TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}.  Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.

Incorrect submission (verdict: Wrong Answer):
H,W = map(int,input().split())
A = [list(input()) for _ in range(H)]

ANS = []
for i in range(H+1):
  t = []
  for j in range(W+1):
    if i == 0 or j == 0 or i == H or j == W:
      t.append("#")
    else:
      t.append(A[i-1][j-1])
  ANS.append("".join(t))
  
print("\n".join(ANS))

Fixed, accepted submission:
H,W = map(int,input().split())
A = [list(input()) for _ in range(H)]

ANS = []
for i in range(H+2):
  t = []
  for j in range(W+2):
    if i == 0 or j == 0 or i == H+1 or j == W+1:
      t.append("#")
    else:
      t.append(A[i-1][j-1])
  ANS.append("".join(t))
  
print("\n".join(ANS))
Problem:
You have an integer variable x. Initially, x=0.  Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`.  Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).

Incorrect submission (verdict: Wrong Answer):
n = int(input())
s = input()
tmp_list = [0]
x = 0
for _ in s:
    if _ == 'I':
        x += 1
    else:
        x -= 1
    tmp_list.append(x)
    print(x)
print(max(tmp_list))

Fixed, accepted submission:
n = int(input())
s = input()
tmp_list = [0]
x = 0
for _ in s:
    if _ == 'I':
        x += 1
    else:
        x -= 1
    tmp_list.append(x)
print(max(tmp_list))
Problem:
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters.  Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way.  You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.

Incorrect submission (verdict: Wrong Answer):
s=input()
a=""
for i in range(len(s)):
    if s[i]=='C':
        a+="c"
    if s[i]=='F':
        a+="f"
if a=="cf":
    print("YES")
else:
    print("NO")

Fixed, accepted submission:
s=input()
a=""
for i in range(len(s)):
    if s[i]=='C':
        a+="c"
    if s[i]=='F':
        a+="f"
if "cf" in a:
    print("Yes")
else:
    print("No")
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
S = list(input())
ops = []
lst = []

for i in range(8):
    ops.append(bin(i))
    ops[i] = ops[i][2:].zfill(3)
    print(ops[i])

for i in range(8):
    res = S[0]
    for j in range(3):
        if ops[i][j] == "0":
            res = res + "+" + S[j+1]
        else:
            res = res + "-" + S[j+1]
    if eval(res) == 7:
        res = res + "=7"
        lst.append(res)
        break

print(lst[0])

Fixed, accepted submission:
S = list(input())
ops = []
lst = []

for i in range(8):
    ops.append(bin(i))
    ops[i] = ops[i][2:].zfill(3)

for i in range(8):
    res = S[0]
    for j in range(3):
        if ops[i][j] == "0":
            res = res + "+" + S[j+1]
        else:
            res = res + "-" + S[j+1]
    if eval(res) == 7:
        res = res + "=7"
        lst.append(res)
        break

print(lst[0])
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())

if g * 10 + b % 4 == 0:
    print('YES')
else:
    print('NO')

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

if (g * 10 + b) % 4 == 0:
    print('YES')
else:
    print('NO')
Problem:
Takahashi loves numbers divisible by 2.  You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.  Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.  For example,    * 6 can be divided by 2 once: 6 -> 3.   * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.   * 3 can be divided by 2 zero times.

Incorrect submission (verdict: Wrong Answer):
from math import log2
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline


def ceil(n):
    return int(-n//1)*(-1)


N = int(input())

ans = ceil(log2(N))
if ans > N:
    ans -= 1
print(2**ans)


Fixed, accepted submission:
from math import log2
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline


def ceil(n):
    return int(-n//1)*(-1)


N = int(input())

ans = ceil(log2(N))
if 2**ans > N:
    ans -= 1
print(2**ans)
Problem:
Joisino is planning to open a shop in a shopping street.  Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.  There are already N stores in the street, numbered 1 through N.  You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.  Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.  Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
f = [list(map(int, input().split())) for _ in range(n)]
p = [list(map(int, input().split())) for _ in range(n)]
benefits = []

for t in range(1,1024):
  bits = list(bin(t))
  a = [0 for _ in range(n)]
  jk = 0
  while jk < 10:
    b = bits.pop()
    if b == "b":
      break
    elif b == "0":
      jk += 1
      continue
    else:
      for i in range(n):
        if f[i][jk]:
          a[i] += 1
  
  s = 0
  for i in range(n):
    s += p[i][a[i]]
  else:
    benefits.append(s)
else:
  print(max(benefits))

Fixed, accepted submission:
n = int(input())
f = [list(map(int, input().split())) for _ in range(n)]
p = [list(map(int, input().split())) for _ in range(n)]

benefits = -1000000000

for t in range(1,1024):
  bits = list(bin(t))
  a = [0 for _ in range(n)]
  jk = 0
  while jk < 10:
    b = bits.pop()
    if b == "b":
      break
    elif b == "0":
      jk += 1
      continue
    else:
      for i in range(n):
        if f[i][jk]:
          a[i] += 1
      jk += 1
  
  s = 0
  for i in range(n):
    s += p[i][a[i]]
  else:
    benefits = max(s,benefits)
else:
  print(benefits)
Problem:
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
print(int(n**0.5))

Fixed, accepted submission:
n = int(input())
m = int(n**0.5)
print(int(m**2))
Problem:
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.  Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Incorrect submission (verdict: Wrong Answer):
N = int(input())
l = 0
count =	0
dlist = []
for i in range(N):
    dlist.append(int(input()))

for i in range(N):
    m =	max(dlist)
    if(l != m):
        count += 1
        dlist[dlist.index(m)] = 0
        l =	m

print(count)

Fixed, accepted submission:
N = int(input())
dlist = []
for i in range(N):
    dlist.append(int(input()))
dlistt = list(set(dlist))
print(len(dlistt))
Problem:
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub

sys.setrecursionlimit(10000)


def read_int():
    return int(input())


def read_int_n():
    return list(map(int, input().split()))


def read_float():
    return float(input())


def read_float_n():
    return list(map(float, input().split()))


def read_str():
    return input().strip()


def read_str_n():
    return list(map(str, input().split()))


def error_print(*args):
    print(*args, file=sys.stderr)


def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.time()
        ret = f(*args, **kwargs)
        e = time.time()

        error_print(e - s, 'sec')
        return ret

    return wrap


@mt
def slv(N):
    if 1 not in N:
        return "No"
    if 9 not in N:
        return "No"
    if 7 not in N:
        return "No"
    if 4 not in N:
        return "No"
    return 'Yes'


def main():
    N = read_int_n()
    print(slv(N))


if __name__ == '__main__':
    main()


Fixed, accepted submission:
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub

sys.setrecursionlimit(10000)


def read_int():
    return int(input())


def read_int_n():
    return list(map(int, input().split()))


def read_float():
    return float(input())


def read_float_n():
    return list(map(float, input().split()))


def read_str():
    return input().strip()


def read_str_n():
    return list(map(str, input().split()))


def error_print(*args):
    print(*args, file=sys.stderr)


def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.time()
        ret = f(*args, **kwargs)
        e = time.time()

        error_print(e - s, 'sec')
        return ret

    return wrap


@mt
def slv(N):
    if 1 not in N:
        return "NO"
    if 9 not in N:
        return "NO"
    if 7 not in N:
        return "NO"
    if 4 not in N:
        return "NO"
    return 'YES'


def main():
    N = read_int_n()
    print(slv(N))


if __name__ == '__main__':
    main()
Problem:
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not.  You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.

Incorrect submission (verdict: Wrong Answer):
s = input()

for n in range(len(s) // 2 - 1, -1, -1):
    if s[:n] == s[n: 2 * n]:
        ans = s[: 2 * n]
        break

print(ans)


Fixed, accepted submission:
s = input()

for n in range(len(s) // 2 - 1, -1, -1):
    if s[:n] == s[n: 2 * n]:
        ans = 2 * n
        break

print(ans)
Problem:
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room.  For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left.  Assume that initially no person lives in the building.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
A = [[[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10]]

print(len(A))
for i in range(N):
    b, f, r, v = [int(j) for j in input().split()]
    A[b-1][f-1][r-1] += v

for a, b in enumerate(A):
    for f in b:
        print(' '.join([str(i) for i in f]))
    else:
        if a != len(A) - 1:
            print('#' * 10)

Fixed, accepted submission:
N = int(input())
A = [[[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10]]

for i in range(N):
    b, f, r, v = [int(j) for j in input().split()]
    A[b-1][f-1][r-1] += v

for a, b in enumerate(A):
    for f in b:
        print(' ' + ' '.join([str(i) for i in f]))
    else:
        if a != len(A) - 1:
            print('#' * 20)
Problem:
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.  Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N.  Find the minimum cost incurred before reaching the goal.

Incorrect submission (verdict: Wrong Answer):
n, m, x = input().split()
n = int(n)
m = int(m)
x = int(x)
c = [0 for _ in range(n+1)]
a = input().split()
for i in range(m):
    c[int(a[i])]=1
min(sum(c[:x]),sum(c[x:]))

Fixed, accepted submission:
n, m, x = input().split()
n = int(n)
m = int(m)
x = int(x)
c = [0 for _ in range(n+1)]
a = input().split()
for i in range(m):
    c[int(a[i])]=1
print(min(sum(c[:x]),sum(c[x:])))
Problem:
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L.  She will concatenate all of the strings in some order, to produce a long string.  Among all strings that she can produce in this way, find the lexicographically smallest one.  Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds:    * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i.   * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-
n, l = map(int, input().split())
a = []
b = ''

for i in range(n):
  a.append(input())
  
sorted(a)

for i in range(len(a)):
  b += a[i]
  
print(b)


Fixed, accepted submission:
# -*- coding: utf-8 -*-
n, l = map(int, input().split())
a = []
b = ''

for i in range(n):
  a.append(input())
  
a.sort()

for i in range(len(a)):
  b += a[i]
print(b)
Problem:
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.  Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)  For example, when n=7, it becomes 0 after two operations, as follows:    * \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.   * \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.  You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).

Incorrect submission (verdict: Wrong Answer):
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 6)


def dfs(n):
    if memo[n] >= 0:
        rst = memo[n]
    else:
        tmp = n
        one = 0
        while tmp > 0:
            one += tmp & 1
            tmp >>= 1
        m = n % one
        rst = 1 if m == 0 else 1 + dfs(m)
        memo[n] = rst
    return rst


N = int(input())
X = list(map(int, list(input())))
one = Counter(X)[1]


if one > 1:
    Sm = 0
    mm = one - 1
    tm = 1
    for i in range(N - 1, -1, -1):
        tm %= mm
        if X[i] == 1:
            Sm += tm
            Sm %= mm
        tm *= 2

Sp = 0
mp = one + 1
tp = 1
for i in range(N - 1, -1, -1):
    tp %= mp
    if X[i] == 1:
        Sp += tp
        Sp %= mp
    tp *= 2


memo = [-1] * (2 * (10 ** 5) + 10)
memo[0] = 0

for i in range(N):
    if X[i] == 0:
        m = (Sp + pow(2, N - 1 - i, mp)) % mp
    elif one > 1:
        m = (Sm - pow(2, N - 1 - i, mm)) % mm
    else:
        print(0)
        continue
    print(dfs(m))


Fixed, accepted submission:
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 6)


def dfs(n):
    if memo[n] >= 0:
        rst = memo[n]
    else:
        tmp = n
        one = 0
        while tmp > 0:
            one += tmp & 1
            tmp >>= 1
        m = n % one
        rst = 1 if m == 0 else 1 + dfs(m)
        memo[n] = rst
    return rst


N = int(input())
X = list(map(int, list(input())))
one = Counter(X)[1]


if one > 1:
    Sm = 0
    mm = one - 1
    tm = 1
    for i in range(N - 1, -1, -1):
        tm %= mm
        if X[i] == 1:
            Sm += tm
            Sm %= mm
        tm *= 2

Sp = 0
mp = one + 1
tp = 1
for i in range(N - 1, -1, -1):
    tp %= mp
    if X[i] == 1:
        Sp += tp
        Sp %= mp
    tp *= 2


memo = [-1] * (2 * (10 ** 5) + 10)
memo[0] = 0

for i in range(N):
    if X[i] == 0:
        m = (Sp + pow(2, N - 1 - i, mp)) % mp
    elif one > 1:
        m = (Sm - pow(2, N - 1 - i, mm)) % mm
    else:
        print(0)
        continue
    print(1 + dfs(m))
Problem:
Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition:    * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = [int(s) for s in input().split(' ')]
counts = [0] * n
for i in a:
  counts[i] += 1
ans = 1 if a[0] == 0 else 0
for u, v in zip(counts, counts[1:]):
  ans = ans * u * v % 998244353
print(ans)

Fixed, accepted submission:
n = int(input())
a = [int(s) for s in input().split(' ')]
counts = [0] * n
for i in a:
  counts[i] += 1
ans = 1 if a[0] == 0 and counts[0] == 1 else 0
for u, v in zip(counts, counts[1:]):
  for _ in range(v):
    ans = ans * u % 998244353
print(ans)
Problem:
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.  To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Incorrect submission (verdict: Wrong Answer):
def haiku(sentence_length_list):

    length_five_cnt=0
    length_seven_cnt=0
    for s in sentence_length_list:
        if s==5:
            length_five_cnt=length_five_cnt+1
        if s==7:
            length_seven_cnt=length_seven_cnt+1

    if length_five_cnt==2 and length_seven_cnt==1:
        return 'YES'
    else:
        return 'NO'

if __name__=='__main__':
    haiku(input().split())

Fixed, accepted submission:
def haiku(sentence_length_list):

    length_five_cnt=0
    length_seven_cnt=0
    for s in sentence_length_list:
        if s=='5':
            length_five_cnt=length_five_cnt+1
        if s=='7':
            length_seven_cnt=length_seven_cnt+1

    if length_five_cnt==2 and length_seven_cnt==1:
        return 'YES'
    else:
        return 'NO'

if __name__=='__main__':
    length_list = input().split()
    print(haiku(length_list))
Problem:
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K.  She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change).  However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money.  Find the amount of money that she will hand to the cashier.

Incorrect submission (verdict: Wrong Answer):
n, k = map(int,input().split())
d = list(map(str,input().split()))

ans=0
for i in range (n,100000):
  s=str(i)
  pay=True
  for j in range(len(s)):
    if s[j] in d:
      pay = False
      break
    
    if pay:
      ans = i
      break
print(ans)
      

Fixed, accepted submission:
n, k = map(int,input().split())
d = list(map(str,input().split()))

ans=0

for i in range (n,100000):
  s= str(i)
  
  pay=True
  
  for j in range (len(s)):
    
    if s[j] in d:
      pay = False
      break
    
    
  if pay:
      ans = i
      break
print(ans)
Problem:
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s).  Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.

Incorrect submission (verdict: Wrong Answer):
S=input()
ans=0
tmp=0
end=0
for i in range(len(S)):
	if S[i]=="A":
		tmp=i
		break
for i in range(1,len(S)):
	if S[-i]=="Z":
		end=len(S)-i
		break
print(tmp,end)
print(end-tmp+1)

Fixed, accepted submission:
S=input()
ans=0
tmp=0
end=0
for i in range(len(S)):
	if S[i]=="A":
		tmp=i
		break
for i in range(1,len(S)):
	if S[-i]=="Z":
		end=len(S)-i
		break
print(end-tmp+1)
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 A<C<B:
    print('YES')
elif B<C<A:
    print('YES')
else:
    print('No')






Fixed, accepted submission:
A, B, C = map(int,input().split(" "))

if A<C<B:
    print('Yes')
elif B<C<A:
    print('Yes')
else:
    print('No')