TnT/CodeNet4Repair huggingface dataset:
Problem:
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1.  Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode:                 BubbleSort(C) 1 for i = 0 to C.length-1 2     for j = C.length-1 downto i+1 3         if C[j].value < C[j-1].value 4             swap C[j] and C[j-1]                 SelectionSort(C) 1 for i = 0 to C.length-1 2     mini = i 3     for j = i to C.length-1 4         if C[j].value < C[mini].value 5             mini = j 6     swap C[i] and C[mini]   Note that, indices for array elements are based on 0-origin.  For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).

Incorrect submission (verdict: Wrong Answer):
import copy
N = int(input())
A = input().split()
B = copy.copy(A)
boo = 1
while boo:
    boo = 0
    for i in range(N-1):
        if A[i][1] > A[i+1][1]:
            A[i], A[i+1] = A[i+1], A[i]
            boo = 1
print(*A)
print("Stable")
for i in range(N-1):
    mi = i
    for j in range(i,N):
        if B[i][1] > B[j][1]:
            mi = j
    B[i], B[mi] = B[mi], B[i]
if A==B:
    print(*B)
    print("Stable")
else:
    print(*B)
    print("Not stable")



Fixed, accepted submission:
import copy
N = int(input())
A = input().split()
B = copy.copy(A)
boo = 1
while boo:
    boo = 0
    for i in range(N-1):
        if A[i][1] > A[i+1][1]:
            A[i], A[i+1] = A[i+1], A[i]
            boo = 1
print(*A)
print("Stable")
for i in range(N):
    mi = i
    for j in range(i,N):
        if B[mi][1] > B[j][1]:
            mi = j
    B[i], B[mi] = B[mi], B[i]
if A==B:
    print(*B)
    print("Stable")
else:
    print(*B)
    print("Not stable")
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, L = map(str, input().split())
P = list(map(str, input().split()))
U = sorted(list(set([str(i) for i in range(10)])-set(P)))
#print(U)
for i in range(len(N)):
    #print(N[i])
    if N[i] in P:
        print(N[i])
        for j in U:
            print(j)
            if j >=N[i]:
                N=N.replace(N[i],j)
                #print(N)
                break
print(N)

Fixed, accepted submission:
import itertools

N, L = map(str, input().split())
P = list(map(str, input().split()))

U = itertools.product(sorted(list(set([str(i) for i in range(10)])-set(P))),repeat=len(N))
F = itertools.product(sorted(list(set([str(i) for i in range(10)])-set(P))),repeat=len(N)+1)
K = str()
cnt=0
for i in U:
	#print(''.join(i))
	K = ''.join(i)
	if int(K)>=int(N):
		cnt = 1
		print(K)
		break
#print(cnt)
if cnt == 0:
    for j in F:
        K=''.join(j)
        
        if int(K)>=int(N):
            print(K)
            break
Problem:
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.  The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates _A_ , _B_ , and _C_ and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of _A_ , _B_ , and _C_ are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for _A_ , _A_ is ensured to be the winner since _A_ already has five votes and _B_ or _C_ can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.  Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.

Incorrect submission (verdict: Wrong Answer):
def secMax(a):
    mx = 0
    smx = 0
    for i in range(0,len(a)):
        if a[i] > mx:
            smx = mx
            mx = a[i]
        elif a[i] > smx:
            smx = a[i]
    else:
        return smx

while True:
    n = int(input())
    if n == 0:
        break
    c = list(map(str,input().split()))
    
    alph = ["A","B","C","D","E","F","G","H",
            "I","J","K","L","M","N","O","P","Q",
            "R","S","T","U","V","W","X","Y","Z"]
    abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(0, n):
        for j in range(0, 26):
            if c[i] == alph[j]:
                abcCount[j] = abcCount[j] + 1
                break
        print(abcCount)
        if max(abcCount) > n-1-i+secMax(abcCount):
            print(alph[abcCount.index(max(abcCount))], sum(abcCount))
            break
    else:
        print("TIE")



Fixed, accepted submission:
def secMax(a):
    mx = 0
    smx = 0
    for i in range(0,len(a)):
        if a[i] > mx:
            smx = mx
            mx = a[i]
        elif a[i] > smx:
            smx = a[i]
    else:
        return smx

while True:
    n = int(input())
    if n == 0:
        break
    c = list(map(str,input().split()))
    
    alph = ["A","B","C","D","E","F","G","H",
            "I","J","K","L","M","N","O","P","Q",
            "R","S","T","U","V","W","X","Y","Z"]
    abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(0, n):
        for j in range(0, 26):
            if c[i] == alph[j]:
                abcCount[j] = abcCount[j] + 1
                break
        #print(abcCount)
        if max(abcCount) > n-1-i+secMax(abcCount):
            print(alph[abcCount.index(max(abcCount))], sum(abcCount))
            break
    else:
        print("TIE")
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("Yes" if A<=X<=A+B else "No")

Fixed, accepted submission:
A, B, X = map(int, input().split())
print("YES" if A<=X<=A+B else "NO")
Problem:
無限に広い平面の上に、無限に長い直線を数本引くと、この平面はいくつかの領域に分割されます。たとえば、直線を１本引くと、平面は２つの領域に分割されます。同じ数の直線を引いても、引き方によって得られる領域の数は異なります。たとえば、2 本の直線を平行に引けば得られる領域は 3 つになり、互いに垂直に引けば得られる領域は 4 つになります。      n 本の直線を引くことで得られる最大の領域の数を出力するプログラムを作成してください。

Incorrect submission (verdict: Wrong Answer):
while True:
    try:
        a = int(input())
        print(((a * a) + a + 2) / 2)
    except EOFError:
        break

Fixed, accepted submission:
while True:
    try:
        a = int(input())
        print(str(int(((a * a) + a + 2) / 2)))
    except EOFError:
        break
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())
a2 = a
if a % x != 0:
    a2 = (x - (a % x)) + a
b2 = b - (b % x)
print((b2 - a2) / x + 1)

Fixed, accepted submission:
a,b,x = map(int,input().split())
def f(n,x):
    if n == -1:
        return 0
    else:
        return n // x + 1
print(int(f(b,x)-f(a-1,x)))
Problem:
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b.  The i-th operation is as follows:    1. Append a_i to the end of b.   2. Reverse the order of the elements in b.  Find the sequence b obtained after these n operations.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
    if i%2: b.insert(0, a[i])
    else: b.append(a[i])
print(b)


Fixed, accepted submission:
n=int(input())
a=input().split()
b1=[a[2*i] for i in range(0,n//2)]
b2=[a[2*i+1] for i in range(0,n//2)]
if n%2: b=[a[-1]]+list(reversed(b1))+b2
else: b=list(reversed(b2))+b1
print(" ".join(b))
Problem:
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode:                 for i = 1 to A.length-1     key = A[i]     /* insert A[i] into the sorted sequence A[0,...,j-1] */     j = i - 1     while j >= 0 and A[j] > key         A[j+1] = A[j]         j--     A[j+1] = key   Note that, indices for array elements are based on 0-origin.  To illustrate the algorithms, your program should trace intermediate result for each step.

Incorrect submission (verdict: Wrong Answer):
def insertionSort(A, N):
    for i in range(1,N):
        v = A[i]
        j = i - 1
        while j>=0 and A[j]>v:
            A[j+1] = A[j]
            j -= 1
        A[j+1] = v
    return A


N = int(input())
A = [int(x) for x in input().split(' ')]

A = insertionSort(A, N)
print(A)


Fixed, accepted submission:
def showList(A, N):
    for i in range(N-1):
        print(A[i],end=' ')
    print(A[N-1])

def insertionSort(A, N):
    showList(A,N)
    for i in range(1,N):
        v = A[i]
        j = i - 1
        while j>=0 and A[j]>v:
            A[j+1] = A[j]
            j -= 1
        A[j+1] = v
        showList(A,N)


N = int(input())
A = [int(x) for x in input().split(' ')]

A = insertionSort(A, N)
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):
n = int(input())
ans = 0
anss = 0
for i in range(1,n+1):
    t = 0
    while i%2 == 0:
        t +=1
        i = i/2
    if t > ans:
        ans = t
        anss = i*(2**t)
    else:
        pass

print(anss)

Fixed, accepted submission:
n = int(input())
ans = 0
anss = 0
for i in range(1,n+1):
    t = 0
    while i%2 == 0:
        t +=1
        i = i/2
    if t > ans:
        ans = t
        anss = i*(2**t)
    else:
        pass

if n == 1:
    anss = 1
print(int(anss))
Problem:
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied:    * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds.  Find all favorite numbers and print the sum of those.

Incorrect submission (verdict: Wrong Answer):
import math
n = int(input())
ans = 0
for i in range(1,math.floor(math.sqrt(n))):
    if n % i == 0:
        ans += (n//i)-1
print(ans)

Fixed, accepted submission:
import math
n = int(input())
ans = 0
for i in range(1,math.floor(math.sqrt(n))+1):
    if n % i == 0:
        num = (n//i)-1
        if num > 0:
            if n // num == n % num:
                ans += num
print(ans)
Problem:
Write a program which solve a simultaneous equation:      ax + by = c   dx + ey = f      The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-

import sys
import os

for s in sys.stdin:
    a, b, c, d, e, f = list(map(int, s.split()))

    delta = a * e - b * d
    x = 1 / delta * (e * c - b * f)
    y = 1 / delta * (-d * c + a * f)
    print(x, y)

Fixed, accepted submission:
# -*- coding: utf-8 -*-

import sys
import os


for s in sys.stdin:
    a, b, c, d, e, f = list(map(int, s.split()))

    delta = a * e - b * d
    if delta == 0:
        break
    else:
        x = 1 / delta * (e * c - b * f)
        y = 1 / delta * (-d * c + a * f)
        print('{:.3f} {:.3f}'.format(x + 0, y + 0))
Problem:
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.  You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.  Find B_i for each i = 1, 2, ..., N.

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

    c = n // 2
    for i in range(n):
        if i < c:
            print(a[c])
        else:
            print(a[c - 1])


if __name__ == '__main__':
    main()


Fixed, accepted submission:
def main():
    c = int(input()) // 2
    elem = list(map(int, input().split()))
    left, right = sorted(elem)[c - 1:c + 1]
    for val in elem:
        print((left, right)[val <= left])


if __name__ == '__main__':
    main()
Problem:
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.

Incorrect submission (verdict: Wrong Answer):
s=list(input())
if len(set(s))==len(list(s)):
  print('Yes')
  
else:print('No')



Fixed, accepted submission:
s=list(input())
if len(set(s))==len(list(s)):
  print('yes')
  
else:print('no')
Problem:
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second.  He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

Incorrect submission (verdict: Wrong Answer):
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if abs(A-B) >= abs((V - W) * T):
    print("YES")
else:
    print("NO")


Fixed, accepted submission:
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if V > W and abs(A - B) <= abs((W - V) * T):
    print("YES")
else:
    print("NO")
Problem:
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.  Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.  Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.

Incorrect submission (verdict: Wrong Answer):
A,B,C,D = [int(x) for x in input().split()]
if(A+B < C+D):
  print('Left')
elif(A+B == C+D):
  print('Balanced')
else:
  print('Right')

Fixed, accepted submission:
A,B,C,D = [int(x) for x in input().split()]
if(A+B > C+D):
	print('Left')
elif(A+B == C+D):
	print('Balanced')
else:
	print('Right')
Problem:
Write a program which reads three integers, and prints them in ascending order.

Incorrect submission (verdict: Wrong Answer):
x=list(map(int, input().split()))
x.sort
print(x[0],x[1],x[2])

Fixed, accepted submission:
x=list(map(int, input().split()))
x.sort()
print(x[0],x[1],x[2])
Problem:
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes:    * Red boxes, each containing R red balls   * Green boxes, each containing G green balls   * Blue boxes, each containing B blue balls  Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?

Incorrect submission (verdict: Wrong Answer):
R, G, B, N = map(int, input().split())

import numpy as np

r = np.arange(0, N, R)
g = np.arange(0, N, G)
r, g = np.meshgrid(r, g)
b = N - r - g

print(np.sum(((b%B)==0) & (b>=0)))



Fixed, accepted submission:
R, G, B, N = map(int, input().split())

import numpy as np

r = np.arange(0, N+1, R, dtype=np.int64)
g = np.arange(0, N+1, G, dtype=np.int64)
r, g = np.meshgrid(r, g)
b = N - r - g

print(np.sum(((b%B)==0) & (b>=0)))
Problem:
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.    An example of a trapezoid  Find the area of this trapezoid.

Incorrect submission (verdict: Wrong Answer):
a=int(input())
b=int(input())
h=int(input())

print((a+b)*h/2)

Fixed, accepted submission:
a=int(input())
b=int(input())
h=int(input())
 
c=(a+b)*h/2
print(int(c))
Problem:
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i.  Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen.  Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.)  At most how much can he earn on balance?  Note that the same string may be written on multiple cards.

Incorrect submission (verdict: Wrong Answer):
from collections import defaultdict
d = defaultdict(int)

n=int(input())
for _ in range(n):
  d[input()]+=1

n=int(input())
for _ in range(n):
  d[input()]-=1
  
print(d[max(d)])

Fixed, accepted submission:
from collections import defaultdict
d = defaultdict(int)
d["D"]=0

n=int(input())
for _ in range(n):
  d[input()]+=1

n=int(input())
for _ in range(n):
  d[input()]-=1
  
print(max(d.values()))
Problem:
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately.  He has very high awareness of safety, and decides to buy two bells, one for each hand.  The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.

Incorrect submission (verdict: Wrong Answer):
lines = input().split()
a = int(lines[0])
b = int(lines[1])
c = int(lines[2])
sum1 = a+b
sum2 = a+c
sum3 = b+c
sum_list = [sum1, sum2,sum3]
print(max(sum_list))


Fixed, accepted submission:
lines = list(input().split())
a = int(lines[0])
b = int(lines[1])
c = int(lines[2])
sum_list = [a+b, a+c, b+c]
print(min(sum_list))
Problem:
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information:  If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest.  Given this information, determine whether TopCoDeer is honest.

Incorrect submission (verdict: Wrong Answer):
a=input();print("DH"[a[0]==a[1]])

Fixed, accepted submission:
a=input();print("DH"[a[0]==a[2]])
Problem:
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N).  For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.  We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:    * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.  Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Incorrect submission (verdict: Wrong Answer):
# coding: utf-8
# Your code here!
import numpy as np

N = int(input())

A = list(map(int,input().split()))
B = A
ct = 0

for i in range(N):
    j = N-i-1
    if(j+1>N/2):
        B[j] = A[j]
        if(B[j]==1):
            ct = ct + 1
    else:
        for k in range(j+1,N):
            sum = 0
            if(k%(j+1)==0 and k!=j):
                sum = sum + B[k];
        if(sum%2==0):
            B[j] = A[j]
            if(B[j]==1):
                ct = ct + 1
        else:
            B[j] = 1 - A[j]
            if(B[j]==1):
                ct = ct + 1

print(ct)
C = []

for i in range(N):
    if(B[i]==1):
        C.append(i)
print(*C)



Fixed, accepted submission:
# coding: utf-8
# Your code here!
import numpy as np

N = int(input())

A = list(map(int,input().split()))
B = A
ct = 0

for i in range(N):
    j = N-i-1
    if(j+1>N/2):
        B[j] = A[j]
        
        if(B[j]==1):
            ct = ct + 1
    else:
        sum = 0
        for k in range(j+1,N+1,j+1):
            if(k%(j+1)==0 and k!=j+1):
                
                sum = sum + B[k-1]
        if(sum%2==0):
            B[j] = A[j]
            if(B[j]==1):
                ct = ct + 1
        else:
            B[j] = 1 - A[j]
            
            if(B[j]==1):
                ct = ct + 1
        

print(ct)
C = []

for i in range(N):
    if(B[i]==1):
        C.append(i+1)
print(*C)
Problem:
Rng is preparing a problem set for a qualification round of CODEFESTIVAL.  He has N candidates of problems. The difficulty of the i-th candidate is D_i.  There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems.  Determine whether Rng can complete the problem set without creating new candidates of problems.

Incorrect submission (verdict: Wrong Answer):
import sys
from collections import Counter
N = int(input())
D = Counter([int(c) for c in input().split()])
M = int(input())
T = Counter([int(c) for c in input().split()])
for t in T.items():
  if D[t[0]] < t[1]:
    print('No')
    sys.exit()
print('Yes')

Fixed, accepted submission:
import sys
from collections import Counter
N = int(input())
D = Counter([int(c) for c in input().split()])
M = int(input())
T = Counter([int(c) for c in input().split()])
for t in T.items():
  if D[t[0]] < t[1]:
    print('NO')
    sys.exit()
print('YES')
Problem:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

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

sum = 0
for i in range(int(a)):
    sum += i

print(a)

Fixed, accepted submission:
a = input()

sum = 0
for i in range(int(a)):
    sum += i + 1

print(sum)