TnT/CodeNet4Repair huggingface dataset:
Problem:
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament.  The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day.  Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.

Incorrect submission (verdict: Wrong Answer):
s = input()
c=0
n = list(map(str, s))
print(n)

for i in n :
    if i == 'o' :
        c +=1
        
if c >= 8 or (8-c) == 15 - len(n) :
    print('YES')
else :
    print('NO')

Fixed, accepted submission:
s = input()
c=0
n = list(map(str, s))

for i in n :
    if i == 'o' :
        c +=1
        
if c >= 8 or (8-c) <= 15 - len(n) :
    print('YES')
else :
    print('NO')
Problem:
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.

Incorrect submission (verdict: Wrong Answer):
k,a,b = map(int,input().split())
if b - a <= 1 or a >= k:
    print(k+1)
else:
    k -= a - 1
    ans = a
    ans += (b - a)*(k//2)
    if k/2 % 2 == 0:
        ans += 1
    print(ans)

Fixed, accepted submission:
k,a,b = map(int,input().split())
if b - a <= 1 or a >= k:
    print(k+1)
else:
    k -= a - 1
    ans = a
    ans += (b - a)*(k//2)
    if k % 2 != 0:
        ans += 1
    print(ans)
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
def main():
    a, b, c = map(int, input().split())

    if a + b == c:
        print('YES')
        return
    if a + c == b:
        print('YES')
        return
    if a == b + c:
        print('YES')
        return

    print('NO')
    return


if __name__ == '__main__':
    main()


Fixed, accepted submission:


def main():
    a, b, c = map(int, input().split())

    if a + b == c:
        print('Yes')
        return
    if a + c == b:
        print('Yes')
        return
    if a == b + c:
        print('Yes')
        return

    print('No')
    return


if __name__ == '__main__':
    main()
Problem:
Niwango-kun is an employee of Dwango Co., Ltd.   One day, he is asked to generate a thumbnail from a video a user submitted.   To generate a thumbnail, he needs to select a frame of the video according to the following procedure:    * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.   * Select t-th frame whose representation a_t is nearest to the average of all frame representations.   * If there are multiple such frames, select the frame with the smallest index.  Find the index t of the frame he should select to generate a thumbnail.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
P = list(map(int,input().split()))
sum = 0
ave = 0
ave_first = 99999
k = []


for i in range(N):
    sum += P[i]
    print(sum)
ave = sum/N
print(ave)

for i in range(0,N-1):
    if abs(P[i]-ave)< ave_first:
            ave_first = abs(P[i]-ave)
            k.clear()
            k.append(i)
    elif abs(P[i]-ave) == ave_first:
        k.append(i)
print(min(k))

Fixed, accepted submission:
N = int(input())
P = list(map(int,input().split()))
sum = 0
ave = 0
ave_first = 99999
k = []


for i in range(N):
    sum += P[i]
ave = sum/N

for i in range(0,N-1):
    if abs(P[i]-ave)< ave_first:
            ave_first = abs(P[i]-ave)
            k.clear()
            k.append(i)
    elif abs(P[i]-ave) == ave_first:
        k.append(i)
print(min(k))
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
_input = list(map(int, input().split(" ")))
a = _input[0]
b = _input[1]
c = _input[2]

check_limit = a * b
ratio = 1
remain_list = []

while (True):
  remain_list.append(a * ratio % b)
  ratio += 1

  if ratio > b:
    break

if c in remain_list:
  print("Yes")
else:
  print("No")



Fixed, accepted submission:
_input = list(map(int, input().split(" ")))
a = _input[0]
b = _input[1]
c = _input[2]

check_limit = a * b
ratio = 1
remain_list = []

while (True):
  remain_list.append(a * ratio % b)
  ratio += 1

  if ratio > b:
    break

if c in remain_list:
  print("YES")
else:
  print("NO")
Problem:
We have a grid with H rows and W columns, where all the squares are initially white.  You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions:    * Choose one row, then paint all the squares in that row black.   * Choose one column, then paint all the squares in that column black.  At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

Incorrect submission (verdict: Wrong Answer):
a = int(input())
b = int(input())
c = int(input())
print(c//max(a,b))

Fixed, accepted submission:
import math
a = int(input())
b = int(input())
c = int(input())
print(math.ceil(c/max(a,b)))
Problem:
AtCoDeer has three cards, one red, one green and one blue.   An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card.   We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer.   Is this integer a multiple of 4?

Incorrect submission (verdict: Wrong Answer):
r,g,b = map(int,input().split())

number = r * 100 + g * 10 + b

if number % 4 == 0:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
r,g,b = map(int,input().split())

number = r * 100 + g * 10 + b

if number % 4 == 0:
    print("YES")
else:
    print("NO")
Problem:
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.  The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)  Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

Incorrect submission (verdict: Wrong Answer):
x=int(input())
i=100
c=0
while i<=x:
    i=i+(i//100)
    c=c+1
print(c)

Fixed, accepted submission:
x=int(input())
i=100
c=0
while i<x:
    i=i+(i//100)
    c=c+1
print(c)
Problem:
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.

Incorrect submission (verdict: Wrong Answer):
print(["No","Yes"]["".join(input())=="abc"])

Fixed, accepted submission:
print(["No","Yes"]["".join(sorted(input()))=="abc"])
Problem:
Takahashi is a user of a site that hosts programming contests.   When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows:    * Let the current rating of the user be a.   * Suppose that the performance of the user in the contest is b.   * Then, the new rating of the user will be the avarage of a and b.  For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000.  Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest.   Find the performance required to achieve it.

Incorrect submission (verdict: Wrong Answer):
r=int(input())
g=int(input())
print((g-r)*2+g)

Fixed, accepted submission:
r=int(input())
g=int(input())
print((g-r)*2+r)
Problem:
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

Incorrect submission (verdict: Wrong Answer):
n,k=map(int,input().split())
if k-n>=1:
    print('YES')
else:
    print('NO')


Fixed, accepted submission:
import sys
n,k=map(int,input().split())
tmp=1
count=1
if n==1 and k==1:
    print('YES')
    sys.exit(0)
if k==0:
    print('YES')
    sys.exit(0)
for i in range(2,n+1):
    if i-tmp>1:
        tmp=i
        count+=1
    if count==k:
        print('YES')
        sys.exit(0)
print('NO')
Problem:
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements?  Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:    * r_1 + c_1 = r_2 + c_2   * r_1 - c_1 = r_2 - c_2  For example, in the following figure, the bishop can move to any of the red squares in one move:  

Incorrect submission (verdict: Wrong Answer):
H,W = map(int, input().split())
if H*W%2 == 0:
  print(H*W/2)
else:
  print(H*W//2+1)
  

Fixed, accepted submission:
H,W = map(int, input().split())
if W == 1 or H == 1:
  print(1)
elif H*W%2 == 0:
  print(H*W//2)
else:
  print(H*W//2+1)
Problem:
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct.  Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X.  Find the duration of time when some song was played while Niwango was asleep.

Incorrect submission (verdict: Wrong Answer):
n = int(input())

s = ['']*n
t = ['']*n
for i in range(n):
    s[i], t[i] = input().split()
    
    
x = input()

p = s.index(x)
total = 0
for i in range(p, n):
    total += int(t[i])
print(total)

Fixed, accepted submission:
n = int(input())

s = ['']*n
t = ['']*n
for i in range(n):
    s[i], t[i] = input().split()
    
    
x = input()

p = s.index(x)
total = 0
for i in range(p+1, n):
    total += int(t[i])
print(total)
Problem:
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Incorrect submission (verdict: Wrong Answer):
n, a, b = map(int, input().split())
ans = 0
for i in range(1, n+1):
    if a <= sum(map(int, str(i))) <= b:
        ans += i
        print(sum(list(map(int,str(i)))))
print(ans)

Fixed, accepted submission:
n, a, b = map(int, input().split())
ans = 0
for i in range(1, n+1):
    if a <= sum(map(int, str(i))) <= b:
        ans += i
print(ans)
Problem:
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.  The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).  Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.  If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Incorrect submission (verdict: Wrong Answer):
n= int(input())
if int(-(-n//1.08)*1.08) == n:
    print(n//1.08)
else:
    print(":(")

Fixed, accepted submission:
n= int(input())
if int(-(-n//1.08)*1.08) == n:
    print(int(-(-n//1.08)))
else:
    print(":(")
Problem:
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.  Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** :    * Add 2 to a_i.   * Add 1 to b_j.

Incorrect submission (verdict: Wrong Answer):
import sys

def solve():
    input = sys.stdin.readline
    N = int(input())
    A = [int(a) for a in input().split()]
    B = [int(b) for b in input().split()]
    A.sort()
    B.sort()
    C = []
    for i in range(N): C.append(A[i] - B[i])
    print(C)
    sumA = sum(A)
    sumB = sum(B)
    if sumA >= sumB: print("No")
    else:
        for i, c in enumerate(C):
            if c > 0: 
                print("No")
                break
        else:
            print("Yes")

    return 0

if __name__ == "__main__":
    solve()

Fixed, accepted submission:
import sys

def solve():
    input = sys.stdin.readline
    N = int(input())
    A = [int(a) for a in input().split()]
    B = [int(b) for b in input().split()]
    C = [A[i] - B[i] for i in range(N)]
    sumA = sum(A)
    sumB = sum(B)
    addA, addB = 0, 0
    for i, c in enumerate(C):
        if c > 0: addB += c
        else: addA += (-1 * c + 1) // 2
    if addA > sumB - sumA or addB > sumB - sumA: print("No")
    else: print("Yes")

    return 0

if __name__ == "__main__":
    solve()
Problem:
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?

Incorrect submission (verdict: Wrong Answer):
a, b, x = map(int, input().split())

c = (a-1) / x
d = b / x

print(d-c)



Fixed, accepted submission:
a, b, x = map(int, input().split())

c = (a-1) // x
d = b // x

print(d-c)
Problem:
There is a cave.  The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.  It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.  Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.    * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.  Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Incorrect submission (verdict: Wrong Answer):

N, M = map(int, input().split())
AB = [list(map(int, input().split())) for i in range(M)]
R = [0] * N
P = 0
S = [1]
Stmp = []
while True:
    for i in range(M):
        for s in S:
            if AB[i][0] == s:
                R[AB[i][1] - 1] = s
                Stmp.append(AB[i][1])
                AB[i] = [0, 0]
                P += 1
            elif AB[i][1] == s:
                R[AB[i][0] - 1] = s
                Stmp.append(AB[i][0])
                AB[i] = [0, 0]
                P += 1
    S = Stmp
    Stmp = []
    print(R, P)
    if P > N - 1:
        break
print('Yes')
for i in range(1, N):
    print(R[i])

Fixed, accepted submission:
from collections import deque
N, M = map(int, input().split())
R = [[] for i in range(N + 1)]
for i in range(M):
    A, B = map(int, input().split())
    R[A].append(B)
    R[B].append(A)
P = [-1] * (N + 1)
P[1] = 0
Q = deque()
Q.append(1)

while Q:
    v = Q.popleft()
    for r in R[v]:
        if P[r] == -1:
            P[r] = v
            Q.append(r)
print('Yes')
for p in P[2:]:
    print(p)
Problem:
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.  The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.  How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?

Incorrect submission (verdict: Wrong Answer):
from sys import stdin
N=int(stdin.readline().rstrip())
H=int(stdin.readline().rstrip())
W=int(stdin.readline().rstrip())
ans=int(N/H)*int(W/H)
print(ans)

Fixed, accepted submission:
from sys import stdin
N=int(stdin.readline().rstrip())
H=int(stdin.readline().rstrip())
W=int(stdin.readline().rstrip())
ans=(N-H+1)*(N-W+1)
print(ans)
Problem:
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.  If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
me = [0,0]
arr = []

for i in range(n):
    arr += [list(map(int,input().split()))]

for t in arr:
    for _ in range(t[0]):
        if me[0] < t[1]:
            me[0] += 1
        elif me[0] > t[1]:
            me[0] -= 1
        elif me[1] < t[2]:
            me[1] += 1
        elif me[1] > t[2]:
            me[1] -= 1
        else:
            me[0] += 1
    if me != t[1:]:
        break
if me == arr[-1][1:]:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
n = int(input())
me = [0,0]
arr = []
tmp = 0
flg = True
for i in range(n):
    arr += [list(map(int,input().split()))]

for t in arr:
    for _ in range(t[0]-tmp):
        if me[0] < t[1]:
            me[0] += 1
        elif me[0] > t[1]:
            me[0] -= 1
        elif me[1] < t[2]:
            me[1] += 1
        elif me[1] > t[2]:
            me[1] -= 1
        else:
            me[0] += 1
        tmp = t[0]
    if me != t[1:]:
        flg = False
        break
if flg:
    print("Yes")
else:
    print("No")
Problem:
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.  Determine if it is possible that there are exactly X cats among these A + B animals.

Incorrect submission (verdict: Wrong Answer):
a,b,x=map(int,input().split());print("NYoe s"[x in range(a,b+1)::2])

Fixed, accepted submission:
a,b,x=map(int,input().split());print("NYOE S"[x in range(a,a+b+1)::2])
Problem:
You start with the number 0 and you want to reach the number N.  You can change the number, paying a certain amount of coins, with the following operations:    * Multiply the number by 2, paying A coins.   * Multiply the number by 3, paying B coins.   * Multiply the number by 5, paying C coins.   * Increase or decrease the number by 1, paying D coins.  You can perform these operations in arbitrary order and an arbitrary number of times.  What is the minimum number of coins you need to reach N?  **You have to solve T testcases.**

Incorrect submission (verdict: Wrong Answer):
from math import ceil
import sys
sys.setrecursionlimit(10**7)
memo=dict()
def dfs(x,a,b,c,d):
    if x==0:
        return 0
    if x==1:
        return d
    if x in memo:
        return memo[x]
    l2=2*(x//2)
    r2=2*ceil(x/2)
    l3=3*(x//3)
    r3=3*ceil(x/3)
    l5=5*(x//5)
    r5=5*ceil(x/5)
    mini=10**18
    mini=min(mini,abs(l2-x)*d+a+dfs(x//2,a,b,c,d))
    mini=min(mini,abs(r2-x)*d+a+dfs(ceil(x/2),a,b,c,d))
    mini=min(mini,abs(l3-x)*d+b+dfs(x//3,a,b,c,d))
    mini=min(mini,abs(r3-x)*d+b+dfs(ceil(x/3),a,b,c,d))
    mini=min(mini,abs(l5-x)*d+c+dfs(x//5,a,b,c,d))
    mini=min(mini,abs(r5-x)*d+c+dfs(ceil(x/5),a,b,c,d))
    memo[x]=mini
    return mini
t=int(input())
for _ in range(t):
    memo=dict()
    n,a,b,c,d=map(int,input().split())
    print(dfs(n,a,b,c,d))

Fixed, accepted submission:
import sys
sys.setrecursionlimit(10**7)
t=int(input())
for _ in range(t):
    memo=dict()
    n,a,b,c,d=map(int,input().split())
    def dfs(x):
        if x==0:
            return 0
        if x==1:
            return d
        if x in memo:
            return memo[x]
        l2=2*(x//2)
        r2=2*(-(-x//2))
        l3=3*(x//3)
        r3=3*(-(-x//3))
        l5=5*(x//5)
        r5=5*(-(-x//5))
        mini=d*x
        mini=min(mini,abs(l2-x)*d+a+dfs(l2//2))
        mini=min(mini,abs(r2-x)*d+a+dfs(r2//2))
        mini=min(mini,abs(l3-x)*d+b+dfs(l3//3))
        mini=min(mini,abs(r3-x)*d+b+dfs(r3//3))
        mini=min(mini,abs(l5-x)*d+c+dfs(l5//5))
        mini=min(mini,abs(r5-x)*d+c+dfs(r5//5))
        memo[x]=mini
        return mini
    print(dfs(n))
Problem:
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:    * 1 \leq x,y,z   * x^2 + y^2 + z^2 + xy + yz + zx = n  Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Incorrect submission (verdict: Wrong Answer):
from math import sqrt
from math import floor
n = int(input())

for i in range(n):
    n2 = floor(sqrt(i))
    ans = 0
    for x in range(1, n2):
        for y in range(1, n2):
            for z in range(1, n2):
                if i == pow(x, 2) + pow(y, 2) + pow(z, 2) + x * y + y * z + z * x:
                    ans += 1
    print(ans)


Fixed, accepted submission:
n = int(input())

ans = [0] * n
for x in range(1, 101):
    for y in range(1, 101):
        for z in range(1, 101):
            a = x**2 + y**2 + z**2 + x*y + y*z + z*x
            if a >= 1 and a <= n:
                ans[a - 1] += 1

for a in ans:
    print(a)
Problem:
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.

Incorrect submission (verdict: Wrong Answer):
a, b, c = map(int, input().split(' '))

if b > c:
    print('yes')
else:
    print('no')

Fixed, accepted submission:
a, b, c = map(int, input().split(' '))

if a > b and b > c and a > c:
    print('No')
if a > c and a > b and c > b:
    print('Yes')
elif b > a and b > c and a > c:
    print('No')
elif b > c and c > a and b > a:
    print('Yes')
elif c > a and c > b and a > b:
    print('No')
elif  c > b and c > a and b > a:
    print('No')