x, y, z, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = sorted(list(map(int, input().split())))[z-k:]
ab, abc = [], []

for i in a:
    for j in b:
        ab.append(i+j)

ab.sort()
s = ab[x*y-k:]
for i in s:
    for j in c:
        abc.append(i+j)

abc.sort(reverse=True)
for i in range(k):
    print(abc[i])

Score: 400 points

Problem Statement

The Patisserie AtCoder sells cakes with number-shaped candles.
There are X, Y and Z kinds of cakes with 1-shaped, 2-shaped and 3-shaped candles, respectively.
Each cake has an integer value called deliciousness, as follows:

The deliciousness of the cakes with 1-shaped candles are A_1, A_2, ..., A_X.

The deliciousness of the cakes with 2-shaped candles are B_1, B_2, ..., B_Y.

The deliciousness of the cakes with 3-shaped candles are C_1, C_2, ..., C_Z.

Takahashi decides to buy three cakes, one for each of the three shapes of the candles, to celebrate ABC 123.

There are X \times Y \times Z such ways to choose three cakes.

We will arrange these X \times Y \times Z ways in descending order of the sum of the deliciousness of the cakes.

Print the sums of the deliciousness of the cakes for the first, second, ..., K-th ways in this list.

Constraints

1 \leq X \leq 1 \ 000

1 \leq Y \leq 1 \ 000

1 \leq Z \leq 1 \ 000

1 \leq K \leq \min(3 \ 000, X \times Y \times Z)

1 \leq A_i \leq 10 \ 000 \ 000 \ 000

1 \leq B_i \leq 10 \ 000 \ 000 \ 000

1 \leq C_i \leq 10 \ 000 \ 000 \ 000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y Z K
A_1 \ A_2 \ A_3 \ ... \ A_X
B_1 \ B_2 \ B_3 \ ... \ B_Y
C_1 \ C_2 \ C_3 \ ... \ C_Z

Output

Print K lines. The i-th line should contain the i-th value stated in the problem statement.

Sample Input 1

2 2 2 8
4 6
1 5
3 8

Sample Output 1

19
17
15
14
13
12
10
8

There are 2 \times 2 \times 2 = 8 ways to choose three cakes, as shown below in descending order of the sum of the deliciousness of the cakes:

(A_2, B_2, C_2): 6 + 5 + 8 = 19

(A_1, B_2, C_2): 4 + 5 + 8 = 17

(A_2, B_1, C_2): 6 + 1 + 8 = 15

(A_2, B_2, C_1): 6 + 5 + 3 = 14

(A_1, B_1, C_2): 4 + 1 + 8 = 13

(A_1, B_2, C_1): 4 + 5 + 3 = 12

(A_2, B_1, C_1): 6 + 1 + 3 = 10

(A_1, B_1, C_1): 4 + 1 + 3 = 8

Sample Input 2

3 3 3 5
1 10 100
2 20 200
1 10 100

Sample Output 2

400
310
310
301
301

There may be multiple combinations of cakes with the same sum of the deliciousness. For example, in this test case, the sum of A_1, B_3, C_3 and the sum of A_3, B_3, C_1 are both 301.
However, they are different ways of choosing cakes, so 301 occurs twice in the output.

Sample Input 3

10 10 10 20
7467038376 5724769290 292794712 2843504496 3381970101 8402252870 249131806 6310293640 6690322794 6082257488
1873977926 2576529623 1144842195 1379118507 6003234687 4925540914 3902539811 3326692703 484657758 2877436338
4975681328 8974383988 2882263257 7690203955 514305523 6679823484 4263279310 585966808 3752282379 620585736

Sample Output 3

23379871545
22444657051
22302177772
22095691512
21667941469
21366963278
21287912315
21279176669
21160477018
21085311041
21059876163
21017997739
20703329561
20702387965
20590247696
20383761436
20343962175
20254073196
20210218542
20150096547

Note that the input or output may not fit into a 32-bit integer type.


n,q = map(int,input().split())
s = list(input())
point = []
mark = []
for i in range(n):
     if s[i] in mark:
          point[mark.index(s[i])].append(i)
     else:
          mark.append(s[i])
          point.append([i])

g = [0] + [1]*n +[0]

def f(j,o,g):
     if o == "L":
          g[j-1] += g[j]
     else:
          g[j+1] += g[j]
     g[j] = 0
     return g

for i in range(q):
     al,o = map(str,input().split())
     if al in mark:
          for j in point[mark.index(al)]:
               if g[j+1] > 0:
                    g = f(j+1,o,g)

g[0],g[-1] = 0,0
print(sum(g))

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

Score : 500 points

Problem Statement

There are N squares numbered 1 to N from left to right.
Each square has a character written on it, and Square i has a letter s_i. Besides, there is initially one golem on each square.

Snuke cast Q spells to move the golems.

The i-th spell consisted of two characters t_i and d_i, where d_i is L or R.
When Snuke cast this spell, for each square with the character t_i, all golems on that square moved to the square adjacent to the left if d_i is L, and moved to the square adjacent to the right if d_i is R.

However, when a golem tried to move left from Square 1 or move right from Square N, it disappeared.

Find the number of golems remaining after Snuke cast the Q spells.

Constraints

1 \leq N,Q \leq 2 \times 10^{5}

|s| = N

s_i and t_i are uppercase English letters.

d_i is L or R.

Input

Input is given from Standard Input in the following format:

N Q
s
t_1 d_1
\vdots
t_{Q} d_Q

Output

Print the answer.

Sample Input 1

3 4
ABC
A L
B L
B R
A R

Sample Output 1

2

Initially, there is one golem on each square.

In the first spell, the golem on Square 1 tries to move left and disappears.

In the second spell, the golem on Square 2 moves left.

In the third spell, no golem moves.

In the fourth spell, the golem on Square 1 moves right.

After the four spells are cast, there is one golem on Square 2 and one golem on Square 3, for a total of two golems remaining.

Sample Input 2

8 3
AABCBDBA
A L
B R
A R

Sample Output 2

5

After the three spells are cast, there is one golem on Square 2, two golems on Square 4 and two golems on Square 6, for a total of five golems remaining.

Note that a single spell may move multiple golems.

Sample Input 3

10 15
SNCZWRCEWB
B R
R R
E R
W R
Z L
S R
Q L
W L
B R
C L
A L
N L
E R
Z L
S L

Sample Output 3

3

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.


b = input()
if b == "A":
  print("T")
elif b == "C":
  print("G")
elif b == "G":
  print("C")
elif b == "T":
  print("A")

Score : 100 points

Problem Statement

On the Planet AtCoder, there are four types of bases: A, C, G and T. A bonds with T, and C bonds with G.

You are given a letter b as input, which is A, C, G or T. Write a program that prints the letter representing the base that bonds with the base b.

Constraints

b is one of the letters A, C, G and T.

Input

Input is given from Standard Input in the following format:

b

Output

Print the letter representing the base that bonds with the base b.

Sample Input 1

A

Sample Output 1

T

Sample Input 2

G

Sample Output 2

C


lngs = input()

ACGT = ["A", "C", "G", "T"]
out_list = []

for lng in lngs:
    if lng in ACGT:
        out_list += lng
out = ''.join(out_list)
print(out)

Score : 200 points

Problem Statement

You are given a string S consisting of uppercase English letters. Find the length of the longest ACGT string that is a substring (see Notes) of S.

Here, a ACGT string is a string that contains no characters other than A, C, G and T.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

S is a string of length between 1 and 10 (inclusive).

Each character in S is an uppercase English letter.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest ACGT string that is a substring of S.

Sample Input 1

ATCODER

Sample Output 1

3

Among the ACGT strings that are substrings of ATCODER, the longest one is ATC.

Sample Input 2

HATAGAYA

Sample Output 2

5

Among the ACGT strings that are substrings of HATAGAYA, the longest one is ATAGA.

Sample Input 3

SHINJUKU

Sample Output 3

0

Among the ACGT strings that are substrings of SHINJUKU, the longest one is (the empty string).


S = list(input())
n = []
a = 0
for s in S :
	if s in ["A","T","G","C"] :
		a += 1
	else :
		n.append(a if a > 0 else 0)
		a = 0
print(max(n))


S = list(input())
m = 0
count = 0

for i in range(len(S)):
    for j in range(i, len(S)):
        count = 0
        for s in list(S[i:j+1]):
            if s in ['A', 'G', 'T', 'C']:
                count += 1
            else:
                if m <= count:
                    m = count
                count = 0
print(m)


n,q=map(int,input().split())
s=input()
lr=[list(map(int,input().split())) for _ in range(q)]
L=[0,0]
c=0
for i in range(1,n):
	if s[i-1]=='A' and s[i]=='C':
		c += 1
	L.append(c)
print(L)
for l,r in lr:
	print(L[r]-L[l])

Score : 300 points

Problem Statement

You are given a string S of length N consisting of A, C, G and T. Answer the following Q queries:

Query i (1 \leq i \leq Q): You will be given integers l_i and r_i (1 \leq l_i < r_i \leq N). Consider the substring of S starting at index l_i and ending at index r_i (both inclusive). In this string, how many times does AC occurs as a substring?

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

2 \leq N \leq 10^5

1 \leq Q \leq 10^5

S is a string of length N.

Each character in S is A, C, G or T.

1 \leq l_i < r_i \leq N

Input

Input is given from Standard Input in the following format:

N Q
S
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

8 3
ACACTACG
3 7
2 3
1 8

Sample Output 1

2
0
3

Query 1: the substring of S starting at index 3 and ending at index 7 is ACTAC. In this string, AC occurs twice as a substring.

Query 2: the substring of S starting at index 2 and ending at index 3 is CA. In this string, AC occurs zero times as a substring.

Query 3: the substring of S starting at index 1 and ending at index 8 is ACACTACG. In this string, AC occurs three times as a substring.


N, Q = map(int, input().split())
S = input()
ns = [-1]*N
n = 0
ns[0] = n
for i in range(N-1):
  if S[i:i+2] == "AC":
    n += 1
  ns[i+1] = n
#print(ns)
for i in range(Q):
  l, r = map(int, input().split())
  l, r = l-1, r-1
  c = ns[r]-ns[l]
  print(c)


n, q = map(int,input().split())
s = input()
ruiseki = [0]*(n+1)
for i in range(n):
    if s[i:i+2] == 'AC':
        ac = 1
    else:
        ac = 0
    ruiseki[i+1] = ruiseki[i] + ac
ans_list = []
for i in range(q):
    l,r = map(int,input().split())
    ans_list.append(ruiseki[r-1]-ruiseki[l-1])
for i in range(q):
    print(ans_list[i])


N,Q=map(int,input().split())
S=input()
t = [0] * (N + 1)
for i in range(N):
    t[i + 1] = t[i] + (1 if S[i : i + 2] == 'AC' else 0)
for i in range(Q):
    l, r = map(int, input().split())
    print(t[r-1] - t[l-1])


[n,q] = [int(i) for i in input().split()]
s = input()
for i in range(q):
    [l,r] = [int(i) for i in input().split()]
    print(s[l-1:r].count("AC"))


# 直前数文字しか影響しないと思えばDPと見当つく
MOD = 10**9 + 7
N = int(input())
dp = [[[[0]*5 for i in range(5)] for i in range(5)] for i in range(N+1)]

dp[0][0][0][0] = 1
for n in range(N):
    for i in range(5):
        for j in range(5):
            for k in range(5):
                for l in range(1,5):
                    if j == 1 and k == 2 and l == 3: continue  # AGC
                    if j == 2 and k == 1 and l == 3: continue  # GAC
                    if j == 1 and k == 3 and l == 2: continue  # ACG
                    if i == 1 and k == 2 and l == 3: continue  # A?GC
                    if i == 1 and j == 2 and l == 3: continue  # AG?C
                    dp[n+1][j][k][l] += dp[n][i][j][k]

ans = 0
for i in range(1,5):
    for j in range(1,5):
        for k in range(1,5):
            ans += dp[N][i][j][k]
            ans %= MOD
print(ans)

Score : 400 points

Problem Statement

You are given an integer N. Find the number of strings of length N that satisfy the following conditions, modulo 10^9+7:

The string does not contain characters other than A, C, G and T.

The string does not contain AGC as a substring.

The condition above cannot be violated by swapping two adjacent characters once.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of strings of length N that satisfy the following conditions, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

61

There are 4^3 = 64 strings of length 3 that do not contain characters other than A, C, G and T. Among them, only AGC, ACG and GAC violate the condition, so the answer is 64 - 3 = 61.

Sample Input 2

4

Sample Output 2

230

Sample Input 3

100

Sample Output 3

388130742

Be sure to print the number of strings modulo 10^9+7.


N = int(input())

if N % 2 == 0:
    #偶数の場合 i , N+1-i ペア以外を出力
    print(N*(N-2)//2)
    for i in range(1,N+1):
        for j in range(i+1,N+1):
            if i + j == N+1:
                continue
            print(i,j)
else:
    #奇数の場合 i, N-i ペア以外を出力
    print(((N-1)**2)//2)
    for i in range(1,N+1):
        for j in range(i+1,N+1):
            if i + j == N:
                continue
            print(i,j)

Score : 700 points

Problem Statement

You are given an integer N.
Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions:

The graph is simple and connected.

There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S.

It can be proved that at least one such graph exists under the constraints of this problem.

Constraints

All values in input are integers.

3 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

In the first line, print the number of edges, M, in the graph you made. In the i-th of the following M lines, print two integers a_i and b_i, representing the endpoints of the i-th edge.

The output will be judged correct if the graph satisfies the conditions.

Sample Input 1

3

Sample Output 1

2
1 3
2 3

For every vertex, the sum of the indices of the vertices adjacent to that vertex is 3.


import sys
sys.setrecursionlimit(10 ** 9)
# input = sys.stdin.readline    ####
def int1(x): return int(x) - 1
def II(): return int(input())
def MI(): return map(int, input().split())
def MI1(): return map(int1, input().split())
def LI(): return list(map(int, input().split()))
def LI1(): return list(map(int1, input().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def MS(): return input().split()
def LS(): return list(input())
def LLS(rows_number): return [LS() for _ in range(rows_number)]
def printlist(lst, k=' '): print(k.join(list(map(str, lst))))
INF = float('inf')
# from math import ceil, floor, log2
from collections import deque, defaultdict
# from itertools import combinations as comb, combinations_with_replacement as comb_w, accumulate, product, permutations
# from heapq import heapify, heappop, heappush
# import numpy as np    # cumsum
# from bisect import bisect_left, bisect_right

def comb(n, r):
    # if n - r < 0: return 0
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n

    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]

    for p in range(2, r + 1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p - 1, r, p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot

    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])

    return result

def solve():
    N = II()
    S = input()

    d = defaultdict(lambda: 0)
    for s in S:
        d[s] += 1

    ans = 1
    for n in d.values():
        ans *= (n+1)
    print(ans-1)

if __name__ == '__main__':
    solve()

Score : 200 points

Problem Statement

You are given a string S of length N.
Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.

Here, a subsequence of a string is a concatenation of one or more characters from the string without changing the order.

Constraints

1 \leq N \leq 100000

S consists of lowercase English letters.

|S|=N

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of the subsequences such that all characters are different, modulo 10^9+7.

Sample Input 1

4
abcd

Sample Output 1

15

Since all characters in S itself are different, all its subsequences satisfy the condition.

Sample Input 2

3
baa

Sample Output 2

5

The answer is five: b, two occurrences of a, two occurrences of ba. Note that we do not count baa, since it contains two as.

Sample Input 3

5
abcab

Sample Output 3

17


H,W = map(int,input().split())
h,w = map(int,input().split())
print((H-h)*(W-w))

Score : 100 points

Problem Statement

There are H rows and W columns of white square cells.

You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.

How many white cells will remain?

It can be proved that this count does not depend on what rows and columns are chosen.

Constraints

All values in input are integers.

1 \leq H, W \leq 20

1 \leq h \leq H

1 \leq w \leq W

Input

Input is given from Standard Input in the following format:

H W
h w

Output

Print the number of white cells that will remain.

Sample Input 1

3 2
2 1

Sample Output 1

1

There are 3 rows and 2 columns of cells. When two rows and one column are chosen and painted in black, there is always one white cell that remains.

Sample Input 2

5 5
2 3

Sample Output 2

6

Sample Input 3

2 4
2 4

Sample Output 3

0


n,m,c=map(int,input().split())
b=list(map(int,input().split()))
count=0
for i in range(n):
    ans=0
    a=list(map(int,input().split()))
    for i in range(m):
        ans+=a[i]*b[i]
    if ans+c>0:
        count+=1
print(count)

Score : 200 points

Problem Statement

There are N pieces of source code. The characteristics of the i-th code is represented by M integers A_{i1}, A_{i2}, ..., A_{iM}.

Additionally, you are given integers B_1, B_2, ..., B_M and C.

The i-th code correctly solves this problem if and only if A_{i1} B_1 + A_{i2} B_2 + ... + A_{iM} B_M + C > 0.

Among the N codes, find the number of codes that correctly solve this problem.

Constraints

All values in input are integers.

1 \leq N, M \leq 20

-100 \leq A_{ij} \leq 100

-100 \leq B_i \leq 100

-100 \leq C \leq 100

Input

Input is given from Standard Input in the following format:

N M C
B_1 B_2 ... B_M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
\vdots
A_{N1} A_{N2} ... A_{NM}

Output

Print the number of codes among the given N codes that correctly solve this problem.

Sample Input 1

2 3 -10
1 2 3
3 2 1
1 2 2

Sample Output 1

1

Only the second code correctly solves this problem, as follows:

Since 3 \times 1 + 2 \times 2 + 1 \times 3 + (-10) = 0 \leq 0, the first code does not solve this problem.

1 \times 1 + 2 \times 2 + 2 \times 3 + (-10) = 1 > 0, the second code solves this problem.

Sample Input 2

5 2 -4
-2 5
100 41
100 40
-3 0
-6 -2
18 -13

Sample Output 2

2

Sample Input 3

3 3 0
100 -100 0
0 100 100
100 100 100
-100 100 100

Sample Output 3

0

All of them are Wrong Answer. Except yours.


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

ab.sort()
sum_money = 0
for i in range(n):
    if m > ab[i][1]:
        sum_money += ab[i][0]*ab[i][1]
        m -= ab[i][1]
    else:
        sum_money += ab[i][0]*m
        break

print(sum_money)

Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.


A,B,K = [int(x) for x in input().split()]
cnt =0
for i in range(1,max(A,B)):
  if A%i==0 and B%i==0:
    cnt+=1
    if cnt==K:
      print(i)

Score : 200 points

Problem Statement

You are given positive integers A and B.

Find the K-th largest positive integer that divides both A and B.

The input guarantees that there exists such a number.

Constraints

All values in input are integers.

1 \leq A, B \leq 100

The K-th largest positive integer that divides both A and B exists.

K \geq 1

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the K-th largest positive integer that divides both A and B.

Sample Input 1

8 12 2

Sample Output 1

2

Three positive integers divides both 8 and 12: 1, 2 and 4.
Among them, the second largest is 2.

Sample Input 2

100 50 4

Sample Output 2

5

Sample Input 3

1 1 1

Sample Output 3

1


A, B, K = map(int, input().split())
l = []
for i in range(1, 101):
    if A % i == 0 and B % i == 0:
        l.append(i)
l.sort(reverse=True)
print(l[K - 1])


s = input()
cnt0 = s.count('0')
cnt1 = s.count('1')
print(min(cnt0, cnt1) * 2)

Score : 300 points

Problem Statement

There are N cubes stacked vertically on a desk.

You are given a string S of length N. The color of the i-th cube from the bottom is red if the i-th character in S is 0, and blue if that character is 1.

You can perform the following operation any number of times: choose a red cube and a blue cube that are adjacent, and remove them. Here, the cubes that were stacked on the removed cubes will fall down onto the object below them.

At most how many cubes can be removed?

Constraints

1 \leq N \leq 10^5

|S| = N

Each character in S is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of cubes that can be removed.

Sample Input 1

0011

Sample Output 1

4

All four cubes can be removed, by performing the operation as follows:

Remove the second and third cubes from the bottom. Then, the fourth cube drops onto the first cube.

Remove the first and second cubes from the bottom.

Sample Input 2

11011010001011

Sample Output 2

12

Sample Input 3

0

Sample Output 3

0


s = input()
cnt = 0
while True:
    idx = s.find("01")
    if idx != -1:
        cnt += 2
        s = s[:idx] + s[idx+2:]
    idxx = s.find("10")
    if idxx != -1:
        cnt += 2
        s = s[:idxx] + s[idxx+2:]
    if idx == -1 and idxx == -1:
        break
print(cnt)


S = list(input())
rm = 0
while True:
    length = len(S)
    for i in range(length - 1):
        # print(S)
        # print(f"1){length} {i} {len(S)}")
        if (int(S[i]) + int(S[i+1])) == 1:
            rm += 2
            del(S[i:i+2])
            # print(f"2){length} {i} {len(S)} {S}")
            break
        else:
            i += 1
    # print(f"3){length} {i} {len(S)}")
    if length == len(S):
        break
    i = 0
print(rm)


import sys
import heapq
import math
import fractions
import bisect
import itertools
from collections import Counter
from collections import deque
from operator import itemgetter
def input(): return sys.stdin.readline().strip()
def mp(): return map(int,input().split())
def lmp(): return list(map(int,input().split()))

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

n,m=mp()
l=[]
for i in range(m):
    l.append(lmp())
ans=[0]*m
ans[m-1]=n*(n-1)//2
uf=UnionFind(n+1)
for i in range(m-1):
    if not uf.same(l[m-1-i][0],l[m-1-i][1]):
        ans[m-2-i]=ans[m-i-1]-uf.size(l[m-1-i][0])*uf.size(l[m-1-i][1])
        uf.union(l[m-1-i][0],l[m-1-i][1])
for a in ans:
    print(a)

Score : 400 points

Problem Statement

There are N islands and M bridges.

The i-th bridge connects the A_i-th and B_i-th islands bidirectionally.

Initially, we can travel between any two islands using some of these bridges.

However, the results of a survey show that these bridges will all collapse because of aging, in the order from the first bridge to the M-th bridge.

Let the inconvenience be the number of pairs of islands (a, b) (a < b) such that we are no longer able to travel between the a-th and b-th islands using some of the bridges remaining.

For each i (1 \leq i \leq M), find the inconvenience just after the i-th bridge collapses.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i < B_i \leq N

All pairs (A_i, B_i) are distinct.

The inconvenience is initially 0.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_M B_M

Output

In the order i = 1, 2, ..., M, print the inconvenience just after the i-th bridge collapses.
Note that the answer may not fit into a 32-bit integer type.

Sample Input 1

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

Sample Output 1

0
0
4
5
6

For example, when the first to third bridges have collapsed, the inconvenience is 4 since we can no longer travel between the pairs (1, 2), (1, 3), (2, 4) and (3, 4).

Sample Input 2

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

Sample Output 2

8
9
12
14
15

Sample Input 3

2 1
1 2

Sample Output 3

1


# coding: utf-8
# Your code here!
s=input()
t="2019/04/30"
if s<=t:
    print("Heisei")
else :
    print("TBD")

Score : 100 points

Problem Statement

You are given a string S as input. This represents a valid date in the year 2019 in the yyyy/mm/dd format. (For example, April 30, 2019 is represented as 2019/04/30.)

Write a program that prints Heisei if the date represented by S is not later than April 30, 2019, and prints TBD otherwise.

Constraints

S is a string that represents a valid date in the year 2019 in the yyyy/mm/dd format.

Input

Input is given from Standard Input in the following format:

S

Output

Print Heisei if the date represented by S is not later than April 30, 2019, and print TBD otherwise.

Sample Input 1

2019/04/30

Sample Output 1

Heisei

Sample Input 2

2019/11/01

Sample Output 2

TBD


s = str(input())

nexts = s.replace("/","")

sint = int(nexts)

print(sint)

if sint <= 20190430:
    print("Heisei")
else:
    print("TBD")


SEIREKI = '2019/04/30'

inputData = input()
ans = ""
if inputData <= SEIREKI:
    ans = 'Heisei'
else:
    ans = 'TBD'
print(ans)


n = int(input())
re = [list(input().split(' ')) for i in range(n)]
def b2y(btc):
  return float(btc) * 380000
ans = 0
for x, u in re:
  if u == 'JPY':
    ans += float(x)
  else:
    ans += b2y(x)
print(ans)

Score : 200 points

Problem Statement

Takahashi received otoshidama (New Year's money gifts) from N of his relatives.

You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either JPY or BTC, and x_i and u_i represent the content of the otoshidama from the i-th relative.

For example, if x_1 = 10000 and u_1 = JPY, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = 0.10000000 and u_2 = BTC, the otoshidama from the second relative is 0.1 bitcoins.

If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Constraints

2 \leq N \leq 10

u_i = JPY or BTC.

If u_i = JPY, x_i is an integer such that 1 \leq x_i \leq 10^8.

If u_i = BTC, x_i is a decimal with 8 decimal digits, such that 0.00000001 \leq x_i \leq 100.00000000.

Input

Input is given from Standard Input in the following format:

N
x_1 u_1
x_2 u_2
:
x_N u_N

Output

If the gifts are worth Y yen in total, print the value Y (not necessarily an integer).

Output will be judged correct when the absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10000 JPY
0.10000000 BTC

Sample Output 1

48000.0

The otoshidama from the first relative is 10000 yen. The otoshidama from the second relative is 0.1 bitcoins, which is worth 38000.0 yen if converted at the rate of 380000.0 JPY per 1.0 BTC. The sum of these is 48000.0 yen.

Outputs such as 48000 and 48000.1 will also be judged correct.

Sample Input 2

3
100000000 JPY
100.00000000 BTC
0.00000001 BTC

Sample Output 2

138000000.0038

In this case, outputs such as 138001000 and 1.38e8 will also be judged correct.


N = int(input())
ans = 0.0
for i in range(N):
    mon, kind = input().split()
    mon = float(mon)
    if kind == 'BTC':
        mon = 380000.0 * mon
    ans += mon
print(ans)


from itertools import product

def solve(NUM, LIST):
    lst = [i for i, li in enumerate(LIST) if li == NUM]
    if len(lst) == 0:
        return 10 ** 5
    ANS = (len(lst) - 1) * 10
    ANS += abs(sum(l[i] for i in lst) - target[NUM])
    return ANS

target = list(map(int, input().split()))
l = [int(input()) for _ in range(target[0])]

ans = 10 ** 5
for pat in product([0, 1, 2, 3], repeat=target[0]):
    tmp = sum(solve(i + 1, pat) for i in range(3))
    ans = min(ans, tmp)
print(ans)

Score : 300 points

Problem Statement

You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.

Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.

Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)

At least how much MP is needed to achieve the objective?

Constraints

3 \leq N \leq 8

1 \leq C < B < A \leq 1000

1 \leq l_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C
l_1
l_2
:
l_N

Output

Print the minimum amount of MP needed to achieve the objective.

Sample Input 1

5 100 90 80
98
40
30
21
80

Sample Output 1

23

We are obtaining three bamboos of lengths 100, 90, 80 from five bamboos 98, 40, 30, 21, 80. We already have a bamboo of length 80, and we can obtain bamboos of lengths 100, 90 by using the magics as follows at the total cost of 23 MP, which is optimal.

Use Extension Magic twice on the bamboo of length 98 to obtain a bamboo of length 100. (MP consumed: 2)

Use Composition Magic on the bamboos of lengths 40, 30 to obtain a bamboo of length 70. (MP consumed: 10)

Use Shortening Magic once on the bamboo of length 21 to obtain a bamboo of length 20. (MP consumed: 1)

Use Composition Magic on the bamboo of length 70 obtained in step 2 and the bamboo of length 20 obtained in step 3 to obtain a bamboo of length 90. (MP consumed: 10)

Sample Input 2

8 100 90 80
100
100
90
90
90
80
80
80

Sample Output 2

0

If we already have all bamboos of the desired lengths, the amount of MP needed is 0. As seen here, we do not necessarily need to use all the bamboos.

Sample Input 3

8 1000 800 100
300
333
400
444
500
555
600
666

Sample Output 3

243


# -*- coding: utf-8 -*-
import sys
from bisect import bisect_left, bisect_right
from operator import itemgetter
from fractions import gcd
from math import ceil, floor, sqrt
from copy import deepcopy
from collections import Counter, deque
import heapq
from functools import reduce
from random import shuffle
import cProfile
sys.setrecursionlimit(200000)
# local only
# if not __debug__:
#     fin = open('text.txt', 'r')
#     sys.stdin = fin
# local only
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
def debug(*args, sep=" ", end="\n"): print("debug:", *args, file=sys.stderr, sep=sep, end=end) if not __debug__ else None
def exit(*arg): print(*arg); sys.exit()
# template

# BEGIN CUT HERE
class Avltree:
    """
    AVL木

    Methods
    -------
        search(x: int) -> Union(int)
            ◯.data == x のものが存在するある場合それを出力、ない場合 None を返す
        insert(x: int) -> None
            ◯.data == x がない場合 ◯.data == x なる頂点を作る。ある場合はなにもしない（上書きしない）
        y.to_s() -> int
            頂点yのdataを返す
        y.left_s() -> Union(int, key=None)
            頂点yのleftを返す(ない場合は key)
        search_lower(x: int, key_lower) -> int
            ○.data が x より小さいものの中で最も大きいものを出力、ない場合 key_lower を返す
        search_higher(x:int , key_higher) -> int
            ○.data が x より大きいものの中で最も小さいものを出力、ない場合 key_higher を返す
    """

    def __init__(self, key=None):
        self.key = key
        self.left = None
        self.right = None
        self.balance = "Even"

    def search(self, key):
        node = self
        while node is not None:
            if node.key == key:
                return node  # node.key == keyの場合
            if node.key > key:
                node = node.left
            elif node.key < key:
                node = node.right
        return None

    def search_lower(self, key, key_lower):
        node = self
        if node.key is None:
            return key_lower
        ans = key_lower
        while node is not None:
            if node.key < key:
                if ans < node.key:
                    ans = node.key
                node = node.right
            elif node.key >= key:
                node = node.left
        return ans

    def search_higher(self, key, key_higher):
        node = self
        if node.key is None:
            return key_higher
        ans = key_higher
        while node is not None:
            if node.key > key:
                if ans > node.key:
                    ans = node.key
                node = node.left
            elif node.key <= key:
                node = node.right
        return ans

    def end_lower(self, end_lower_key):
        node = self
        while node.left is not None:
            node = node.left
        return node.key

    def end_higher(self, end_higher_key):
        node = self
        while node.right is not None:
            node = node.right
        return node.key

    def DoubleRightRotation(self):
        tl = self.left
        self.left = tl.right.right
        tl.right.right = self  # selfはそのノード
        tlr = tl.right
        tl.right = tlr.left
        tlr.left = tl
        if tlr.balance == "Left":
            tlr.right.balance = "Right"
            tlr.left.balance = "Even"
        elif tlr.balance == "Right":
            tlr.right.balance = "Even"
            tlr.left.balance = "Left"
        elif tlr.balance == "Even":
            tlr.right.balance = "Even"
            tlr.left.balance = "Even"
        tlr.balance = "Even"
        return tlr
    def DoubleLeftRotation(self):
        tr = self.right
        self.right = tr.left.left
        tr.left.left = self
        trl = tr.left
        tr.left = trl.right
        trl.right = tr
        if trl.balance == "Right":
            trl.left.balance = "Left"
            trl.right.balance = "Even"
        elif trl.balance == "Left":
            trl.left.balance = "Even"
            trl.right.balance = "Right"
        elif trl.balance == "Even":
            trl.left.balance = "Even"
            trl.right.balance = "Even"
        trl.balance = "Even"
        return trl
    def SingleLeftRotation(self):
        tr = self.right
        tr.balance = "Even"
        self.balance = "Even"
        self.right = tr.left
        tr.left = self
        return tr
    def SingleRightRotation(self):
        tl = self.left
        tl.balance = "Even"
        self.balance = "Even"
        self.left = tl.right
        tl.right = self
        return tl
    def replace(self, p, v):  # 親ノードpの下にある自分(self)をvに置き換える。
        if p.left == self:
            p.left = v
        else:
            p.right = v
    def insert(self, key):  # rootでのみ呼ばれる挿入
        if self.key is None:  # rootを含むrotationはしないことにする。
            self.key = key
            return self
        if key < self.key:
            if self.left is None:
                self.left = Avltree(key)
            else:
                self.left.insertx(self, key)
        elif key > self.key:
            if self.right is None:
                self.right = Avltree(key)
            else:
                self.right.insertx(self, key)
        else:  # key == self.key:
            pass  # do not overwrite
    def insertx(self, p, key):  # replaceを呼ぶために一つ上の親を持っているinsert
        node = self
        s = []
        while True:
            if node.key > key:
                s.append((node, -1))
                if node.left is None:
                    node.left = Avltree(key)
                    node = node.left
                    break
                else:
                    node = node.left
            elif node.key < key:
                s.append((node, 1))
                if node.right is None:
                    node.right = Avltree(key)
                    node = node.right
                    break
                else:
                    node = node.right
        while len(s) != 0:
            node, direct = s.pop()
            if len(s) != 0:
                par = s[-1][0]
            else:
                par = p
            if direct == -1:
                if node.balance == "Right":
                    node.balance = "Even"
                    break
                elif node.balance == "Even":
                    node.balance = "Left"
                elif node.balance == "Left":
                    if node.left.balance == "Right":
                        node.replace(par, node.DoubleRightRotation())
                    elif node.left.balance == "Left":
                        node.replace(par, node.SingleRightRotation())
                    break
            elif direct == 1:
                if node.balance == "Left":
                    node.balance = "Even"
                    break
                elif node.balance == "Even":
                    node.balance = "Right"
                elif node.balance == "Right":
                    if node.right.balance == "Left":
                        node.replace(par, node.DoubleLeftRotation())
                    elif node.right.balance == "Right":
                        node.replace(par, node.SingleLeftRotation())
                    break

    def to_s(self):
        return self.key
    def left_s(self):
        if self.left is None:
            return None
        return (self.left).key
    def right_s(self):
        if self.right is None:
            return None
        return (self.right).key
# END CUT HERE

def main():
    rb1 = Avltree()
    a, b, q = map(int, input().split())

    rb2 = Avltree()
    for i in range(a):
        s = ii()
        rb1.insert(s)
    for i in range(b):
        t = ii()
        rb2.insert(t)
    # rb1.output()
    # rb2.output()
    for i in range(q):
        x = ii()
        s1 = rb1.search_lower(x, -10**18)
        s2 = rb1.search_higher(x, 10**18)
        t1 = rb2.search_lower(x, -10**18)
        t2 = rb2.search_higher(x, 10**18)
        # print(s1, s2, t1, t2)
        res = 10**18
        for S in [s1, s2]:
            for T in [t1, t2]:
                d1, d2 = abs(S - x) + abs(T - S), abs(T - x) + abs(S - T)
                res = min(res, d1, d2)
        print(res)

if __name__ == '__main__':
    # cProfile.run('main()')
    main()

Score : 400 points

Problem Statement

Along a road running in an east-west direction, there are A shrines and B temples.
The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.

Answer the following Q queries:

Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)

Constraints

1 \leq A, B \leq 10^5

1 \leq Q \leq 10^5

1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}

1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}

1 \leq x_i \leq 10^{10}

s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

2 3 4
100
600
400
900
1000
150
2000
899
799

Sample Output 1

350
1400
301
399

There are two shrines and three temples. The shrines are located at distances of 100, 600 meters from the west end of the road, and the temples are located at distances of 400, 900, 1000 meters from the west end of the road.

Query 1: If we start from a point at a distance of 150 meters from the west end of the road, the optimal move is first to walk 50 meters west to visit a shrine, then to walk 300 meters east to visit a temple.

Query 2: If we start from a point at a distance of 2000 meters from the west end of the road, the optimal move is first to walk 1000 meters west to visit a temple, then to walk 400 meters west to visit a shrine. We will pass by another temple on the way, but it is fine.

Query 3: If we start from a point at a distance of 899 meters from the west end of the road, the optimal move is first to walk 1 meter east to visit a temple, then to walk 300 meters west to visit a shrine.

Query 4: If we start from a point at a distance of 799 meters from the west end of the road, the optimal move is first to walk 199 meters west to visit a shrine, then to walk 200 meters west to visit a temple.

Sample Input 2

1 1 3
1
10000000000
2
9999999999
5000000000

Sample Output 2

10000000000
10000000000
14999999998

The road is quite long, and we may need to travel a distance that does not fit into a 32-bit integer.


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

Score : 100 points

Problem Statement

You are given positive integers A and B.

If A is a divisor of B, print A + B; otherwise, print B - A.

Constraints

All values in input are integers.

1 \leq A \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

If A is a divisor of B, print A + B; otherwise, print B - A.

Sample Input 1

4 12

Sample Output 1

16

As 4 is a divisor of 12, 4 + 12 = 16 should be printed.

Sample Input 2

8 20

Sample Output 2

12

Sample Input 3

1 1

Sample Output 3

2

1 is a divisor of 1.


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

list_score = [ list(map(int,input().split(" "))) for i in range(n)]
ans = set(list_score[0][1:])

for c in list_score:
    ans = ans & set(c[1:])

print(len(ans))

Score : 200 points

Problem Statement

Katsusando loves omelette rice.

Besides, he loves crème brûlée, tenderloin steak and so on, and believes that these foods are all loved by everyone.

To prove that hypothesis, he conducted a survey on M kinds of foods and asked N people whether they like these foods or not.

The i-th person answered that he/she only likes the A_{i1}-th, A_{i2}-th, ..., A_{iK_i}-th food.

Find the number of the foods liked by all the N people.

Constraints

All values in input are integers.

1 \leq N, M \leq 30

1 \leq K_i \leq M

1 \leq A_{ij} \leq M

For each i (1 \leq i \leq N), A_{i1}, A_{i2}, ..., A_{iK_i} are distinct.

Constraints

Input is given from Standard Input in the following format:

N M
K_1 A_{11} A_{12} ... A_{1K_1}
K_2 A_{21} A_{22} ... A_{2K_2}
:
K_N A_{N1} A_{N2} ... A_{NK_N}

Output

Print the number of the foods liked by all the N people.

Sample Input 1

3 4
2 1 3
3 1 2 3
2 3 2

Sample Output 1

1

As only the third food is liked by all the three people, 1 should be printed.

Sample Input 2

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

Sample Output 2

0

Katsusando's hypothesis turned out to be wrong.

Sample Input 3

1 30
3 5 10 30

Sample Output 3

3


n,m = map(int, input().split())
l = [0]*m

for i in range(n):
	k,*p = map(int, open(0).read().split())
	for j in range(k):
		l[p[j]-1] += 1
ans = 0
for i in range(m):
	if l[i] == 0:
		ans += 1
print ans


from fractions import gcd

N = int(input())

As = list(map(int, input().split()))

d = As[0]
for a in As[1:]:
    d = gcd(d, a)

print(d)

Score : 300 points

Problem Statement

There are N monsters, numbered 1, 2, ..., N.

Initially, the health of Monster i is A_i.

Below, a monster with at least 1 health is called alive.

Until there is only one alive monster, the following is repeated:

A random alive monster attacks another random alive monster.

As a result, the health of the monster attacked is reduced by the amount equal to the current health of the monster attacking.

Find the minimum possible final health of the last monster alive.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

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 possible final health of the last monster alive.

Sample Input 1

4
2 10 8 40

Sample Output 1

2

When only the first monster keeps on attacking, the final health of the last monster will be 2, which is minimum.

Sample Input 2

4
5 13 8 1000000000

Sample Output 2

1

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

1000000000


N = int(raw_input())
health = map(int, raw_input().split())

def gcd(a, b):
	while b:
		a, b = b, a % b
	return a

def gcd_driver(lst):
	if len(lst) <= 1:
		return lst[0]
	if len(lst) > 2:
		lst[1] = gcd(lst[0], lst[1])
		return gcd_driver(lst[1:])
	else:
		return gcd(lst[0], lst[1])

print gcd_driver(health)


N,K = map(int,input().split())

if (N+1)/2 >= K:
    print('YES')
else:
    print('NO')

Score : 100 points

Problem Statement

Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

Constraints

1\leq N,K\leq 100

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

If we can choose K integers as above, print YES; otherwise, print NO.

Sample Input 1

3 2

Sample Output 1

YES

We can choose 1 and 3.

Sample Input 2

5 5

Sample Output 2

NO

Sample Input 3

31 10

Sample Output 3

YES

Sample Input 4

10 90

Sample Output 4

NO


Score : 400 points

Problem Statement

Snuke has one biscuit and zero Japanese yen (the currency) in his pocket.
He will perform the following operations exactly K times in total, in the order he likes:

Hit his pocket, which magically increases the number of biscuits by one.

Exchange A biscuits to 1 yen.

Exchange 1 yen to B biscuits.

Find the maximum possible number of biscuits in Snuke's pocket after K operations.

Constraints

1 \leq K,A,B \leq 10^9

K,A and B are integers.

Input

Input is given from Standard Input in the following format:

K A B

Output

Print the maximum possible number of biscuits in Snuke's pocket after K operations.

Sample Input 1

4 2 6

Sample Output 1

7

The number of biscuits in Snuke's pocket after K operations is maximized as follows:

Hit his pocket. Now he has 2 biscuits and 0 yen.

Exchange 2 biscuits to 1 yen. his pocket. Now he has 0 biscuits and 1 yen.

Hit his pocket. Now he has 1 biscuits and 1 yen.

Exchange 1 yen to 6 biscuits. his pocket. Now he has 7 biscuits and 0 yen.

Sample Input 2

7 3 4

Sample Output 2

8

Sample Input 3

314159265 35897932 384626433

Sample Output 3

48518828981938099


hours, multi_time = map(int, input().split())

result = hours / multi_time
print(result)

Score : 100 points

Problem Statement

In order to pass the entrance examination tomorrow, Taro has to study for T more hours.

Fortunately, he can leap to World B where time passes X times as fast as it does in our world (World A).

While (X \times t) hours pass in World B, t hours pass in World A.

How many hours will pass in World A while Taro studies for T hours in World B?

Constraints

All values in input are integers.

1 \leq T \leq 100

1 \leq X \leq 100

Input

Input is given from Standard Input in the following format:

T X

Output

Print the number of hours that will pass in World A.

The output will be regarded as correct when its absolute or relative error from the judge's output is at most 10^{-3}.

Sample Input 1

8 3

Sample Output 1

2.6666666667

While Taro studies for eight hours in World B where time passes three times as fast, 2.6666... hours will pass in World A.

Note that an absolute or relative error of at most 10^{-3} is allowed.

Sample Input 2

99 1

Sample Output 2

99.0000000000

Sample Input 3

1 100

Sample Output 3

0.0100000000


T,X = map(int,input().split())
print(T/X)


t,x = map(int,input().split())
print(t/x)


# ABC 117: A – Entrance Examination
t, x = [int(s) for s in input().split()]
print('{:.4f}'.format(t / x))


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

l_s = list(sorted(l))
print( 'Yes' if (l_s[-1] < sum(l_s[:-1])) else 'No')

Score : 200 points

Problem Statement

Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.

You can use the following theorem:

Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.

Constraints

All values in input are integers.

3 \leq N \leq 10

1 \leq L_i \leq 100

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Output

If an N-sided polygon satisfying the condition can be drawn, print Yes; otherwise, print No.

Sample Input 1

4
3 8 5 1

Sample Output 1

Yes

Since 8 < 9 = 3 + 5 + 1, it follows from the theorem that such a polygon can be drawn on a plane.

Sample Input 2

4
3 8 4 1

Sample Output 2

No

Since 8 \geq 8 = 3 + 4 + 1, it follows from the theorem that such a polygon cannot be drawn on a plane.

Sample Input 3

10
1 8 10 5 8 12 34 100 11 3

Sample Output 3

No


input()
L = sorted([int(x) for x in input().split()])

l = L[-1]
if l < sum(L[:-1]):
    print("Yes")
else:
    print("No")


n = int(input())
nums = list(map(int, input().split()))
nums.sort()
if nums[n-1] < sum(nums[:n]):
    print("Yes")
else:
    print("No")


n,m,*x = map(int,open(0).read().split())
x.sort()
if n >= m:
  print(0)
  exit()
a = sorted([x[i+1]-x[i] for i in range(m-1)],reverse=True)
ans = x[-1]-x[0]
for i in range(n-1):
  ans -= a[i]
print(ans)

Score : 300 points

Problem Statement

We will play a one-player game using a number line and N pieces.

First, we place each of these pieces at some integer coordinate.

Here, multiple pieces can be placed at the same coordinate.

Our objective is to visit all of the M coordinates X_1, X_2, ..., X_M with these pieces, by repeating the following move:

Move: Choose a piece and let x be its coordinate. Put that piece at coordinate x+1 or x-1.

Note that the coordinates where we initially place the pieces are already regarded as visited.

Find the minimum number of moves required to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

-10^5 \leq X_i \leq 10^5

X_1, X_2, ..., X_M are all different.

Input

Input is given from Standard Input in the following format:

N M
X_1 X_2 ... X_M

Output

Find the minimum number of moves required to achieve the objective.

Sample Input 1

2 5
10 12 1 2 14

Sample Output 1

5

The objective can be achieved in five moves as follows, and this is the minimum number of moves required.

Initially, put the two pieces at coordinates 1 and 10.

Move the piece at coordinate 1 to 2.

Move the piece at coordinate 10 to 11.

Move the piece at coordinate 11 to 12.

Move the piece at coordinate 12 to 13.

Move the piece at coordinate 13 to 14.

Sample Input 2

3 7
-10 -3 0 9 -100 2 17

Sample Output 2

19

Sample Input 3

100 1
-100000

Sample Output 3

0


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

X=list(map(int,input().split()))
X.sort()

L=[0]*(M-1)
for i in range(M-1):
 L[i] = X[i+1]-X[i]

L.sort()

print(sum(L[:-N+1]))


n=int(input())
a=input()
b=input()
c=input()

ret = 0
for i in range(n):
    if a[i] == b[i] == c[i]:
        continue
    if a[i] == b[i] or b[i] == c[i] or a[i] == c[i] :
        ret += 1
    else:
        ret += 2

print(ret)

Score : 200 points

Problem Statement

You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters.

Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation:

Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter.

What is the minimum number of operations required to achieve the objective?

Constraints

1 \leq N \leq 100

Each of the strings A, B and C is a string of length N.

Each character in each of the strings A, B and C is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

N
A
B
C

Output

Print the minimum number of operations required.

Sample Input 1

4
west
east
wait

Sample Output 1

3

In this sample, initially A = west、B = east、C = wait. We can achieve the objective in the minimum number of operations by performing three operations as follows:

Change the second character in A to a. A is now wast.

Change the first character in B to w. B is now wast.

Change the third character in C to s. C is now wast.

Sample Input 2

9
different
different
different

Sample Output 2

0

If A, B and C are already equal in the beginning, the number of operations required is 0.

Sample Input 3

7
zenkoku
touitsu
program

Sample Output 3

13


s=[list(map(int,input().split()))for i in range(int(input()))];print(sum(sorted([x[1]+x[0] for x in s])[::-2])-sum([x[1]for x in s]))

Score : 400 points

Problem Statement

There are N dishes of cuisine placed in front of Takahashi and Aoki.
For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N.

When Takahashi eats Dish i, he earns A_i points of happiness; when Aoki eats Dish i, she earns B_i points of happiness.

Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat.
Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end".

Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Sample Input 1

3
10 10
20 20
30 30

Sample Output 1

20

In this sample, both of them earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since Takahashi and Aoki have the same "taste", each time they will choose the dish with which they can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (30 + 10) - 20 = 20.

Sample Input 2

3
20 10
20 20
20 30

Sample Output 2

20

In this sample, Takahashi earns 20 points of happiness by eating any one of the dishes 1, 2 and 3, but Aoki earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since only Aoki has likes and dislikes, each time they will choose the dish with which Aoki can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (20 + 20) - 20 = 20.

Sample Input 3

6
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000

Sample Output 3

-2999999997

Note that the answer may not fit into a 32-bit integer.


# 有効グラフ上で、近道に当たる点はすべて余計な辺
# ある点に入ってくる最も距離が長い辺のみが有効となる
# 一番距離が長い人版ダイクストラやってみる
# 入力のあった頂点を記録することで、入力の無かった頂点を根とみなせる

import sys
readline = sys.stdin.readline

N,M = map(int,readline().split())
G = [[] for i in range(N)]
grandparent = set(range(N))
for i in range(N - 1 + M):
  a,b = map(int,readline().split())
  G[a-1].append(b-1)
  grandparent -= {b - 1}

start = list(grandparent)[0]
import heapq

# 距離, 頂点, 親
q = [(0, start, -1)]
heapq.heapify(q)

# ans[頂点] = (到達コスト,親)
ans = [(-1, -1) for i in range(N)]
while q:
  dist, v, parent = heapq.heappop(q)
  dist = -dist
  if ans[v][0] >= dist:
    continue
  ans[v] = (dist, parent)
  for child in G[v]:
    heapq.heappush(q,(-(dist + 1), child, v))

for a in ans:
  print(a[1] + 1)

Score : 500 points

Problem Statement

There is a rooted tree (see Notes) with N vertices numbered 1 to N.
Each of the vertices, except the root, has a directed edge coming from its parent.
Note that the root may not be Vertex 1.

Takahashi has added M new directed edges to this graph.
Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.

You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges.
More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.

Restore the original rooted tree.

Notes

For "tree" and other related terms in graph theory, see the article in Wikipedia, for example.

Constraints

3 \leq N

1 \leq M

N + M \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

If i \neq j, (A_i, B_i) \neq (A_j, B_j).

The graph in input can be obtained by adding M edges satisfying the condition in the problem statement to a rooted tree with N vertices.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_{N-1+M} B_{N-1+M}

Output

Print N lines.
In the i-th line, print 0 if Vertex i is the root of the original tree, and otherwise print the integer representing the parent of Vertex i in the original tree.

Note that it can be shown that the original tree is uniquely determined.

Sample Input 1

3 1
1 2
1 3
2 3

Sample Output 1

0
1
2

The graph in this input is shown below:

It can be seen that this graph is obtained by adding the edge 1 \rightarrow 3 to the rooted tree 1 \rightarrow 2 \rightarrow 3.

Sample Input 2

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

Sample Output 2

6
4
2
0
6
2


N, M = [int(_) for _ in input().split()]
AB = [[int(_) for _ in input().split()] for i in range(N - 1 + M)]

from collections import defaultdict

nodes = list(range(1, N + 1))
edges = defaultdict(set)
rev_edges = defaultdict(set)

for a, b in AB:
    edges[a].add(b)
    rev_edges[b].add(a)

#print(vs)
#print(ws)

def TopologicalSort(nodes, edges, rev_edges):
    '''
    nodes     ... Node[]
    edges     ... Dict[From Node](To Node)
    rev_edges ... Dict[To Node](From Node)
    '''
    rev_edges_num = dict((n, len(rev_edges[n])) for n in nodes)
    L = []
    S = [n for n in nodes if not rev_edges_num[n]]
    while S:
        n = S.pop()
        L.append(n)
        for m in edges[n]:
            rev_edges_num[m] -= 1
            if not rev_edges_num[m]:
                S.append(m)

    # プロコン用なのでエラーチェックはしない

    return L

L = TopologicalSort(nodes, edges, rev_edges)

nrank = dict((n, r) for r, n in enumerate(L))

for n in range(1, N + 1):
    r = 0
    if rev_edges[n]:
        r = sorted(rev_edges[n], key=lambda x:nrank[x], reverse=True)[0]
    print(r)


def f(n):
    if n % 2 == 0:
        return n // 2
    return 3 * n + 1

def main():
    s = int(input())
    aa = {s}
    a = f(s)
    for i in range(2, 1000001):
        if a in aa:
            print(i)
            break
        aa.add(a)
        a = f(a)

if __name__ == "__main__":
    main()

Score : 200 points

Problem Statement

A sequence a=\{a_1,a_2,a_3,......\} is determined as follows:

The first term s is given as input.

Let f(n) be the following function: f(n) = n/2 if n is even, and f(n) = 3n+1 if n is odd.

a_i = s when i = 1, and a_i = f(a_{i-1}) when i > 1.

Find the minimum integer m that satisfies the following condition:

There exists an integer n such that a_m = a_n (m > n).

Constraints

1 \leq s \leq 100

All values in input are integers.

It is guaranteed that all elements in a and the minimum m that satisfies the condition are at most 1000000.

Input

Input is given from Standard Input in the following format:

s

Output

Print the minimum integer m that satisfies the condition.

Sample Input 1

8

Sample Output 1

5

a=\{8,4,2,1,4,2,1,4,2,1,......\}. As a_5=a_2, the answer is 5.

Sample Input 2

7

Sample Output 2

18

a=\{7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,......\}.

Sample Input 3

54

Sample Output 3

114


s = int(input())
def func(n):
    if n % 2 == 0:
        return n // 2
    else:
        return 3*n + 1
a = [0,s]
for i in range(2,10**6+1):
    a_i = func(a[i-1])
    if a_i in a:
        print(i)
        break
    a.append(a_i)


import sys

sys.setrecursionlimit(10 ** 6)
INF = float("inf")
MOD = 10 ** 9 + 7

def input():
    return sys.stdin.readline().strip()

def check(H):
    N = len(H)
    for i in range(N):
        if H[i] != 0:
            return False

    return True

def main():
    N = int(input())
    H = list(map(int, input().split()))

    ans = 0
    while True:
        if check(H):
            break

        # 区間分割
        i = 0
        while i < N:
            if H[i] == 0:
                i += 1
            else:
                # 区間が始まる
                ans += 1
                while i < N and H[i] > 0:
                    H[i] -= 1
                    i += 1

    print(ans)

if __name__ == "__main__":
    main()

Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221


n=int(input())
ii=input().split(" ")
flower=[]
count=0
end=n

for i in ii:
  flower.append(int(i))

#print (flower)

while True:
  end=n
  if flower[0]==0:
    del flower[0]
    n=n-1
    if n==0:
      print(str(count))
      exit()
    else:
      continue
  for j in range(0,n):
    if flower[j]==0:
      end=j
      break

  for m in range (0,end):
    flower[m]=flower[m]-1
  #print(end)
  count=count+1
  #print(flower)
  continue


n, k = map(int, input().split())
td = sorted([list(map(int, input().split())) for i in range(n)], reverse=True, key=lambda x: x[1])

type = set()
L = []
Sum = 0
for x in td[:k]:
    Sum += x[1]
    if x[0] not in type:
        type.add(x[0])
    else:
        L.append(x[1])

L = L[::-1]
type_cnt = len(type)
ans = Sum + type_cnt ** 2
for x in td[k:]:
    if len(L)==0:break
    if x[0] not in type:
        type.add(x[0])
        type_cnt += 1
        Sum = Sum - L.pop(0) + x[1]
        ans = max(ans, Sum + type_cnt ** 2)
print(ans)

Score : 400 points

Problem Statement

There are N pieces of sushi. Each piece has two parameters: "kind of topping" t_i and "deliciousness" d_i.
You are choosing K among these N pieces to eat.
Your "satisfaction" here will be calculated as follows:

The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".

The base total deliciousness is the sum of the deliciousness of the pieces you eat.

The variety bonus is x*x, where x is the number of different kinds of toppings of the pieces you eat.

You want to have as much satisfaction as possible.
Find this maximum satisfaction.

Constraints

1 \leq K \leq N \leq 10^5

1 \leq t_i \leq N

1 \leq d_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
t_1 d_1
t_2 d_2
.
.
.
t_N d_N

Output

Print the maximum satisfaction that you can obtain.

Sample Input 1

5 3
1 9
1 7
2 6
2 5
3 1

Sample Output 1

26

If you eat Sushi 1,2 and 3:

The base total deliciousness is 9+7+6=22.

The variety bonus is 2*2=4.

Thus, your satisfaction will be 26, which is optimal.

Sample Input 2

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

Sample Output 2

25

It is optimal to eat Sushi 1,2,3 and 4.

Sample Input 3

6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000

Sample Output 3

4900000016

Note that the output may not fit into a 32-bit integer type.


S = input()

import re
s = "keyence"
ans = s in S
for i in range(1, len(s)):
    s1 = s[:i]
    s2 = s[i:]
    ans = ans or re.match(".*{}.*{}".format(s1, s2), S)

print("YES" if ans else "NO")

Score : 200 points

Problem Statement

A string is called a KEYENCE string when it can be changed to keyence by removing its contiguous substring (possibly empty) only once.

Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

Constraints

The length of S is between 7 and 100 (inclusive).

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is a KEYENCE string, print YES; otherwise, print NO.

Sample Input 1

keyofscience

Sample Output 1

YES

keyence is an abbreviation of key of science.

Sample Input 2

mpyszsbznf

Sample Output 2

NO

Sample Input 3

ashlfyha

Sample Output 3

NO

Sample Input 4

keyence

Sample Output 4

YES


N, D = map(int, input().split())
*A, = map(int, input().split())

N0 = 2**(N-1).bit_length()
INF = (10**18, -1)
ldata = [INF]*(2*N0)
rdata = [INF]*(2*N0)

for i in range(N0-2, -1, -1):
    ldata[i] = min(ldata[2*i+1], ldata[2*i+2])
    rdata[i] = min(rdata[2*i+1], rdata[2*i+2])
def update(data, k, x):
    k += N0-1
    data[k] = x
    while k >= 0:
        k = (k - 1) // 2
        data[k] = min(data[2*k+1], data[2*k+2])
def query(data, l, r):
    if l >= r:
        return INF
    L = l + N0; R = r + N0
    s = INF
    while L < R:
        if R & 1:
            R -= 1
            s = min(s, data[R-1])

        if L & 1:
            s = min(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s

*I, = range(N)
I.sort(key=A.__getitem__)

E = []
for i in I:
    c, k = query(ldata, 0, i)
    if k != -1:
        c -= (N-1-i)*D
        E.append((c + A[i], k, i))
    c, k = query(rdata, i+1, N)
    if k != -1:
        c -= i*D
        E.append((c + A[i], i, k))
    update(ldata, i, ((N-1-i)*D + A[i], i))
    update(rdata, i, (i*D + A[i], i))

*p, = range(N)
def root(x):
    if x == p[x]:
        return x
    p[x] = y = root(p[x])
    return y
def unite(x, y):
    px = root(x); py = root(y)
    if px == py:
        return 0
    if px < py:
        p[py] = px
    else:
        p[px] = py
    return 1

E.sort()
ans = 0
for c, i, j in E:
    if unite(i, j):
        ans += c
print(ans)

Score : 600 points

Problem Statement

There are N cities in Republic of AtCoder. The size of the i-th city is A_{i}.
Takahashi would like to build N-1 bidirectional roads connecting two cities so that any city can be reached from any other city by using these roads.

Assume that the cost of building a road connecting the i-th city and the j-th city is |i-j| \times D + A_{i} + A_{j}.
For Takahashi, find the minimum possible total cost to achieve the objective.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq D \leq 10^9

1 \leq A_{i} \leq 10^9

A_{i} and D are integers.

Input

Input is given from Standard Input in the following format:

N D
A_1 A_2 ... A_N

Output

Print the minimum possible total cost.

Sample Input 1

3 1
1 100 1

Sample Output 1

106

This cost can be achieved by, for example, building roads connecting City 1, 2 and City 1, 3.

Sample Input 2

3 1000
1 100 1

Sample Output 2

2202

Sample Input 3

6 14
25 171 7 1 17 162

Sample Output 3

497

Sample Input 4

12 5
43 94 27 3 69 99 56 25 8 15 46 8

Sample Output 4

658


# D knap sack 2
from math import isinf

N,W = map(int, input().split())
w,v = [0]*N, [0]*N
for i in range(N):
    w[i],v[i] = map(int, input().split())

inf = float('inf')
# dp tabel -> dp list
dp = [inf]*(1000+1)
dp[0]  = 0
for i in range(N):
    for j in reversed(range(1000+1)):
        if j-v[i]<0 or dp[j-v[i]] + w[i] > W:
            dp[j] = dp[j]
        else:
            dp[j] = min(dp[j],
                        dp[j-v[i]] + w[i])
    # print(dp[:100])

ans = 0
for value,dpi in enumerate(dp):
    if dpi<=W:
        ans = max(ans, value)
print(ans)

Score : 100 points

Problem Statement

There are N items, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i.

Taro has decided to choose some of the N items and carry them home in a knapsack.
The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W.

Find the maximum possible sum of the values of items that Taro takes home.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq W \leq 10^9

1 \leq w_i \leq W

1 \leq v_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible sum of the values of items that Taro takes home.

Sample Input 1

3 8
3 30
4 50
5 60

Sample Output 1

90

Items 1 and 3 should be taken.
Then, the sum of the weights is 3 + 5 = 8, and the sum of the values is 30 + 60 = 90.

Sample Input 2

1 1000000000
1000000000 10

Sample Output 2

10

Sample Input 3

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3

17

Items 2, 4 and 5 should be taken.
Then, the sum of the weights is 5 + 6 + 3 = 14, and the sum of the values is 6 + 6 + 5 = 17.


import sys
def ii(): return int(sys.stdin.readline())
def mi(): return map(int, sys.stdin.readline().split())
def li(): return list(map(int, sys.stdin.readline().split()))
def li2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]
def dp2(ini, i, j): return [[ini]*i for i2 in range(j)]
def dp3(ini, i, j, k): return [[[ini]*i for i2 in range(j)] for i3 in range(k)]
#import bisect #bisect.bisect_left(B, a)
#from collections import defaultdict #d = defaultdict(int) d[key] += value

from collections import deque
N, M = mi()
x = li2(M)

adj = [[] for i in range(N)]
for i in range(M):
    x[i][0] -= 1
    x[i][1] -= 1
    adj[x[i][0]].append(x[i][1])

flag = [0]*N
ans = deque([])

def dfs(x):
    if flag[x]:
        return False
    else:
        flag[x] = 1
        for a in adj[x]:
            dfs(a)
        global ans
        ans.appendleft(x)

for i in range(N):
    dfs(i)

#print(ans)

dp = [0]*N
for a in ans:
    for b in adj[a]:
        dp[b] = max(dp[b], dp[a] + 1)

print(max(dp))

Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

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

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

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

Sample Output 3

3

The red directed path in the following figure is one of the longest:


n = list(input())
c = 0
for i in n:
    if i == "2":
        c += 1
print(c)

Score : 100 points

Problem Statement

You are given an integer N that has exactly four digits in base ten.
How many times does 2 occur in the base-ten representation of N?

Constraints

1000 \leq N \leq 9999

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

1222

Sample Output 1

3

2 occurs three times in 1222. By the way, this contest is held on December 22 (JST).

Sample Input 2

3456

Sample Output 2

0

Sample Input 3

9592

Sample Output 3

1


from math import sqrt

def fact(n):
    insu = []
    for i in range(2,int(sqrt(n))+1):
        if n % i == 0:
            while n % i == 0:
                n //= i
                insu.append(i)
        if n == 1:
            break
    if n > 1:
        insu.append(n)
    return insu

def main():
    N,P = map(int,input().split())
    insu = fact(P)
    insu_set,insu_list,pt = set(),[],0
    for i in range(len(insu)):
        if insu[i] not in insu_set:
            insu_set.add(insu[i])
            insu_list.append([insu[i],1])
            pt += 1
        else:
            insu_list[pt-1][1] += 1
    for i in range(len(insu_list)):
        insu_list[i][1] //= N

    ans = 1
    for a,b in insu_list:
        ans *= a**b
    print(ans)

if __name__ =="__main__":
    main()

Score : 300 points

Problem Statement

There are N integers a_1, a_2, ..., a_N not less than 1.
The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.

Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Constraints

1 \leq N \leq 10^{12}

1 \leq P \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N P

Output

Print the answer.

Sample Input 1

3 24

Sample Output 1

2

The greatest common divisor would be 2 when, for example, a_1=2, a_2=6 and a_3=2.

Sample Input 2

5 1

Sample Output 2

1

As a_i are positive integers, the only possible case is a_1 = a_2 = a_3 = a_4 = a_5 = 1.

Sample Input 3

1 111

Sample Output 3

111

Sample Input 4

4 972439611840

Sample Output 4

206


import math
N,P=map(int,input().split())
L=[]
for i in range(1,int(math.sqrt(P))+1):
	if P%i==0:
		L.append(i)
L=set(L)
L=sorted(L)
S=[]
for i in L:
	if P%(i**N)==0:
		S.append(i)
print(max(S))


src = input()
count = 0

while src.find('BW') > -1 :
    src = src.replace('BW', 'WB', 1)
    count = count + 1

print( count )

Score : 300 points

Problem Statement

There are N Reversi pieces arranged in a row. (A Reversi piece is a disc with a black side and a white side.)
The state of each piece is represented by a string S of length N.
If S_i=B, the i-th piece from the left is showing black;
If S_i=W, the i-th piece from the left is showing white.

Consider performing the following operation:

Choose i (1 \leq i < N) such that the i-th piece from the left is showing black and the (i+1)-th piece from the left is showing white, then flip both of those pieces. That is, the i-th piece from the left is now showing white and the (i+1)-th piece from the left is now showing black.

Find the maximum possible number of times this operation can be performed.

Constraints

1 \leq |S| \leq 2\times 10^5

S_i=B or W

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum possible number of times the operation can be performed.

Sample Input 1

BBW

Sample Output 1

2

The operation can be performed twice, as follows:

Flip the second and third pieces from the left.

Flip the first and second pieces from the left.

Sample Input 2

BWBWBW

Sample Output 2

6


def main():
    n = int(input())
    nums = list(map(int, input().split()))
    nums.sort()
    used = [False] * n

    num_bin_digits = [1 << len("{:b}".format(x)) for x in nums]
    cnt = 0

    for cur in reversed(range(n)):
        if used[cur]:
            continue

        num = nums[cur]
        power_of_2 = num_bin_digits[cur]
        pair = power_of_2 - num
        idx = -1
        if pair in nums[:cur]:
            idx = nums.index(pair)

        if idx != -1 and not used[idx]:
            used[idx] = True
            cnt += 1

        #print(idx, num, power_of_2, pair)
        #print(used)

    print(cnt)

    return

main()

Score : 600 points

Problem Statement

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i.
He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2.
Note that a ball cannot belong to multiple pairs.
Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

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 pairs such that the sum of the integers written on each pair of balls is a power of 2.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2


def main():
    N = int(input())
    prices = [int(input()) for _ in range(N)]
    r = sum(prices) - max(prices) / 2
    print(int(r))

if __name__ == '__main__':
    main()

Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.


N = int(input())

kakaku = [int(input()) for _ in range(N)]
total = int(sum(kakaku) - (max(kakaku) / 2))
print(total)


n, k = map(int, input().split())
h = [int(input()) for i in range(n)]

h.sort()

ans = 100000000

for i in range(n-k+1):
  if ans > h[i+k-1]-h[i]:
    ans = h[i+k-1]-h[i]

print(ans)

Score : 300 points

Problem Statement

In some other world, today is Christmas Eve.

There are N trees planted in Mr. Takaha's garden. The height of the i-th tree (1 \leq i \leq N) is h_i meters.

He decides to choose K trees from these trees and decorate them with electric lights. To make the scenery more beautiful, the heights of the decorated trees should be as close to each other as possible.

More specifically, let the height of the tallest decorated tree be h_{max} meters, and the height of the shortest decorated tree be h_{min} meters. The smaller the value h_{max} - h_{min} is, the better. What is the minimum possible value of h_{max} - h_{min}?

Constraints

2 \leq K < N \leq 10^5

1 \leq h_i \leq 10^9

h_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
h_1
h_2
:
h_N

Output

Print the minimum possible value of h_{max} - h_{min}.

Sample Input 1

5 3
10
15
11
14
12

Sample Output 1

2

If we decorate the first, third and fifth trees, h_{max} = 12, h_{min} = 10 so h_{max} - h_{min} = 2. This is optimal.

Sample Input 2

5 3
5
7
5
7
7

Sample Output 2

0

If we decorate the second, fourth and fifth trees, h_{max} = 7, h_{min} = 7 so h_{max} - h_{min} = 0. This is optimal.

There are not too many trees in these sample inputs, but note that there can be at most one hundred thousand trees (we just can't put a sample with a hundred thousand lines here).


N,K = map(int,input().split())
H = sorted(int(input()) for _ in range(N))
ans = float("INF")
for i in range(N-K+1):
    ans = min(H[i+K-1]-H[i],ans)
print(ans)


N,K=map(int,input().split())
H=[int(input()) for i in range(N)]

H.sort()
ANS=10**10
for i in range(K-1,N):
    if ANS>H[i]-H[i-K+1]:
        ANS=H[i]-H[i-K+1]
print(ANS)


N, X = map(int, input().split())
a = [1]
p = [1]
for i in range(N):
	a.append(a[i]*2+3)
	p.append(p[i]*2+1)

def f(N, X):
	if N == 0:
		return 0 if X <= 0 else 1
	elif 1 <= X <= a[N-1]+1:
		return f(N-1, X-1)
	elif X == a[N-1]+2:
		return p[N-1] + 1
	elif a[N-1]+3 <= X <= 2*a[N-1]+2:
		return f(N-1, X-2-a[N-1]) + p[N-1] + 1
	else:
		return 2*p[N-1] + 1

print(f(N, X))

Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

A level-0 burger is a patty.

A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

1 \leq N \leq 50

1 \leq X \leq ( the total number of layers in a level-N burger )

N and X are integers.

Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.


if input() in ['7', '5', '3']:
  print('YES')
else:
  print('NO')

Score : 100 points

Problem Statement

Shichi-Go-San (literally "Seven-Five-Three") is a traditional event in a certain country to celebrate the growth of seven-, five- and three-year-old children.

Takahashi is now X years old. Will his growth be celebrated in Shichi-Go-San this time?

Constraints

1 ≤ X ≤ 9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

If Takahashi's growth will be celebrated, print YES; if it will not, print NO.

Sample Input 1

5

Sample Output 1

YES

The growth of a five-year-old child will be celebrated.

Sample Input 2

6

Sample Output 2

NO

See you next year.


X = int(input())
l = [7,5,3]

if X in l:
  print('Yes')
else:
  print('No')


S = input()
ans = 753
for i in range(len(S)-2):
    ans = min(ans, abs(753-int(S[i:i+3])))
print(ans)

Score : 200 points

Problem Statement

There is a string S consisting of digits 1, 2, ..., 9.
Lunlun, the Dachshund, will take out three consecutive digits from S, treat them as a single integer X and bring it to her master. (She cannot rearrange the digits.)

The master's favorite number is 753. The closer to this number, the better.
What is the minimum possible (absolute) difference between X and 753?

Constraints

S is a string of length between 4 and 10 (inclusive).

Each character in S is 1, 2, ..., or 9.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum possible difference between X and 753.

Sample Input 1

1234567876

Sample Output 1

34

Taking out the seventh to ninth characters results in X = 787, and the difference between this and 753 is 787 - 753 = 34. The difference cannot be made smaller, no matter where X is taken from.

Note that the digits cannot be rearranged. For example, taking out 567 and rearranging it to 765 is not allowed.

We cannot take out three digits that are not consecutive from S, either. For example, taking out the seventh digit 7, the ninth digit 7 and the tenth digit 6 to obtain 776 is not allowed.

Sample Input 2

35753

Sample Output 2

0

If 753 itself can be taken out, the answer is 0.

Sample Input 3

1111111111

Sample Output 3

642

No matter where X is taken from, X = 111, with the difference 753 - 111 = 642.


S=input()
S_len = len(S)
S_num = [0]*(S_len-2)
for i in range(S_len - 2):
    S_num[i] = int(S[i]+S[i+1]+S[i+2])
    S_num[i] = abs(753-S_num[i])
print(min(S_num))


def function(n):
    digit=len(str(n))
    N=str(n)
    k=int(N[0])
    if digit<3:
        return 0
    else:
        if k<3:
            return function(10**(digit-1)-1)
        elif k==3:
            return function(10**(digit-1)-1)+function(n%(10**(digit-1)))-function(10**(digit-2))+subfunction(5,7,n%(10**(digit-1)))
        elif k==4:
            return function(4*(10**(digit-1))-1)
        elif k==5:
            return function(4*(10**(digit-1))-1)+function(n%(10**(digit-1)))-function(10**(digit-2))+subfunction(3,7,n%(10**(digit-1)))
        elif k==6:
            return function(6*(10**(digit-1))-1)
        elif k==7:
            return function(6*(10**(digit-1))-1)+function(n%(10**(digit-1)))-function(10**(digit-2))+subfunction(3,5,n%(10**(digit-1)))
        else:
            return function(8*(10**(digit-1))-1)

def subfunction(a,b,n):
    digit=len(str(n))
    N=str(n)
    k=int(N[0])
    if digit<2:
        return 0
    else:
        if k<a:
            return 0
        elif k==a:
            if n%(10**(digit-1))<b*(10**(digit-1)-1)//9:
                s=0
            else:
                s=1
            return subfunction(a,b,n%(10**(digit-1)))+s
        elif a<k<b:
            return subfunction(a,b,(a+1)*(10**(digit-1))-1)
        elif k==b:
            if n%(10**(digit-1))<a*(10**(digit-1)-1)//9:
                s=0
            else:
                s=1
            return subfunction(a,b,(a+1)*(10**(digit-1))-1)+subfunction(a,b,n%(10**(digit-1)))+s
        else:
            return subfunction(a,b,(b+1)*(10**(digit-1))-1)

N=int(input())
print(function(N))

Score : 300 points

Problem Statement

You are given an integer N. Among the integers between 1 and N (inclusive), how many Shichi-Go-San numbers (literally "Seven-Five-Three numbers") are there?

Here, a Shichi-Go-San number is a positive integer that satisfies the following condition:

When the number is written in base ten, each of the digits 7, 5 and 3 appears at least once, and the other digits never appear.

Constraints

1 \leq N < 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go-San numbers between 1 and N (inclusive).

Sample Input 1

575

Sample Output 1

4

There are four Shichi-Go-San numbers not greater than 575: 357, 375, 537 and 573.

Sample Input 2

3600

Sample Output 2

13

There are 13 Shichi-Go-San numbers not greater than 3600: the above four numbers, 735, 753, 3357, 3375, 3537, 3557, 3573, 3575 and 3577.

Sample Input 3

999999999

Sample Output 3

26484


Score : 400 points

Problem Statement

You are given an integer N. Among the divisors of N! (= 1 \times 2 \times ... \times N), how many Shichi-Go numbers (literally "Seven-Five numbers") are there?

Here, a Shichi-Go number is a positive integer that has exactly 75 divisors.

Note

When a positive integer A divides a positive integer B, A is said to a divisor of B.
For example, 6 has four divisors: 1, 2, 3 and 6.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of the Shichi-Go numbers that are divisors of N!.

Sample Input 1

9

Sample Output 1

0

There are no Shichi-Go numbers among the divisors of 9! = 1 \times 2 \times ... \times 9 = 362880.

Sample Input 2

10

Sample Output 2

1

There is one Shichi-Go number among the divisors of 10! = 3628800: 32400.

Sample Input 3

100

Sample Output 3

543


#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int main()
{
	int a,b,c;
	cin>>a>>b;
	c=a+b/2;
	cout<<c<<endl;
}

Score: 100 points

Problem Statement

There is a train going from Station A to Station B that costs X yen (the currency of Japan).

Also, there is a bus going from Station B to Station C that costs Y yen.

Joisino got a special ticket. With this ticket, she can take the bus for half the fare if she travels from Station A to Station B by train and then travels from Station B to Station C by bus.

How much does it cost to travel from Station A to Station C if she uses this ticket?

Constraints

1 \leq X,Y \leq 100

Y is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

If it costs x yen to travel from Station A to Station C, print x.

Sample Input 1

81 58

Sample Output 1

110

The train fare is 81 yen.

The train fare is 58 ⁄ 2=29 yen with the 50% discount.

Thus, it costs 110 yen to travel from Station A to Station C.

Sample Input 2

4 54

Sample Output 2

31


x, y = map(int,input().split())
#x = int(input())
print(x + y//2)


X, Y = map(int,input().split())

print(X + Y//2)


def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))

n = ii()
t,a = iim()
H = iil()
mdif = 10**9
ans = 0
for i,item in enumerate(H):
    tem = t - item*0.006
    if abs(tem - a) < mdif:
        ans = i+1
        mdif = abs(tem - a)
print(ans)

Score: 200 points

Problem Statement

A country decides to build a palace.

In this country, the average temperature of a point at an elevation of x meters is T-x \times 0.006 degrees Celsius.

There are N places proposed for the place. The elevation of Place i is H_i meters.

Among them, Princess Joisino orders you to select the place whose average temperature is the closest to A degrees Celsius, and build the palace there.

Print the index of the place where the palace should be built.

It is guaranteed that the solution is unique.

Constraints

1 \leq N \leq 1000

0 \leq T \leq 50

-60 \leq A \leq T

0 \leq H_i \leq 10^5

All values in input are integers.

The solution is unique.

Input

Input is given from Standard Input in the following format:

N
T A
H_1 H_2 ... H_N

Output

Print the index of the place where the palace should be built.

Sample Input 1

2
12 5
1000 2000

Sample Output 1

1

The average temperature of Place 1 is 12-1000 \times 0.006=6 degrees Celsius.

The average temperature of Place 2 is 12-2000 \times 0.006=0 degrees Celsius.

Thus, the palace should be built at Place 1.

Sample Input 2

3
21 -11
81234 94124 52141

Sample Output 2

3


N = int(input())
T,A =map(int, input().split())
H = list(map(int, input().split()))
dif = [A-i for i in [T-h*0.006 for h in H]]
dif_min = dif.index(min(dif))
print(dif_min+1)


N=int(input())
T,A=map(int,input().split())
H=[abs(A-(T-i*0.006)) for i in map(int,input().split())]
print(H)
print(H.index(min(H))+1)


N = int(input())
T,A = (int(i) for i in input().split())
p = float('inf')
for i,h in enumerate([int(i) for i in input().split()]):
    tmp = T-h*0.006
    if p > abs(A-tmp):
        p = A-tmp
        ans = i+1
print(ans)


from sys import stdin

def main():
    N = int(input())
    T, A = map(int, input().split())
    H = map(int, input().split())

    temp = [T - x * 0.006 for x in H]
    diff = float('inf')
    ans = 0
    for i, a in enumerate(temp, start=1):
        b = abs(A - a)
        if not b:
            print(i)
            return
        elif diff > b:
            ans, diff = i, b
    print(ans)

input = lambda: stdin.readline()
main()


def main():
    n, m = map(int, input().split())
    query = []
    result =  [[] for i in range(n)]
    for i in range(m):
        qList = list(map(int, input().split()))
        query.append(qList)
        result[qList[0]-1].append(qList[1])

    for i in result:
        i.sort()

    for city in query:
        rank = bisect.bisect_left(result[city[0]-1], city[1])
        ans = "%06d%06d"%(city[0],rank+1)
        print(ans)

if __name__ == "__main__":
    main()

Score: 300 points

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Y_i and belongs to Prefecture P_i.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is P_i, and the last six digits of the ID number is x.

Here, if P_i or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq P_i \leq N

1 \leq Y_i \leq 10^9

Y_i are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
P_1 Y_1
:
P_M Y_M

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, ...).

Sample Input 1

2 3
1 32
2 63
1 12

Sample Output 1

000001000002
000002000001
000001000001

As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.

As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.

As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

Sample Input 2

2 3
2 55
2 77
2 99

Sample Output 2

000002000001
000002000002
000002000003


N, M = map(int, input().split())
p = list()
y = list()
d = dict()
for i in range(M):
    inp_p, inp_y = map(int, input().split())
    p.append(inp_p)
    y.append(inp_y)
    if inp_p in d:
        d[inp_p].append(inp_y)
    else:
        d[inp_p] = [inp_y]
for i in d:
    d[i] = sorted(d[i])
for i, j in zip(p, y):
    print("{:06d}{:06d}".format(i, d[i].index(j)+1))


n,*a=map(int,open(0).readlines())
a.sort()
if n%2:
  b=a[:n//2]
  c=a[n//2:]
  for i in range(n//2):
    b[i]*=2
  for i in range(2,n//2+1):
    c[i]*=2
  m=sum(c)-sum(b)
  b=a[:n//2+1]
  c=a[n//2+1:]
  for i in range(n//2-1):
    b[i]*=2
  for i in range(n//2):
    c[i]*=2
  m=max(m,sum(c)-sum(b))
else:
  b=a[:n//2]
  c=a[n//2:]
  for i in range(n//2-1):
    b[i]*=2
  for i in range(1,n//2):
    c[i]*=2
  m=sum(c)-sum(b)
print(m)

Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8