, and has a weight of W_i.
Additionally, you are given an integer X.

Find the number of ways to paint each edge in this graph either white or black such that the following condition is met, modulo 10^9 + 7:

The graph has a spanning tree that contains both an edge painted white and an edge painted black. Furthermore, among such spanning trees, the one with the smallest weight has a weight of X.

Here, the weight of a spanning tree is the sum of the weights of the edges contained in the spanning tree.

Constraints

1 \leq N \leq 1 000

1 \leq M \leq 2 000

1 \leq U_i, V_i \leq N (1 \leq i \leq M)

1 \leq W_i \leq 10^9 (1 \leq i \leq M)

If i \neq j, then (U_i, V_i) \neq (U_j, V_j) and (U_i, V_i) \neq (V_j, U_j).

U_i \neq V_i (1 \leq i \leq M)

The given graph is connected.

1 \leq X \leq 10^{12}

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
X
U_1 V_1 W_1
U_2 V_2 W_2
:
U_M V_M W_M

Output

Print the answer.

Sample Input 1

3 3
2
1 2 1
2 3 1
3 1 1

Sample Output 1

6

Sample Input 2

3 3
3
1 2 1
2 3 1
3 1 2

Sample Output 2

2

Sample Input 3

5 4
1
1 2 3
1 3 3
2 4 6
2 5 8

Sample Output 3

0

Sample Input 4

8 10
49
4 6 10
8 4 11
5 8 9
1 8 10
3 8 128773450
7 8 10
4 2 4
3 4 1
3 1 13
5 2 2

Sample Output 4

4


n=int(input())
s=[input() for i in range(n)]
m=int(input())
t=[input() for j in range(m)]
c=0
for k in s:
  a=s.count(k)-t.count(k)
  if a>c:c=a
print(c)

Score : 200 points

Problem Statement

Takahashi has N blue cards and M red cards.
A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i.

Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen.

Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces atcoder, he will not earn money even if there are blue cards with atcoderr, atcode, btcoder, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.)

At most how much can he earn on balance?

Note that the same string may be written on multiple cards.

Constraints

N and M are integers.

1 \leq N, M \leq 100

s_1, s_2, ..., s_N, t_1, t_2, ..., t_M are all strings of lengths between 1 and 10 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N
M
t_1
t_2
:
t_M

Output

If Takahashi can earn at most X yen on balance, print X.

Sample Input 1

3
apple
orange
apple
1
grape

Sample Output 1

2

He can earn 2 yen by announcing apple.

Sample Input 2

3
apple
orange
apple
5
apple
apple
apple
apple
apple

Sample Output 2

1

If he announces apple, he will lose 3 yen. If he announces orange, he can earn 1 yen.

Sample Input 3

1
voldemort
10
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort

Sample Output 3

0

If he announces voldemort, he will lose 9 yen. If he announces orange, for example, he can avoid losing a yen.

Sample Input 4

6
red
red
blue
yellow
yellow
red
5
red
red
yellow
green
blue

Sample Output 4

1


n = int(input())
ab = [list(map(int, input().split())) for _ in range(n)]
cd = [list(map(int, input().split())) for _ in range(n)]

cd.sort()
remain = set(range(n))

ans = 0
for c,d in cd:
    cand = [-1, -1]
    for i in remain:
        a,b = ab[i]
        if((a<c) & (b<d)):
            if(cand[1] < b):
                cand = [i,b]

    if(cand[0] == -1):
        continue

    remain.remove(cand[0])
    ans += 1

print(ans)

Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points.
The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

All input values are integers.

1 \leq N \leq 100

0 \leq a_i, b_i, c_i, d_i < 2N

a_1, a_2, ..., a_N, c_1, c_2, ..., c_N are all different.

b_1, b_2, ..., b_N, d_1, d_2, ..., d_N are all different.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_N b_N
c_1 d_1
c_2 d_2
:
c_N d_N

Output

Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4


from collections import namedtuple

N = int(input())
ab = [tuple(map(int, input().split())) for i in range(N)]
cd = [tuple(map(int, input().split())) for i in range(N)]

G = [[] for i in range(N * 2 + 2)]

Edge = namedtuple("Edge", ("to", "cap", "rev"))

def add_edge(frm, to, cap):
    G[frm].append(Edge(to, cap, len(G[to])))
    G[to].append(Edge(frm, 0, len(G[frm]) - 1))

# 現在の頂点, 終点, 増加フロー
def dfs(v, t, f):
    if v == t:
        return f
    used[v] = True
    for i, e in enumerate(G[v]):
        if not used[e.to] and e.cap > 0:
            d = dfs(e.to, t, min(f, e.cap))
            if d > 0:
                G[v][i] = Edge(e.to, e.cap - d, e.rev)
                G[e.to][e.rev] = Edge(v, G[e.to][e.rev].cap + d, i)
                return d
    return 0

for i in range(1, N + 1):
    add_edge(0, i, 1)
    add_edge(N + i, N * 2 + 1, 1)

for i in range(1, N + 1):
    a, b = ab[i - 1]
    for j in range(1, N + 1):
        c, d = cd[j - 1]
        if a < c and b < d:
            add_edge(i, N + j, 1)

flow = 0
while 1:
    used = [False for i in range(N * 2 + 2)]
    f = dfs(0, N * 2 + 1, 114514)
    if f == 0:
        break
    flow += f

print(flow)


Score : 100 points

Problem Statement

We have a 3×3 square grid, where each square contains a lowercase English letters.
The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.

Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Constraints

Input consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

c_{11}c_{12}c_{13}
c_{21}c_{22}c_{23}
c_{31}c_{32}c_{33}

Output

Print the string of length 3 that can be obtained by concatenating the letters on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Sample Input 1

ant
obe
rec

Sample Output 1

abc

The letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid are a, b and c from top-right to bottom-left. Concatenate these letters and print abc.

Sample Input 2

edu
cat
ion

Sample Output 2

ean


a,b = map(int,input().split())
res = 0
for i in range(a,b+1):
  i = str(i)
  if i == i[::-1]:
    res += 1
print(res)

Score : 200 points

Problem Statement

Find the number of palindromic numbers among the integers between A and B (inclusive).
Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.

Constraints

10000 \leq A \leq B \leq 99999

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the number of palindromic numbers among the integers between A and B (inclusive).

Sample Input 1

11009 11332

Sample Output 1

4

There are four integers that satisfy the conditions: 11011, 11111, 11211 and 11311.

Sample Input 2

31415 92653

Sample Output 2

612


x, y = map(int,input().split())
ans = 0

for i in range(x,y+1):
    if str(i) == str(i)[::-1]:
        ans += 1

print(ans)


a = input().split(" ")
start = a[0]
start = list(map(int, start))
end = a[1]
end = list(map(int, end))

if (end[0] - start[0]) != 0:
    ans = (end[0] - start[0]  - 1) * 100 + end[1] * 10 + end[2] + 1 + (9 - start[1]) * 10 + (10 - start[2])
    if start[1]  < start[3] :
        ans -=   1
    else:
        if start[0] < start[4]:
            ans -= 1
else:
    if (end[1] - start[1]) != 0:
        ans = (end[1] - start[1] - 1)  * 10 + end[2] + (10 - start[2])
        if start[1]  < start[3] :
            ans -=   1
        else:
            if start[0] < start[4]:
                ans -= 1

    else:
        ans = end[2] - start[2] + 1

print(ans)


N,M = map(int, input().split())
if N<3 and M<3:
    print(0)
elif N==2:
    print(M-2)
elif M==2:
    print(N-2)
else:
    print((N-2)*(M-2))

Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region.
The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed.
Find the number of cards that face down after all the operations.

Constraints

1 \leq N,M \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.

Sample Input 1

2 2

Sample Output 1

0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.

Sample Input 2

1 7

Sample Output 2

5

After all the operations, all cards except at both ends face down.

Sample Input 3

314 1592

Sample Output 3

496080


from math import ceil,floor,factorial,gcd,sqrt,log2,cos,sin,tan,acos,asin,atan,degrees,radians,pi,inf,comb
from itertools import accumulate,groupby,permutations,combinations,product,combinations_with_replacement
from collections import deque,defaultdict,Counter
from bisect import bisect_left,bisect_right
from operator import itemgetter
from heapq import heapify,heappop,heappush
from queue import Queue,LifoQueue,PriorityQueue
from copy import deepcopy
from time import time
from functools import reduce
import string
import sys
sys.setrecursionlimit(10 ** 7)
def input() : return sys.stdin.readline().strip()
def INT()   : return int(input())
def MAP()   : return map(int,input().split())
def LIST()  : return list(MAP())

n, k = MAP()
ans = 0
if k == 0:
    ans = n**2
else:
    for b in range(k+1, n+1):
        x = n//b
        y = n % b
        ans += (b-k)*x + (y-k+1 if y >= k else 0)
print(ans)

Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten.
He remembers that the remainder of a divided by b was greater than or equal to K.
Find the number of possible pairs that he may have had.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N-1

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.

Sample Input 1

5 2

Sample Output 1

7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).

Sample Input 2

10 0

Sample Output 2

100

Sample Input 3

31415 9265

Sample Output 3

287927211


Score : 100 points

Problem Statement

There are N students in a school.

We will divide these students into some groups, and in each group they will discuss some themes.

You think that groups consisting of two or less students cannot have an effective discussion, so you want to have as many groups consisting of three or more students as possible.

Divide the students so that the number of groups consisting of three or more students is maximized.

Constraints

1 \leq N \leq 1000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N

Output

If you can form at most x groups consisting of three or more students, print x.

Sample Input 1

8

Sample Output 1

2

For example, you can form a group of three students and another of five students.

Sample Input 2

2

Sample Output 2

0

Sometimes you cannot form any group consisting of three or more students, regardless of how you divide the students.

Sample Input 3

9

Sample Output 3

3


h,w,d = map(int, input().split())

from collections import defaultdict
G = defaultdict(list)
for i in range(h):
    l = list(map(int, input().split()))
    for n, item in enumerate(l):
        G[item] = [n,i]

# magic is just distance by 1

# if d is 4,
# 1,5,9 or 2,6,10.. make surplus to categorize
# query comes 10^5, so O(1) or O(logN) is limit -> cumulativesum

dp = [0 for i in range(h*w+1)]
for i in range((h*w)%d+1): # 1,2,...d
    current_place = i+1
    while current_place + d <= h*w:
        x,y = G[current_place]
        next_place = current_place + d
        x1,y1 = G[next_place]
        magic = abs(x-x1) + abs(y-y1)
        dp[next_place] = magic + dp[current_place]
        current_place = next_place

q = int(input())
for i in range(q):
    l, r = map(int, input().split())
    print( dp[r]-dp[l], flush=True)

Score : 400 points

Problem Statement

We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).

The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is A_{i,j}.

You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |x-i|+|y-j| magic points.

You now have to take Q practical tests of your ability as a magical girl.

The i-th test will be conducted as follows:

Initially, a piece is placed on the square where the integer L_i is written.

Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not R_i. The test ends when x=R_i.

Here, it is guaranteed that R_i-L_i is a multiple of D.

For each test, find the sum of magic points consumed during that test.

Constraints

1 \leq H,W \leq 300

1 \leq D \leq H×W

1 \leq A_{i,j} \leq H×W

A_{i,j} \neq A_{x,y} ((i,j) \neq (x,y))

1 \leq Q \leq 10^5

1 \leq L_i \leq R_i \leq H×W

(R_i-L_i) is a multiple of D.

Input

Input is given from Standard Input in the following format:

H W D
A_{1,1} A_{1,2} ... A_{1,W}
:
A_{H,1} A_{H,2} ... A_{H,W}
Q
L_1 R_1
:
L_Q R_Q

Output

For each test, print the sum of magic points consumed during that test.

Output should be in the order the tests are conducted.

Sample Input 1

3 3 2
1 4 3
2 5 7
8 9 6
1
4 8

Sample Output 1

5

4 is written in Square (1,2).

6 is written in Square (3,3).

8 is written in Square (3,1).

Thus, the sum of magic points consumed during the first test is (|3-1|+|3-2|)+(|3-3|+|1-3|)=5.

Sample Input 2

4 2 3
3 7
1 4
5 2
6 8
2
2 2
2 2

Sample Output 2

0
0

Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.

Sample Input 3

5 5 4
13 25 7 15 17
16 22 20 2 9
14 11 12 1 19
10 6 23 8 18
3 21 5 24 4
3
13 13
2 10
13 13

Sample Output 3

0
5
0


n = int(input())
cards = sorted(list(map(int, input().split())), reverse=True)

alice = 0
bob = 0
for i in range(n):
    if i % 2 == 0:
        alice += cards[i]
    else:
        bob += cards[i]

print(alice - bob)

Score: 100 points

Problem Statement

E869120 has A 1-yen coins and infinitely many 500-yen coins.

Determine if he can pay exactly N yen using only these coins.

Constraints

N is an integer between 1 and 10000 (inclusive).

A is an integer between 0 and 1000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
A

Output

If E869120 can pay exactly N yen using only his 1-yen and 500-yen coins, print Yes; otherwise, print No.

Sample Input 1

2018
218

Sample Output 1

Yes

We can pay 2018 yen with four 500-yen coins and 18 1-yen coins, so the answer is Yes.

Sample Input 2

2763
0

Sample Output 2

No

When we have no 1-yen coins, we can only pay a multiple of 500 yen using only 500-yen coins. Since 2763 is not a multiple of 500, we cannot pay this amount.

Sample Input 3

37
514

Sample Output 3

Yes


N = int(input())
score = list(map(int, input().split()))
total = 0

score.sort(reverse=True)
for i in range(0, N, 2):
  if score[i+1] == None:
    total += score[i]
  else:
    total += score[i]-score[i+1]

print(total)

Score: 200 points

Problem Statement

We have N cards. A number a_i is written on the i-th card.

Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first.

The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

Constraints

N is an integer between 1 and 100 (inclusive).

a_i \ (1 \leq i \leq N) is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 a_3 ... a_N

Output

Print Alice's score minus Bob's score when both players take the optimal strategy to maximize their scores.

Sample Input 1

2
3 1

Sample Output 1

2

First, Alice will take the card with 3. Then, Bob will take the card with 1.
The difference of their scores will be 3 - 1 = 2.

Sample Input 2

3
2 7 4

Sample Output 2

5

First, Alice will take the card with 7. Then, Bob will take the card with 4. Lastly, Alice will take the card with 2. The difference of their scores will be 7 - 4 + 2 = 5. The difference of their scores will be 3 - 1 = 2.

Sample Input 3

4
20 18 2 18

Sample Output 3

18


n = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
ans = sum(A[::2]) - sum(A[1::2])
print(ans)


Score: 300 points

Problem Statement

We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.

Determine if he is correct.

Constraints

c_{i, j} \ (1 \leq i \leq 3, 1 \leq j \leq 3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c_{1,1} c_{1,2} c_{1,3}
c_{2,1} c_{2,2} c_{2,3}
c_{3,1} c_{3,2} c_{3,3}

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a_1=0,a_2=1,a_3=0,b_1=1,b_2=0,b_3=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No


import numpy as np
c = np.array([[int(i) for i in input().split()] for _ in range(3)])
print(np.sum(c) % 3 == 0)


from collections import deque
import sys
h,w = map(int,input().split())
mp = [list(input()) for _ in range(h)]

visited = [[0]*w for _ in range(h)]

dxdy = [[1,0],[0,1],[-1,0],[0,-1]]

queue = deque()
queue.append([0,0,1])
visited[0][0]=1

while queue:
    x,y,d = queue.popleft()
    for dx,dy in dxdy:
        nx,ny = x+dx,y+dy
        if -1<nx<h and -1<ny<w and visited[nx][ny]==0 and mp[nx][ny]!="#":
            visited[nx][ny]=d+1
            queue.append([nx,ny,d+1])
            #[print(*visited[l]) for l in range(h)]
            #print("---------------")

if visited[-1][-1]==0:
    print(-1)
    sys.exit()

cnt = 0
for i in range(h):
    for j in range(w):
        if mp[i][j]!="#":
            cnt += 1

print(cnt-visited[-1][-1])

Score: 400 points

Problem statement

We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j).

Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares.

Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game.

When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares.

The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is .; if square (i, j) is initially painted by black, s_{i, j} is #.

Constraints

H is an integer between 2 and 50 (inclusive).

W is an integer between 2 and 50 (inclusive).

s_{i, j} is . or # (1 \leq i \leq H, 1 \leq j \leq W).

s_{1, 1} and s_{H, W} are ..

Input

Input is given from Standard Input in the following format:

H W
s_{1, 1}s_{1, 2}s_{1, 3} ... s_{1, W}
s_{2, 1}s_{2, 2}s_{2, 3} ... s_{2, W}
: :
s_{H, 1}s_{H, 2}s_{H, 3} ... s_{H, W}

Output

Print the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed.

Sample Input 1

3 3
..#
#..
...

Sample Output 1

2

The score 2 can be achieved by changing the color of squares as follows:

Sample Input 2

10 37
.....................................
...#...####...####..###...###...###..
..#.#..#...#.##....#...#.#...#.#...#.
..#.#..#...#.#.....#...#.#...#.#...#.
.#...#.#..##.#.....#...#.#.###.#.###.
.#####.####..#.....#...#..##....##...
.#...#.#...#.#.....#...#.#...#.#...#.
.#...#.#...#.##....#...#.#...#.#...#.
.#...#.####...####..###...###...###..
.....................................

Sample Output 2

209


import sys

sys.setrecursionlimit(10 ** 8)

input = sys.stdin.readline

def main():
    N = int(input())
    A = [int(x) for x in input().split()]
    B = [int(x) for x in input().split()]

    if sum(A) > sum(B):
        print("No")
        return

    bamari = 0
    for i in range(N):
        if A[i] == B[i]:
            continue
        elif A[i] > B[i]:
            bamari -= A[i] - B[i]
        else:
            bamari += -(-(B[i] - A[i])) // 2
            bamari -= (B[i] - A[i]) % 2

    if bamari >= 0:
        print("Yes")
    else:
        print("No")

if __name__ == '__main__':
    main()

Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No


(n,),aa,bb = [list(map(int, s.split())) for s in open(0)]

up = down = 0

for i in range(n):
    a = aa[i]
    b = bb[i]
    if a > b:
        down += a-b
    if b > a:
        d,m = divmod(b-a,2)
        up += d+m
        down += m

diff = sum(bb) - sum(aa)

if up <= diff and down <= diff:
    print('Yes')
else:
    print('No')


X = int(input())
A = int(input())
B = int(input())

print((X-A)%B)

Score : 100 points

Problem Statement

You went shopping to buy cakes and donuts with X yen (the currency of Japan).

First, you bought one cake for A yen at a cake shop.
Then, you bought as many donuts as possible for B yen each, at a donut shop.

How much do you have left after shopping?

Constraints

1 \leq A, B \leq 1 000

A + B \leq X \leq 10 000

X, A and B are integers.

Input

Input is given from Standard Input in the following format:

X
A
B

Output

Print the amount you have left after shopping.

Sample Input 1

1234
150
100

Sample Output 1

84

You have 1234 - 150 = 1084 yen left after buying a cake.
With this amount, you can buy 10 donuts, after which you have 84 yen left.

Sample Input 2

1000
108
108

Sample Output 2

28

Sample Input 3

579
123
456

Sample Output 3

0

Sample Input 4

7477
549
593

Sample Output 4

405


n = int(input())
grid = [list(map(int, input().split())), list(map(int, input().split()))]
dist = [[0 for _ in range(n)] for _ in range(2)]
dist[0][0] = grid[0][0]
for i in range(2):
  for j in range(n):
    if j < n - 1:
      dist[i][j + 1] = max(dist[i][j + 1], dist[i][j] + grid[i][j + 1])
    if i == 0:
      dist[1][j] = max(dist[1][j], dist[0][j] + grid[1][j])
print(dist[-1][-1])

Score : 300 points

Problem Statement

We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).

You are initially in the top-left square, (1, 1).
You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.

The square (i, j) contains A_{i, j} candies.
You will collect all the candies you visit during the travel.
The top-left and bottom-right squares also contain candies, and you will also collect them.

At most how many candies can you collect when you choose the best way to travel?

Constraints

1 \leq N \leq 100

1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}

Output

Print the maximum number of candies that can be collected.

Sample Input 1

5
3 2 2 4 1
1 2 2 2 1

Sample Output 1

14

The number of collected candies will be maximized when you:

move right three times, then move down once, then move right once.

Sample Input 2

4
1 1 1 1
1 1 1 1

Sample Output 2

5

You will always collect the same number of candies, regardless of how you travel.

Sample Input 3

7
3 3 4 5 4 5 3
5 3 4 4 2 3 2

Sample Output 3

29

Sample Input 4

1
2
3

Sample Output 4

5


n = int(''.join(input().split()))
if isinstance(n ** 0.5, int):
    print('Yes')
else:
    print('No')

Score : 200 points

Problem Statement

AtCoDeer the deer has found two positive integers, a and b.
Determine whether the concatenation of a and b in this order is a square number.

Constraints

1 ≤ a,b ≤ 100

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

If the concatenation of a and b in this order is a square number, print Yes; otherwise, print No.

Sample Input 1

1 21

Sample Output 1

Yes

As 121 = 11 × 11, it is a square number.

Sample Input 2

100 100

Sample Output 2

No

100100 is not a square number.

Sample Input 3

12 10

Sample Output 3

No


import math

a,b = input().split()

r = math.sqrt(int(a+b))
if r - int(r) > 0:
    print('No')
else:
    print('Yes')


a,b=map(int, input().split())
for i in range(1,101):
  if i*i==a*100+b:
    print("Yes")
    exit()
print("No")


n = int(input())
for i in range(n):
    t,x,y=map(int,input().split())
    if (x + y) > t or (x + y + t) % 2:
        print("No")
        exit()
print("Yes")

Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane.
In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1).
Note that he cannot stay at his place.
Determine whether he can carry out his plan.

Constraints

1 ≤ N ≤ 10^5

0 ≤ x_i ≤ 10^5

0 ≤ y_i ≤ 10^5

1 ≤ t_i ≤ 10^5

t_i < t_{i+1} (1 ≤ i ≤ N-1)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 x_1 y_1
t_2 x_2 y_2
:
t_N x_N y_N

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No


# -*- coding: utf-8 -*-
"""
Created on Thu Aug  9 01:39:56 2018

@author: Yuki
"""

N = int(input())
plans = [[0, 0, 0]]
for i in range(N):
    plans.append([int(ii) for ii in input().split()])

Yes_flag = True
# 移動距離が遠すぎないかのチェック
for i in range(N):
    if (abs(plans[i+1][1] - plans[i][1]) + abs(plans[i+1][2] - plans[i][2])) > (plans[i+1][0] - plans[i][0]):
        Yes_flag = False
        break
for i in range(1, N+1):
    if plans[i][0] % 2 == 0:
        if (plans[i][1] + plans[i][2]) % 2 != 0:
            Yes_flag = False
            break
    else:
        if (plans[i][1] + plans[i][2]) % 2 == 0:
            Yes_flag = False
            break
if Yes_flag:
    print('Yes')
else:
    print('No')


Score : 100 points

Problem Statement

On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in yyyy/mm/dd format. For example, January 23, 2018 should be written as 2018/01/23.

After finishing the document, she noticed that she had mistakenly wrote 2017 at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to 2018 and prints it.

Constraints

S is a string of length 10.

The first eight characters in S are 2017/01/.

The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Replace the first four characters in S with 2018 and print it.

Sample Input 1

2017/01/07

Sample Output 1

2018/01/07

Sample Input 2

2017/01/31

Sample Output 2

2018/01/31


a = set()
for i in range(int(input())):
    a.add(int(input()))
print(len(a))

Score : 200 points

Problem Statement

An X-layered kagami mochi (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.

Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Constraints

1 ≤ N ≤ 100

1 ≤ d_i ≤ 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
d_1
:
d_N

Output

Print the maximum number of layers in a kagami mochi that can be made.

Sample Input 1

4
10
8
8
6

Sample Output 1

3

If we stack the mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, we have a 3-layered kagami mochi, which is the maximum number of layers.

Sample Input 2

3
15
15
15

Sample Output 2

1

When all the mochi have the same diameter, we can only have a 1-layered kagami mochi.

Sample Input 3

7
50
30
50
100
50
80
30

Sample Output 3

4


Score : 100 points

Problem Statement

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

Constraints

1≤M≤23

M is an integer.

Input

Input is given from Standard Input in the following format:

M

Output

If we have x hours until New Year at M o'clock on 30th, December, print x.

Sample Input 1

21

Sample Output 1

27

We have 27 hours until New Year at 21 o'clock on 30th, December.

Sample Input 2

12

Sample Output 2

36


a,b=map(int,input().split())
s=input()
flag=True
for i in range(0,a+b+1):
  if i<a and not s[i].isdecimal():
    flag=False
    break
  elif i==a and not s[i]=="-":
    flag=False
    break
  elif a<i and not s[i].isdecimal():
    flag=False
    break
if flag:
  print("Yes")
else:
  print("No")

Score : 200 points

Problem Statement

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen -, and the other characters are digits from 0 through 9.

You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Constraints

1≤A,B≤5

|S|=A+B+1

S consists of - and digits from 0 through 9.

Input

Input is given from Standard Input in the following format:

A B
S

Output

Print Yes if S follows the postal code format in AtCoder Kingdom; print No otherwise.

Sample Input 1

3 4
269-6650

Sample Output 1

Yes

The (A+1)-th character of S is -, and the other characters are digits from 0 through 9, so it follows the format.

Sample Input 2

1 1
---

Sample Output 2

No

S contains unnecessary -s other than the (A+1)-th character, so it does not follow the format.

Sample Input 3

1 2
7444

Sample Output 3

No


import sys
from collections import Counter

a, b = map(int, input().split())
s = input()

try:
    fs = int(s[:a])
    if s[a] != '-':
        raise NotHyphenError()
    ls = int(s[a+1:-1])
except:
    print('No')
    sys.exit()

print('Yes')


#!/usr/bin/env python3
import sys
import math

def solve(N: int, C: "List[int]", S: "List[int]", F: "List[int]"):
    # 駅間をC秒ではしる
    for i in range(N-1):
        cur = 0
        for j in range(i,N-1):
            if cur <= S[j]:
                cur = S[j]
            else:
                cur = -(-(cur-S[j])//F[j])*F[j]+S[j]
            cur += C[j]
        print(cur)
    print(0)
    return

def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    C = [int()] * (N - 1)  # type: "List[int]"
    S = [int()] * (N - 1)  # type: "List[int]"
    F = [int()] * (N - 1)  # type: "List[int]"
    for i in range(N - 1):
        C[i] = int(next(tokens))
        S[i] = int(next(tokens))
        F[i] = int(next(tokens))
    solve(N, C, S, F)

if __name__ == '__main__':
    main()

Score : 300 points

Problem Statement

A railroad running from west to east in Atcoder Kingdom is now complete.

There are N stations on the railroad, numbered 1 through N from west to east.

Tomorrow, the opening ceremony of the railroad will take place.

On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.

The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.

Here, it is guaranteed that F_i divides S_i.

That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.

For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Constraints

1≤N≤500

1≤C_i≤100

1≤S_i≤10^5

1≤F_i≤10

S_i％F_i=0

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 S_1 F_1
:
C_{N-1} S_{N-1} F_{N-1}

Output

Print N lines. Assuming that we are at Station i (1≤i≤N) when the ceremony begins, if the earliest possible time we can reach Station N is x seconds after the ceremony begins, the i-th line should contain x.

Sample Input 1

3
6 5 1
1 10 1

Sample Output 1

12
11
0

We will travel from Station 1 as follows:

5 seconds after the beginning: take the train to Station 2.

11 seconds: arrive at Station 2.

11 seconds: take the train to Station 3.

12 seconds: arrive at Station 3.

We will travel from Station 2 as follows:

10 seconds: take the train to Station 3.

11 seconds: arrive at Station 3.

Note that we should print 0 for Station 3.

Sample Input 2

4
12 24 6
52 16 4
99 2 2

Sample Output 2

187
167
101
0

Sample Input 3

4
12 13 1
44 17 17
66 4096 64

Sample Output 3

4162
4162
4162
0


def sieve(n):
    is_prime = [True for _ in range(n+1)]
    is_prime[0] = False

    for i in range(2, n+1):
        if is_prime[i-1]:
            j = 2 * i
            while j <= n:
                is_prime[j-1] = False
                j += i
    table = [ i for i in range(n+1) if is_prime[i-1]]
    return table

N = int(input())
L = [list(map(int, input().split())) for _ in range(N)]

prime = sieve(10 ** 5)

result = []

for i in range(N):
    res = 0
    for j in range(L[i][0], L[i][1]+1):
        if j in prime and (j+1)//2:
            res += 1
    result.append(res)

for res in result:
    print(res)

Score : 400 points

Problem Statement

We say that a odd number N is similar to 2017 when both N and (N+1)/2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Constraints

1≤Q≤10^5

1≤l_i≤r_i≤10^5

l_i and r_i are odd.

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2


a, b, c, d=map(int, input().split())
print("Left" if a+b > c+d else "Right" if a+b < c+d else "Balanced")

Score : 100 points

Problem Statement

A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.

Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.

Constraints

1\leq A,B,C,D \leq 10

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.

Sample Input 1

3 8 7 1

Sample Output 1

Left

The total weight of the masses on the left pan is 11, and the total weight of the masses on the right pan is 8. Since 11>8, we should print Left.

Sample Input 2

3 4 5 2

Sample Output 2

Balanced

The total weight of the masses on the left pan is 7, and the total weight of the masses on the right pan is 7. Since 7=7, we should print Balanced.

Sample Input 3

1 7 6 4

Sample Output 3

Right

The total weight of the masses on the left pan is 8, and the total weight of the masses on the right pan is 10. Since 8<10, we should print Right.


x,y = map(int,input().split())
cnt = 0
while True:
    if x > y:
        break
    else:
        cnt+=1
        x *= 2
print(cnt)

Score : 300 points

Problem Statement

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence.
The sequence A needs to satisfy the conditions below:

A consists of integers between X and Y (inclusive).

For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i.

Find the maximum possible length of the sequence.

Constraints

1 \leq X \leq Y \leq 10^{18}

All input values are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the maximum possible length of the sequence.

Sample Input 1

3 20

Sample Output 1

3

The sequence 3,6,18 satisfies the conditions.

Sample Input 2

25 100

Sample Output 2

3

Sample Input 3

314159265 358979323846264338

Sample Output 3

31


import math
a,b=map(int,input().split())
print(math.ceil((a+b)/2))

Score : 100 points

Problem Statement

You are given two positive integers a and b.
Let x be the average of a and b.
Print x rounded up to the nearest integer.

Constraints

a and b are integers.

1 \leq a, b \leq 100

Input

Input is given from Standard Input in the following format:

a b

Output

Print x rounded up to the nearest integer.

Sample Input 1

1 3

Sample Output 1

2

The average of 1 and 3 is 2.0, and it will be rounded up to the nearest integer, 2.

Sample Input 2

7 4

Sample Output 2

6

The average of 7 and 4 is 5.5, and it will be rounded up to the nearest integer, 6.

Sample Input 3

5 5

Sample Output 3

5


a,b = map(int,input().split())
if (a+b) % 2 == 0:
    print((a+b)//2)
else:
    print((a+b)//2+1)


s = raw_input()
t = raw_input()

sd = "".join(sorted(list(s)))
td = "".join(sorted(list(t), reverse=True))
if sd < td:
	print "Yes"
else:
	print "No"

Score : 200 points

Problem Statement

You are given strings s and t, consisting of lowercase English letters.
You will create a string s' by freely rearranging the characters in s.
You will also create a string t' by freely rearranging the characters in t.
Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Notes

For a string a = a_1 a_2 ... a_N of length N and a string b = b_1 b_2 ... b_M of length M, we say a < b for the lexicographic order if either one of the following two conditions holds true:

N < M and a_1 = b_1, a_2 = b_2, ..., a_N = b_N.

There exists i (1 \leq i \leq N, M) such that a_1 = b_1, a_2 = b_2, ..., a_{i - 1} = b_{i - 1} and a_i < b_i. Here, letters are compared using alphabetical order.

For example, xy < xya and atcoder < atlas.

Constraints

The lengths of s and t are between 1 and 100 (inclusive).

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If it is possible to satisfy s' < t', print Yes; if it is not, print No.

Sample Input 1

yx
axy

Sample Output 1

Yes

We can, for example, rearrange yx into xy and axy into yxa. Then, xy < yxa.

Sample Input 2

ratcode
atlas

Sample Output 2

Yes

We can, for example, rearrange ratcode into acdeort and atlas into tslaa. Then, acdeort < tslaa.

Sample Input 3

cd
abc

Sample Output 3

No

No matter how we rearrange cd and abc, we cannot achieve our objective.

Sample Input 4

w
ww

Sample Output 4

Yes

Sample Input 5

zzz
zzz

Sample Output 5

No


N = int(input())
A = list(map(int,input().split()))

d = {}
for a in A:
  if(a in d):
    d[a] += 1
  else:
    d[a] = 1
cnt = 0
for k, v in d.items():
  if(k > v):
    cnt += v
  elif(k < v):
    cnt += v - k
print(cnt)

Score : 300 points

Problem Statement

You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N).
Your objective is to remove some of the elements in a so that a will be a good sequence.

Here, an sequence b is a good sequence when the following condition holds true:

For each element x in b, the value x occurs exactly x times in b.

For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not.

Find the minimum number of elements that needs to be removed so that a will be a good sequence.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of elements that needs to be removed so that a will be a good sequence.

Sample Input 1

4
3 3 3 3

Sample Output 1

1

We can, for example, remove one occurrence of 3. Then, (3, 3, 3) is a good sequence.

Sample Input 2

5
2 4 1 4 2

Sample Output 2

2

We can, for example, remove two occurrences of 4. Then, (2, 1, 2) is a good sequence.

Sample Input 3

6
1 2 2 3 3 3

Sample Output 3

0

Sample Input 4

1
1000000000

Sample Output 4

1

Remove one occurrence of 10^9. Then, () is a good sequence.

Sample Input 5

8
2 7 1 8 2 8 1 8

Sample Output 5

5


require 'prime'
require 'set'
require 'tsort'
include Math
ALP = ('a'..'z').to_a
INF = 0xffffffffffffffff
def max(a,b);              a > b ? a : b                              end
def min(a,b);              a < b ?  a : b                             end
def swap(a,b);             a, b = b, a                                end
def gif;                   gets.to_i                                  end
def gff;                   gets.to_f                                  end
def gsf;                   gets.chomp                                 end
def gi;                    gets.split.map(&:to_i)                     end
def gf;                    gets.split.map(&:to_f)                     end
def gs;                    gets.chomp.split.map(&:to_s)               end
def gc;                    gets.chomp.split('')                       end
def pr(num);               num.prime_division                         end
def pr?(num);              Prime.prime?(num)                          end
def digit(num);            num.to_s.length                            end
def array(s,ini=nil);      Array.new(s){ini}                          end
def darray(s1,s2,ini=nil); Array.new(s1){Array.new(s2){ini}}          end
def rep(num);              num.times{|i|yield(i)}                     end
def repl(st,en,n=1);       st.step(en,n){|i|yield(i)}                 end

n = gif

a = gi

b = Hash.new
a.each do |aa|
b[aa] ||= 0
b[aa] += 1
end
ans = 0
b.each do |key,value|
  if value >= key
  ans += value - key
else
  ans += value
end
end

puts ans


n = int(input())
a = list(map(int, input().split()))
cnt = 0
while True:
	is_odd = False
	for i in range(len(a)):
		if a[i] % 2 == 1:
			is_odd = True
			break
		else:
			a[i] /= 2
	if is_odd:
		break
	cnt += 1

print(cnt)

Score : 200 points

Problem Statement

There are N positive integers written on a blackboard: A_1, ..., A_N.

Snuke can perform the following operation when all integers on the blackboard are even:

Replace each integer X on the blackboard by X divided by 2.

Find the maximum possible number of operations that Snuke can perform.

Constraints

1 \leq N \leq 200

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of operations that Snuke can perform.

Sample Input 1

3
8 12 40

Sample Output 1

2

Initially, [8, 12, 40] are written on the blackboard.
Since all those integers are even, Snuke can perform the operation.

After the operation is performed once, [4, 6, 20] are written on the blackboard.
Since all those integers are again even, he can perform the operation.

After the operation is performed twice, [2, 3, 10] are written on the blackboard.
Now, there is an odd number 3 on the blackboard, so he cannot perform the operation any more.

Thus, Snuke can perform the operation at most twice.

Sample Input 2

4
5 6 8 10

Sample Output 2

0

Since there is an odd number 5 on the blackboard already in the beginning, Snuke cannot perform the operation at all.

Sample Input 3

6
382253568 723152896 37802240 379425024 404894720 471526144

Sample Output 3

8


num = input()
l = input().split()
count = 0
for k in l:
    x = int(k) % 2
    if x == 0:
        for i in l:
            y = int(i) % 2
            if y == 0:
                count += 1
print(int((count - 3 )/ 3))


n = input()
nList = list(map(int,input().split()))
count = 0
while True:
  for i in range(len(nList)):
    nList[i] = nList[i]/2
    if not(nList[i].is_integer()):
      print(count)
      exit()
  count += 1


A=list(map(int,input().split()))
flag=0
count=0
ans=[]
for a in A:
  if a%2==0:
    count=a/2
  else:
    count=(a-1)/2
    flag=1
    ans.append(count)
    if flag==1:
      print(int(min(ans)))
    else:
      print(0)


N = int(input())
number = input().split()
number = [int(s) for s in number]
for i in range(1, 27):
   frag = 0
   for j in range(N):
      if number[j] % (2**i) != 0:
         frag = 1
         break
   if frag == 1:
      print(i - 1)
      break


n=int(input())
dat=list(map(int, input().split()))

for e in dat:
       e = bin(e)[2:]
       e = list(e).reverse()

re = []

for e in dat:
      try:
          w=e.index('1')
          re.append(w)
      except Exception:
          re.append(0)

print(min(re))


n = input()
a = list(map(int, input().split()))
ans = float("inf")
for i in a:
    ans = min(ans, len(bin(i)) - (bin(i).rfind("1") + 1))
print(ans)


import bisect,collections,copy,heapq,itertools,math,string
import sys
def S(): return sys.stdin.readline().rstrip()
def M(): return map(int,sys.stdin.readline().rstrip().split())
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LS(): return list(sys.stdin.readline().rstrip().split())
n, a, b = M()
print(min(n*a, b))

Score : 100 points

Problem Statement

You are parking at a parking lot. You can choose from the following two fee plans:

Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours.

Plan 2: The fee will be B yen, regardless of the duration.

Find the minimum fee when you park for N hours.

Constraints

1≤N≤20

1≤A≤100

1≤B≤2000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

When the minimum fee is x yen, print the value of x.

Sample Input 1

7 17 120

Sample Output 1

119

If you choose Plan 1, the fee will be 7×17=119 yen.

If you choose Plan 2, the fee will be 120 yen.

Thus, the minimum fee is 119 yen.

Sample Input 2

5 20 100

Sample Output 2

100

The fee might be the same in the two plans.

Sample Input 3

6 18 100

Sample Output 3

100


def main():
    N = input()
    print('YNeos'[not int(N) % sum(int(n) for n in N) == 0::2])

main()

Score : 200 points

Problem Statement

An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.

Given an integer N, determine whether it is a Harshad number.

Constraints

1?N?10^8

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print Yes if N is a Harshad number; print No otherwise.

Sample Input 1

12

Sample Output 1

Yes

f(12)=1+2=3. Since 12 is divisible by 3, 12 is a Harshad number.

Sample Input 2

57

Sample Output 2

No

f(57)=5+7=12. Since 57 is not divisible by 12, 12 is not a Harshad number.

Sample Input 3

148

Sample Output 3

No


import itertools
N = int(input())
F = []
for i in range(N):
    F_next = input().split()
    F_next = [int(i) for i in F_next]
    F.append(F_next)
P = []
for i in range(N):
    P_next = input().split()
    P_next = [int(i) for i in P_next]
    P.append(P_next)

#営業する時間を入力してそれぞれのお店との被る日数を出力
def count_number(shop):
    count = [0]*N
    for i in range(N):
        for j in range(10):
            if(shop[j]==1 and F[i][j]==1):
                count[i] = count[i]+1
    return (count)
#お店と被る日数を入力して利益を出力
def benefit(count):
    #一日も店を開かない日があってはダメなので利益を最小にしておく
    if(sum(count)==0):
        return(-100000000)
    bene_sum = 0
    for i in range(N):
        bene_sum = bene_sum + P[i][count[i]]
    return(bene_sum)

bene = []
for shop in itertools.product(range(2),repeat = 10):
    bene_next = benefit(count_number(shop))
    bene.append(bene_next)
print(max(bene))

Score : 300 points

Problem Statement

Joisino is planning to open a shop in a shopping street.

Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.

There are already N stores in the street, numbered 1 through N.

You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.

Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.

Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

Constraints

1≤N≤100

0≤F_{i,j,k}≤1

For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.

-10^7≤P_{i,j}≤10^7

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}

Output

Print the maximum possible profit of Joisino's shop.

Sample Input 1

1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2

Sample Output 1

8

If her shop is open only during the periods when Shop 1 is opened, the profit will be 8, which is the maximum possible profit.

Sample Input 2

2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1

Sample Output 2

-2

Note that a shop must be open during at least one period, and the profit may be negative.

Sample Input 3

3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7

Sample Output 3

23


n,c=list(map(int,input().split()))
s=[0 for i in range(n)]
t=[0 for i in range(n)]
c=[0 for i in range(n)]

for i in range(n):
  s[i],t[i],c[i]=list(map(int,input().split()))
  s[i]=s[i]-0.5

u=s+t
u.sort()

cnt=[0 for i in range(len(u)+1)]

for i in range(1,len(u)+1):
  cnt[i]=cnt[i-1]
  if isinstance(u[i-1],float):
    cnt[i]+=1
  else:
    cnt[i]-=1

print(max(cnt))

Score : 400 points

Problem Statement

Joisino is planning to record N TV programs with recorders.

The TV can receive C channels numbered 1 through C.

The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i.

Here, there will never be more than one program that are broadcast on the same channel at the same time.

When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T).

Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.

Constraints

1≤N≤10^5

1≤C≤30

1≤s_i<t_i≤10^5

1≤c_i≤C

If c_i=c_j and i≠j, either t_i≤s_j or s_i≥t_j.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N C
s_1 t_1 c_1
:
s_N t_N c_N

Output

When the minimum required number of recorders is x, print the value of x.

Sample Input 1

3 2
1 7 2
7 8 1
8 12 1

Sample Output 1

2

Two recorders can record all the programs, for example, as follows:

With the first recorder, record Channel 2 from time 1 to time 7. The first program will be recorded. Note that this recorder will be unable to record other channels from time 0.5 to time 7.

With the second recorder, record Channel 1 from time 7 to time 12. The second and third programs will be recorded. Note that this recorder will be unable to record other channels from time 6.5 to time 12.

Sample Input 2

3 4
1 3 2
3 4 4
1 4 3

Sample Output 2

3

There may be a channel where there is no program to record.

Sample Input 3

9 4
56 60 4
33 37 2
89 90 3
32 43 1
67 68 3
49 51 3
31 32 3
70 71 1
11 12 3

Sample Output 3

2


import bisect
N,Q = map(int,raw_input().split())
MAXN = 10**9
borders = [1]
while borders[-1] < MAXN:
  borders.append(borders[-1] + N**(len(borders) - 1))

def pos(a):
  rank = bisect.bisect_right(borders, a) - 1
  idx = a - borders[rank]
  return (rank,idx)

def parent(p):
  rank,idx = p
  return (rank-1, idx//N)

for i in range(Q):
  a,b = map(int,raw_input().split())
  if a > b: a,b = b,a
  p1,p2 = pos(a),pos(b)
  while p2[0] > p1[0]:
    p2 = parent(p2)
  while p1 != p2:
    p1,p2 = parent(p1),parent(p2)
  rank,idx = p1
  print(rank**N + idx + 1)

Score : 100 points

Problem Statement

You are given an integer N. Consider an infinite N-ary tree as shown below:

Figure: an infinite N-ary tree for the case N = 3

As shown in the figure, each vertex is indexed with a unique positive integer, and for every positive integer there is a vertex indexed with it. The root of the tree has the index 1. For the remaining vertices, vertices in the upper row have smaller indices than those in the lower row. Among the vertices in the same row, a vertex that is more to the left has a smaller index.

Regarding this tree, process Q queries. The i-th query is as follows:

Find the index of the lowest common ancestor (see Notes) of Vertex v_i and w_i.

Notes

In a rooted tree, the lowest common ancestor (LCA) of Vertex v and w is the farthest vertex from the root that is an ancestor of both Vertex v and w. Here, a vertex is considered to be an ancestor of itself. For example, in the tree shown in Problem Statement, the LCA of Vertex 5 and 7 is Vertex 2, the LCA of Vertex 8 and 11 is Vertex 1, and the LCA of Vertex 3 and 9 is Vertex 3.

Constraints

1 ≤ N ≤ 10^9

1 ≤ Q ≤ 10^5

1 ≤ v_i < w_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N Q
v_1 w_1
:
v_Q w_Q

Output

Print Q lines. The i-th line (1 ≤ i ≤ Q) must contain the index of the lowest common ancestor of Vertex v_i and w_i.

Sample Input 1

3 3
5 7
8 11
3 9

Sample Output 1

2
1
3

The queries in this case correspond to the examples shown in Notes.

Sample Input 2

100000 2
1 2
3 4

Sample Output 2

1
1


from collections import Counter
S = input()
counter = Counter(S)
for c in ["a", "b", "c"]:
    if c not in counter:
        counter[c] = 0
if (max(counter.values()) - min(counter.values())) <= 1:
    print("YES")
else:
    print("NO")

Score : 400 points

Problem Statement

Snuke has a string S consisting of three kinds of letters: a, b and c.

He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.

Constraints

1 \leq |S| \leq 10^5

S consists of a, b and c.

Input

Input is given from Standard Input in the following format:

S

Output

If the objective is achievable, print YES; if it is unachievable, print NO.

Sample Input 1

abac

Sample Output 1

YES

As it stands now, S contains a palindrome aba, but we can permute the characters to get acba, for example, that does not contain a palindrome of length 2 or more.

Sample Input 2

aba

Sample Output 2

NO

Sample Input 3

babacccabab

Sample Output 3

YES


N = input()
if N[0] ==N[1] and N[1]==N[2]:
  print("Yes")
elif N[1] ==N[2] and N[2]==N[3]:
  print("Yes")
else:
  print("No")

Score : 100 points

Problem Statement

We call a 4-digit integer with three or more consecutive same digits, such as 1118, good.

You are given a 4-digit integer N. Answer the question: Is N good?

Constraints

1000 ≤ N ≤ 9999

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is good, print Yes; otherwise, print No.

Sample Input 1

1118

Sample Output 1

Yes

N is good, since it contains three consecutive 1.

Sample Input 2

7777

Sample Output 2

Yes

An integer is also good when all the digits are the same.

Sample Input 3

1234

Sample Output 3

No


def main():
    a, b, c, d = input()
    print('YNeos'[not (a == b == c or b == c == d)::2])

main()


n=int(input())
flag=0
for i in range(1,10):
  for j in range(1,10):
    if n == i*1000+i*100+i*10+j or n == j*1000+j*100+j*10+i:
      flag = 1
if flag or n%1000==0:
  print("Yes")
else:
  print("No")


if __name__ == '__main__':

    n = int(input())

    A = [2,1]
    if n == 1:
        print(A[1])
    else:
        for i in range(2,n+1):
            A.append(A[i-2] + A[i-1])
    print(A[n])

Score : 200 points

Problem Statement

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.

You are given an integer N. Find the N-th Lucas number.

Here, the i-th Lucas number L_i is defined as follows:

L_0=2

L_1=1

L_i=L_{i-1}+L_{i-2} (i≥2)

Constraints

1≤N≤86

It is guaranteed that the answer is less than 10^{18}.

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the N-th Lucas number.

Sample Input 1

5

Sample Output 1

11

L_0=2

L_1=1

L_2=L_0+L_1=3

L_3=L_1+L_2=4

L_4=L_2+L_3=7

L_5=L_3+L_4=11

Thus, the 5-th Lucas number is 11.

Sample Input 2

86

Sample Output 2

939587134549734843


def inputs(num_of_input):
    ins = [input() for i in range(num_of_input)]
    return ins

def solve(inputs):
    nums = list(map(lambda x: int(x), inputs[0]))
    OPS = ["+", "-"]
    for temp_op1 in OPS:
        for temp_op2 in OPS:
            for temp_op3 in OPS:
                sum = nums[0]
                if temp_op1 == "+":
                    sum += nums[1]
                else:
                    sum -= nums[1]

                if temp_op2 == "+":
                    sum += nums[2]
                else:
                    sum -= nums[2]

                if temp_op3 == "+":
                    sum += nums[3]
                else:
                    sum -= nums[3]

                if sum == 7:
                    return "%d%s%d%s%d%s%d=7" % (
                        nums[0],
                        temp_op1,
                        nums[1],
                        temp_op2,
                        nums[2],
                        temp_op3,
                        nums[3],
                    )
    return ""

if __name__ == "__main__":
    ret = solve(inputs(1))
    print(ret)

Score : 300 points

Problem Statement

Sitting in a station waiting room, Joisino is gazing at her train ticket.

The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).

In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with + or - so that the formula holds.

The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Constraints

0≤A,B,C,D≤9

All input values are integers.

It is guaranteed that there is a solution.

Input

Input is given from Standard Input in the following format:

ABCD

Output

Print the formula you made, including the part =7.

Use the signs + and -.

Do not print a space between a digit and a sign.

Sample Input 1

1222

Sample Output 1

1+2+2+2=7

This is the only valid solution.

Sample Input 2

0290

Sample Output 2

0-2+9+0=7

0 - 2 + 9 - 0 = 7 is also a valid solution.

Sample Input 3

3242

Sample Output 3

3+2+4-2=7


def warshall_floyd():
    '''
    すべての頂点間の最短距離を求める
    '''
    for k in range(10):
        for i in range(10):
            for j in range(10):
                c[i][j] = min(c[i][j], c[i][k] + c[k][j])

h, w = map(int, input().split())
c = [list(map(int, input().split())) for _ in range(10)]
a = [list(map(int, input().split())) for _ in range(h)]

warshall_floyd()

res = 0
for i in range(h):
    for j in range(w):
        if a[i][j] >= 0:
            res += c[a[i][j]][1]

print(res)

Score : 400 points

Problem Statement

Joisino the magical girl has decided to turn every single digit that exists on this world into 1.

Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points).

She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive).

You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows:

If A_{i,j}≠-1, the square contains a digit A_{i,j}.

If A_{i,j}=-1, the square does not contain a digit.

Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

Constraints

1≤H,W≤200

1≤c_{i,j}≤10^3 (i≠j)

c_{i,j}=0 (i=j)

-1≤A_{i,j}≤9

All input values are integers.

There is at least one digit on the wall.

Input

Input is given from Standard Input in the following format:

H W
c_{0,0} ... c_{0,9}
:
c_{9,0} ... c_{9,9}
A_{1,1} ... A_{1,W}
:
A_{H,1} ... A_{H,W}

Output

Print the minimum total amount of MP required to turn every digit on the wall into 1 in the end.

Sample Input 1

2 4
0 9 9 9 9 9 9 9 9 9
9 0 9 9 9 9 9 9 9 9
9 9 0 9 9 9 9 9 9 9
9 9 9 0 9 9 9 9 9 9
9 9 9 9 0 9 9 9 9 2
9 9 9 9 9 0 9 9 9 9
9 9 9 9 9 9 0 9 9 9
9 9 9 9 9 9 9 0 9 9
9 9 9 9 2 9 9 9 0 9
9 2 9 9 9 9 9 9 9 0
-1 -1 -1 -1
8 1 1 8

Sample Output 1

12

To turn a single 8 into 1, it is optimal to first turn 8 into 4, then turn 4 into 9, and finally turn 9 into 1, costing 6 MP.

The wall contains two 8s, so the minimum total MP required is 6×2=12.

Sample Input 2

5 5
0 999 999 999 999 999 999 999 999 999
999 0 999 999 999 999 999 999 999 999
999 999 0 999 999 999 999 999 999 999
999 999 999 0 999 999 999 999 999 999
999 999 999 999 0 999 999 999 999 999
999 999 999 999 999 0 999 999 999 999
999 999 999 999 999 999 0 999 999 999
999 999 999 999 999 999 999 0 999 999
999 999 999 999 999 999 999 999 0 999
999 999 999 999 999 999 999 999 999 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1

Sample Output 2

0

Note that she may not need to change any digit.

Sample Input 3

3 5
0 4 3 6 2 7 2 5 3 3
4 0 5 3 7 5 3 7 2 7
5 7 0 7 2 9 3 2 9 1
3 6 2 0 2 4 6 4 2 3
3 5 7 4 0 6 9 7 6 7
9 8 5 2 2 0 4 7 6 5
5 4 6 3 2 3 0 5 4 3
3 6 2 3 4 2 4 0 8 9
4 6 5 4 3 5 3 2 0 8
2 1 3 4 5 7 8 6 4 0
3 5 2 6 1
2 5 3 2 1
6 9 2 5 6

Sample Output 3

47


import sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, acos, atan, asin, log, log10
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from fractions import gcd
def input(): return sys.stdin.readline().strip()
def STR(): return input()
def INT(): return int(input())
def FLOAT(): return float(input())
def MAP(): return map(int, input().split())
def S_MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
def S_LIST(): return list(map(str, input().split()))
def lcm(a, b): return a * b // gcd(a, b)
sys.setrecursionlimit(10 ** 9)
inf = sys.maxsize
mod = 10 ** 9 + 7
dx = [0, 0, 1, -1, 1, -1, -1, 1]
dy = [1, -1, 0, 0, 1, -1, 1, -1]

def func(num, goto, ignore):
    if num == goto:
        return 0
    mn = inf
    retnum = 0
    for j in range(10):
        if j == num or j in ignore:
            continue
        if c[num][j] < mn:
            mn = c[num][j]
            retnum = j
    ignore.append(num)
    res = func(retnum, goto, ignore) + mn
    return res

h, w = MAP()
c = [LIST() for _ in range(10)]
a = [LIST() for _ in range(h)]
for i in range(10):
    for j in range(10):
        c[i][j] = min(c[i][j], func(i, j, []))
ans = 0
for i in range(h):
    for j in range(w):
        if a[i][j] == -1 or a[i][j] == 1:
            continue
        ans += c[a[i][j]][1]
print(ans)


h,w = map(int, input().split())
c = [list(map(int, input().split())) for i in range(10)]
a = [list(map(int, input().split())) for x in range(h)]

for k in range(10):
    for i in range(10):
        for j in range(10):
            c[i][j] = min(c[i][j], c[i][k]+c[k][j])

ans = 0

for i in range(h):
    for j in range(w):
        if a[i][j] != 0 and a[i][j] != -1:
            ans += c[a[i][j]][1]

print(ans)


Score : 200 points

Problem Statement

We have a long seat of width X centimeters.
There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.

We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.

At most how many people can sit on the seat?

Constraints

All input values are integers.

1 \leq X, Y, Z \leq 10^5

Y+2Z \leq X

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

13 3 1

Sample Output 1

3

There is just enough room for three, as shown below:

Figure

Sample Input 2

12 3 1

Sample Output 2

2

Sample Input 3

100000 1 1

Sample Output 3

49999

Sample Input 4

64146 123 456

Sample Output 4

110

Sample Input 5

64145 123 456

Sample Output 5

109


x, y, z = map(int, input().split())
print(int((x - z) / (y + z)))


def MI(): return map(int, input().split())
def II(): return int(input())
def IS(): return input()
def LI(): return list(map(int, input().split()))

n, m = MI()
p = (0.5) ** m
t = 1900*m + 100*(n-m)
print(int(t//p))

Score : 300 points

Problem Statement

Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is YES or NO.

When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.

Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.

Now, he goes through the following process:

Submit the code.

Wait until the code finishes execution on all the cases.

If the code fails to correctly solve some of the M cases, submit it again.

Repeat until the code correctly solve all the cases in one submission.

Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).

Constraints

All input values are integers.

1 \leq N \leq 100

1 \leq M \leq {\rm min}(N, 5)

Input

Input is given from Standard Input in the following format:

N M

Output

Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.

Sample Input 1

1 1

Sample Output 1

3800

In this input, there is only one case. Takahashi will repeatedly submit the code that correctly solves this case with 1/2 probability in 1900 milliseconds.

The code will succeed in one attempt with 1/2 probability, in two attempts with 1/4 probability, and in three attempts with 1/8 probability, and so on.

Thus, the answer is 1900 \times 1/2 + (2 \times 1900) \times 1/4 + (3 \times 1900) \times 1/8 + ... = 3800.

Sample Input 2

10 2

Sample Output 2

18400

The code will take 1900 milliseconds in each of the 2 cases, and 100 milliseconds in each of the 10-2=8 cases. The probability of the code correctly solving all the cases is 1/2 \times 1/2 = 1/4.

Sample Input 3

100 5

Sample Output 3

608000


#!python3

# input
N, M = list(map(int, input().split()))

def main():
    t = 1900 * M + 100 * (N - M)
    e = 2 ** M
    ans = t * e
    print(ans)

if __name__ == "__main__":
    main()


Score : 200 points

Problem Statement

Find the largest square number not exceeding N. Here, a square number is an integer that can be represented as the square of an integer.

Constraints

1 \leq N \leq 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the largest square number not exceeding N.

Sample Input 1

10

Sample Output 1

9

10 is not square, but 9 = 3 × 3 is. Thus, we print 9.

Sample Input 2

81

Sample Output 2

81

Sample Input 3

271828182

Sample Output 3

271821169


def solve():
    N = int(input())
    ans1 = 0
    n = 0
    while ans1 <= N:
        ans1 = n ** 2
        n += 1

    print((n - 2) **  2)

solve()


import bisect

N=int(input())
A=[int(i) for i in input().split()]
B=[int(i) for i in input().split()]
C=[int(i) for i in input().split()]
A.sort()
B.sort()
C.sort()
# print(A)
# print(B)
# print(C)

# ans=0
# for i in range(N):
#     part1=A[i]
#     part2_pos=bisect.bisect_right(B,part1)
#     for j in range(part2_pos,N):
#         part2=B[j]
#         part3_pos=bisect.bisect_right(C,part2)
#         # print(i,j,part3_spos,": ",A[i],B[j],C[part3_pos])
#         ans+=(N-part3_pos)
# print(ans)

AtoB_Pos=[]
BtoC_Pos=[]
ans=0
for i in range(N):
    B_Pos=bisect.bisect_right(B,A[i])
    AtoB_Pos.append(B_Pos)

for i in range(N):
    C_Pos=bisect.bisect_right(C,B[i])
    BtoC_Pos.append(C_Pos)
# print(AtoB_Pos)
# print(BtoC_Pos)
for i in range(N):
    for j in range(AtoB_Pos[i],N):
        # print(i,j,": ",A[i],B[j])
        # print(N-AtoB_Pos[i],N-BtoC_Pos[j])
        ans+=(N-BtoC_Pos[j])
print(ans)

Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

1 \leq C_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N
B_1 ... B_N
C_1 ... C_N

Output

Print the number of different altars that Ringo can build.

Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

3

The following three altars can be built:

Upper: 1-st part, Middle: 1-st part, Lower: 1-st part

Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part

Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

Sample Input 3

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

87


import bisect
n=int(input())
a=sorted(list(map(int,input().split())))
b=sorted(list(map(int,input().split())))
c=sorted(list(map(int,input().split())))
# print(a)
# print(b)
# print(c)

ans=0
acount=[0]*n
bcount=[0]*n
ruiseki=[0]*(n+1)
for i in range(n):
    acount[i]=n-bisect.bisect(b,a[i])
    bcount[i]=n-bisect.bisect(c,b[i])
    ruiseki[i+1]=ruiseki[i]+bcount[i]
# print(acount)
# print(bcount)
# print(ruiseki)

for i in range(n):
    # print(ruiseki[n]-ruiseki[n-acount[i]])
    ans+=ruiseki[n]-ruiseki[n-acount[i]]
print(ans)


n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
a.sort()
a.reverse()
b.sort()
b.reverse()
c.sort()
c.reverse()
c.append(0)

i=0
l=[]
for j in b:
  while c[i]>j:
    i+=1
  l.append(i)
b.append(0)
s=0
cnt=0
i=0
for j in a:
  while b[i]>j:
    s+=l[i]
    i+=1
  cnt+=s
print(cnt)


r = int(input())
g = int(input())

print(g*2 - r)

Score : 100 points

Problem Statement

Takahashi is a user of a site that hosts programming contests.

When a user competes in a contest, the rating of the user (not necessarily an integer) changes according to the performance of the user, as follows:

Let the current rating of the user be a.

Suppose that the performance of the user in the contest is b.

Then, the new rating of the user will be the avarage of a and b.

For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.

Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.

Find the performance required to achieve it.

Constraints

0 \leq R, G \leq 4500

All input values are integers.

Input

Input is given from Standard Input in the following format:

R
G

Output

Print the performance required to achieve the objective.

Sample Input 1

2002
2017

Sample Output 1

2032

Takahashi's current rating is 2002.

If his performance in the contest is 2032, his rating will be the average of 2002 and 2032, which is equal to the desired rating, 2017.

Sample Input 2

4500
0

Sample Output 2

-4500

Although the current and desired ratings are between 0 and 4500, the performance of a user can be below 0.


N = int(input())
K = int(input())

a = 1

for i in range(N):
  if K >= a:
    a *= 2
  else:
    a += K

print(a)

Score : 200 points

Problem Statement

Square1001 has seen an electric bulletin board displaying the integer 1.
He can perform the following operations A and B to change this value:

Operation A: The displayed value is doubled.

Operation B: The displayed value increases by K.

Square1001 needs to perform these operations N times in total.
Find the minimum possible value displayed in the board after N operations.

Constraints

1 \leq N, K \leq 10

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the minimum possible value displayed in the board after N operations.

Sample Input 1

4
3

Sample Output 1

10

The value will be minimized when the operations are performed in the following order: A, A, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 7 → 10.

Sample Input 2

10
10

Sample Output 2

76

The value will be minimized when the operations are performed in the following order: A, A, A, A, B, B, B, B, B, B.

In this case, the value will change as follows: 1 → 2 → 4 → 8 → 16 → 26 → 36 → 46 → 56 → 66 → 76.

By the way, this contest is AtCoder Beginner Contest 076.


nums = input()
set_nums = list(set(nums))
other = sum(nums) - sum(set_nums)
set_nums.remove(other)
print(set_nums[0])

Score : 100 points

Problem Statement

You are given three integers, A, B and C.

Among them, two are the same, but the remaining one is different from the rest.

For example, when A=5,B=7,C=5, A and C are the same, but B is different.

Find the one that is different from the rest among the given three integers.

Constraints

-100 \leq A,B,C \leq 100

A, B and C are integers.

The input satisfies the condition in the statement.

Input

Input is given from Standard Input in the following format:

A B C

Output

Among A, B and C, print the integer that is different from the rest.

Sample Input 1

5 7 5

Sample Output 1

7

This is the same case as the one in the statement.

Sample Input 2

1 1 7

Sample Output 2

7

In this case, C is the one we seek.

Sample Input 3

-100 100 100

Sample Output 3

-100


n,m,k = map(int, input().split())

for i in range(n+1):
  for j in range(m+1):
    if i*(m-j)+j*(n-1):
      print("Yes")
      exit()
print("No")

Score : 200 points

Problem Statement

We have a grid with N rows and M columns of squares. Initially, all the squares are white.

There is a button attached to each row and each column.
When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa.
When a button attached to a column is pressed, the colors of all the squares in that column are inverted.

Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Constraints

1 \leq N,M \leq 1000

0 \leq K \leq NM

Input

Input is given from Standard Input in the following format:

N M K

Output

If Takahashi can have exactly K black squares in the grid, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

Press the buttons in the order of the first row, the first column.

Sample Input 2

2 2 1

Sample Output 2

No

Sample Input 3

3 5 8

Sample Output 3

Yes

Press the buttons in the order of the first column, third column, second row, fifth column.

Sample Input 4

7 9 20

Sample Output 4

No


a = list(map(int,input().split()))
s = []
if a[0] <= a[1]:
	small = a[0]
	big = a[1]
else:
	small = a[1]
	big = a[0]
s.append(a[0] * a[1] - 1)

for i in range(1,small + 1):
	s.append(s[i-1] + s[0] - (2 * i) * 2)
if a[2] in s:
	print('Yes')
else:
	print('No')


from collections import Counter

def main():
    H, W = map(int, input().split())
    A = [input() for _ in range(H)]
    space = {}
    space[1] = (H % 2) * (W % 2)
    space[2] = (H % 2) * (W//2*2) + (W % 2) * (H//2*2)
    space[4] = (H//2*2) * (W//2*2)
    cnt = Counter("".join(A))
    for alpha in cnt:
        c = cnt[alpha]
        for k in (4, 2, 1):
            while c//k >= 1 and space[k]:
                y = min(c//k*k, space[k])
                space[k] -= y
                c -= y
        if c:
            print("No")
            return
    print("Yes")

if __name__ == "__main__":
    main()

Score : 400 points

Problem Statement

We have an H-by-W matrix.
Let a_{ij} be the element at the i-th row from the top and j-th column from the left.
In this matrix, each a_{ij} is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A.
Here, he wants to satisfy the following condition:

Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

3 4
aabb
aabb
aacc

Sample Output 1

Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

2 2
aa
bb

Sample Output 2

No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

5 1
t
w
e
e
t

Sample Output 3

Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

2 5
abxba
abyba

Sample Output 4

No

Sample Input 5

1 1
z

Sample Output 5

Yes


n = int(input())
a = int(input())
print(n**2-a)

Score : 100 points

Problem Statement

We have an N \times N square grid.

We will paint each square in the grid either black or white.

If we paint exactly A squares white, how many squares will be painted black?

Constraints

1 \leq N \leq 100

0 \leq A \leq N^2

Inputs

Input is given from Standard Input in the following format:

N
A

Outputs

Print the number of squares that will be painted black.

Sample Input 1

3
4

Sample Output 1

5

There are nine squares in a 3 \times 3 square grid.
Four of them will be painted white, so the remaining five squares will be painted black.

Sample Input 2

19
100

Sample Output 2

261

Sample Input 3

10
0

Sample Output 3

100

As zero squares will be painted white, all the squares will be painted black.


n = int(input())
k = int(input())
x = list(map(int,input().split()))
ans = 0
for i in x:
    ans += 2 * min(i,k-i)
print(ans)

Score : 200 points

Problem Statement

There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i).
Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N.

In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B.
Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i).
Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N.

When activated, each type of robot will operate as follows.

When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.

Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.

Constraints

1 \leq N \leq 100

1 \leq K \leq 100

0 < x_i < K

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
K
x_1 x_2 ... x_N

Outputs

Print the minimum possible total distance covered by robots.

Sample Input 1

1
10
2

Sample Output 1

4

There are just one ball, one type-A robot and one type-B robot.

If the type-A robot is used to collect the ball, the distance from the robot to the ball is 2, and the distance from the ball to the original position of the robot is also 2, for a total distance of 4.

Similarly, if the type-B robot is used, the total distance covered will be 16.

Thus, the total distance covered will be minimized when the type-A robot is used. The output should be 4.

Sample Input 2

2
9
3 6

Sample Output 2

12

The total distance covered will be minimized when the first ball is collected by the type-A robot, and the second ball by the type-B robot.

Sample Input 3

5
20
11 12 9 17 12

Sample Output 3

74


from collections import Counter,defaultdict,deque
from heapq import heappop,heappush,heapify
import sys,bisect,math,itertools,fractions
sys.setrecursionlimit(10**8)
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))

a,b,c,d,e,f = inpl()
mx = 0
res = 0
ans = 0
for i in range(40):
    for j in range(40):
        water = 100*a*i + 100*b*j
        if water == 0:
            continue
        for k in range(300):
            for l in range(300):
                sugar = c*k + d*l
                if water+sugar > f or sugar > (i+j)*e:
                    continue
                if sugar/(water+sugar) >= mx:
                    mx = sugar/(water+sugar)
                    res = water+sugar
                    ans = sugar
print(res,ans)

Score : 300 points

Problem Statement

Snuke is making sugar water in a beaker.
Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations.

Operation 1: Pour 100A grams of water into the beaker.

Operation 2: Pour 100B grams of water into the beaker.

Operation 3: Put C grams of sugar into the beaker.

Operation 4: Put D grams of sugar into the beaker.

In our experimental environment, E grams of sugar can dissolve into 100 grams of water.

Snuke will make sugar water with the highest possible density.

The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker.
Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it.
If there is more than one candidate, any of them will be accepted.

We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent.
Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.

Constraints

1 \leq A < B \leq 30

1 \leq C < D \leq 30

1 \leq E \leq 100

100A \leq F \leq 3 000

A, B, C, D, E and F are all integers.

Inputs

Input is given from Standard Input in the following format:

A B C D E F

Outputs

Print two integers separated by a space.
The first integer should be the mass of the desired sugar water, and the second should be the mass of the sugar dissolved in it.

Sample Input 1

1 2 10 20 15 200

Sample Output 1

110 10

In this environment, 15 grams of sugar can dissolve into 100 grams of water, and the beaker can contain at most 200 grams of substances.

We can make 110 grams of sugar water by performing Operation 1 once and Operation 3 once.
It is not possible to make sugar water with higher density.
For example, the following sequences of operations are infeasible:

If we perform Operation 1 once and Operation 4 once, there will be undissolved sugar in the beaker.

If we perform Operation 2 once and Operation 3 three times, the mass of substances in the beaker will exceed 200 grams.

Sample Input 2

1 2 1 2 100 1000

Sample Output 2

200 100

There are other acceptable outputs, such as:

400 200

However, the output below is not acceptable:

300 150

This is because, in order to make 300 grams of sugar water containing 150 grams of sugar, we need to pour exactly 150 grams of water into the beaker, which is impossible.

Sample Input 3

17 19 22 26 55 2802

Sample Output 3

2634 934