TnT/CodeNet4Repair huggingface dataset:
Problem:
Snuke has N hats. The i-th hat has an integer a_i written on it.  There are N camels standing in a circle. Snuke will put one of his hats on each of these camels.  If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`.    * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.  What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int, input().split()))
if n%3!=0:
    print("No")
else:
    bina = []
    maxketa=0
    for i in range(len(a)):
        bina.append(format(a[i], 'b'))
        if maxketa<len(bina[i]):
            maxketa=len(bina[i])
    flag=1
    for i in range(maxketa):
        onecount=0
        for j in range(len(bina)):
            k = i-maxketa+len(bina[j])
            print(k)
            if k>=0:
                if bina[j][k]=="1":
                    onecount+=1
        print(onecount)
        if len(bina)*2/3!=onecount:
            if onecount!=0:
                flag=0
                print("No")
                break
       
    if flag==1:
        print("Yes")
        

Fixed, accepted submission:
N=int(input())
A=list(map(int,input().split()))
if A.count(0)==N:
    print("Yes")
elif N%3!=0:
    print("No")
elif len(set(A))==2:
    if 0 in set(A):
        if A.count(0)==N/3:
            print("Yes")
        else:
            print("No")
    else:
        print("No")
elif len(set(A))==3:
    xor=0
    for a in set(A):
        if A.count(a)==N/3:
            xor^=a
        else:
            print("No")
            exit()
    if xor==0:
        print("Yes")
    else:
        print("No")
else:
    print("No")
Problem:
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions:    * Question 1: Are you subscribing to Newspaper X?   * Question 2: Are you subscribing to Newspaper Y?  As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2.  What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y?  Write a program to answer this question.

Incorrect submission (verdict: Wrong Answer):
n,a,b=map(int,input().split());print(min(a,b),min(0,a+b-n))

Fixed, accepted submission:
n,a,b=map(int,input().split());print(min(a,b),max(0,a+b-n))
Problem:
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

Incorrect submission (verdict: Wrong Answer):
A, B, C = [int(n) for n in input().split()]

if C >= A and C <= B:
    print('YES')
else:
    print('NO')

Fixed, accepted submission:
A, B, C = [int(n) for n in input().split()]

if C >= A and C <= B:
    print('Yes')
else:
    print('No')
Problem:
The goal of the matrix-chain multiplication problem is to find the most efficient way to multiply given $n$ matrices $M_1, M_2, M_3,...,M_n$.  Write a program which reads dimensions of $M_i$, and finds the minimum number of scalar multiplications to compute the maxrix-chain multiplication $M_1M_2...M_n$.

Incorrect submission (verdict: Wrong Answer):
import sys

def get_min_cost(n, p):
	m = [[0 for x in range(n+1)] for x in range(n+1)]
	for s in range(n):
		i = 1
		for j in range(s+2, n):
			m[i][j] = sys.maxsize
			for k in range(i, j+1):
				m[i][j] = min(m[i][j], m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j])
			i+=1	
	return m[1][n]

n = int(input())
p = [0] * (n+1)
j = 0
for i in range(n):
	a = list(map(int, input().split()))
	if i == 0:
		p[j] = a[0]
		j += 1
	p[j] = a[1]
	j += 1
print(get_min_cost(n, p))
		
		


Fixed, accepted submission:

def mcm(p,n):
	m=[[0 for x in range(n+1)] for x in range(n+1)]
	for s in range(n):
		for i,j in zip(range(1,n+1),range(s+2,n+1)):
			m[i][j] = 1 << 30
			for k in range(i,j):
				m[i][j]=min(m[i][j],m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j])
	return m[1][n]

n=int(input())
p=[0]*(n+1)
for i in range(n-1):
	t=list(map(int,input().split()))
	p[i]=t[0];
t=list(map(int,input().split()))
p[n-1]=t[0]
p[n]=t[1]

ans=mcm(p,n)
print(ans)
Problem:
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.  According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

Incorrect submission (verdict: Wrong Answer):
NY = input().split()
N_Y = [int(s)for s in NY]

counter_Y = 0
counter = 0
listx = []
for i in range(N_Y[0]):
    for j in range(N_Y[0]):
        for k in range(N_Y[0]):
            counter_Y = 10000 * i + 5000 * j + 1000 * k
            if counter_Y == N_Y[1] and i+j+k <= N_Y[0]:
                listx = [i,j,k]
                counter += 1

print(counter)
if counter == 0:
    print("-1 -1 -1")
else:
    print("{} {} {}".format(listx[0],listx[1],listx[2]))

Fixed, accepted submission:
NY = input().split()
N_Y = [int(s)for s in NY]

counter_Y = 0
counter = 0
listx = []
for i in range(N_Y[0]+1):
    for j in range(N_Y[0]+1):
        k = N_Y[0] - i - j
        counter_Y = 10000 * i + 5000 * j + 1000 * k
        if counter_Y == N_Y[1] and k >= 0:
            listx = [i,j,k]
            counter += 1

if counter == 0:
    print("-1 -1 -1")
else:
    print("{} {} {}".format(listx[0],listx[1],listx[2]))
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):
x, y, z = map(int, input().split())
print('Yes') if x <= y and y <= z else print('No')

Fixed, accepted submission:
x, y, z = map(int, input().split())
print('Yes') if (x - z) * (y - z) < 0 else print('No')
Problem:
There is data on sales of your company. Your task is to write a program which identifies good workers.  The program should read a list of data where each item includes the employee ID _i_ , the amount of sales _q_ and the corresponding unit price _p_. Then, the program should print IDs of employees whose total sales proceeds (i.e. sum of p × q) is greater than or equal to 1,000,000 in the order of inputting. If there is no such employees, the program should print "NA". You can suppose that _n_ < 4000, and each employee has an unique ID. The unit price _p_ is less than or equal to 1,000,000 and the amount of sales _q_ is less than or equal to 100,000.

Incorrect submission (verdict: Wrong Answer):

if __name__ == '__main__':

    list = []

    while (True):
        n = int(input())
        if (n == 0):
            break

        isNA = True

        for i in range(n):
            ds = input().split()
            e = ds[0]
            p = int(ds[1])
            q = int(ds[2])

            sales = p * q

            if (1000000 <= sales):
                list.append(e)
                isNA = False

        if isNA:
            list.append("NA")

    for l in list:
        print("%s" % l)

Fixed, accepted submission:
while True:
    no = {}
    n = int(input())
    if (n == 0):
        break
    
    for i in range(n):
        datas = map(int, input().split())
        li = list(datas) 
        e = li[0]
        p = li[1]
        q = li[2]

        s = p * q
        if (e not in no):
            no[e] = s
        else:
            no[e] += s

    na = True    
    for k in no.keys():
        if (1000000 <= no[k]):
            print(k)
            na = False
    
    if (na == True):
        print("NA")
Problem:
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_.  When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC.  What is the abbreviation for the N-th round of ABC? Write a program to output the answer.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: <encoding name> -*-

n = int(input())
print("ABC" + 'n')

Fixed, accepted submission:
# -*- coding: <encoding name> -*-

n = input()
print("ABC" + n)
Problem:
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.  He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.  Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Incorrect submission (verdict: Wrong Answer):
from collections import Counter

N, K = map(int, input().split())
A = [int(x) for x in input().split()]

lenght = len(set(A))-K
if lenght<0:print(0)
else:print(len(Counter(A).most_common()[::-1][lenght:]))




Fixed, accepted submission:
from collections import Counter

N, K = map(int, input().split())
A = [int(x) for x in input().split()]

lenght = len(set(A))-K
ans = 0
if lenght<=0:print(ans), exit()
else:
    for k, v in Counter(A).most_common()[::-1][:lenght]:
        ans += v
print(ans)
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())
for i in range(7):
  if 2**i < N and 2**i > N:
    print(2**i)
    break


Fixed, accepted submission:
N = int(input())
for i in range(7):
  if 2**i <= N and 2**(i+1) > N:
    print(2**i)
    break
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

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

if a * b / 2 == 0:
           print("Even")
else:
           print("Odd")

Fixed, accepted submission:
a, b = map(int, input().split())
if a*b % 2 == 0:
    print("Even")
else:
    print("Odd")
Problem:
Snuke is giving cookies to his three goats.  He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).  Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Incorrect submission (verdict: Wrong Answer):
a,b=map(int,input().split())
print('Possible' if a%3==0 or b%3==0 or a+b %3==0 else 'Impossible')

Fixed, accepted submission:
a,b=map(int,input().split())
print('Possible' if a%3==0 or b%3==0 or (a+b )%3==0 else 'Impossible')
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(At, Nt):
	for i in range(Nt):
		v = At[i]
		j = i - 1
		while j >= 0 and At[j] > v:
			A[j + 1] = A[j]
			j = j - 1
		A[j + 1] = v
		print(At)
	return At

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

Fixed, accepted submission:
def insertionSort(At, Nt):
	for i in range(Nt):
		v = At[i]
		j = i - 1
		while j >= 0 and At[j] > v:
			A[j + 1] = A[j]
			j = j - 1
		A[j + 1] = v
		print(" ".join(map(str,At)))
	return At

if __name__ == '__main__':
	N = int(input())
	A = list(map(int, input().split()))
	A = insertionSort(A, N)
	#print(" ".join(map(str,A)))
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):
a = int(input())
i = 1
while(i<=a):
	i *= 2
print(i/2)


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

breakNum = 1

while breakNum*2 <= n:
	breakNum *= 2

print(breakNum)
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):
N = int(input())

x,y,z = 1,1,1

def is_true(x,y,z,i):
    if x**2 + y**2 + z**2 + x*y + y*z + z*x == i:
        return True
    return False

loop = int(N ** (1/2))
for i in range(N):
    count = 0

    if i < 6:
        print(count)
        continue
    for a in range(loop):
        for b in range(loop):
            for c in range(loop):
                X = x+a
                Y = y+b
                Z = z+c
                if is_true(X,Y,Z,i):
                    count += 1
    print(count)


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

ans = [0 for _ in range(10050)]

for x in range(1,100):
    for y in range(1,100):
        for z in range(1,100):
            a = x**2 + y**2 + z**2 + x*y + y*z + z*x
            if a < 10050:
                ans[a] += 1
for i in range(1,N+1):
    print(ans[i])
Problem:
We will buy a product for N yen (the currency of Japan) at a shop.  If we use only 1000-yen bills to pay the price, how much change will we receive?  Assume we use the minimum number of bills required.

Incorrect submission (verdict: Wrong Answer):
i = int(input())
print(i%1000)


Fixed, accepted submission:
i = int(input())
if i%1000 == 0:
    print(0)
else:
    print(1000 - (i%1000))
Problem:
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets.  Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

Incorrect submission (verdict: Wrong Answer):
#Repetitive Processing - Print Test Cases

x=list(map(int,input().split()))
for i in range(len(x)-1):
    print("Case " +str(i)+": "+str(x[i]))


Fixed, accepted submission:
### Repetitive Processing - Print Test Cases

i = 0
while True:
    x = int(input());
    i += 1
    if x == 0: break
    print("Case {}: {}".format(i, x))
Problem:
We have a tree with N vertices. The vertices are numbered 1, 2, ..., N. The i-th (1 ≦ i ≦ N - 1) edge connects the two vertices A_i and B_i.  Takahashi wrote integers into K of the vertices. Specifically, for each 1 ≦ j ≦ K, he wrote the integer P_j into vertex V_j. The remaining vertices are left empty. After that, he got tired and fell asleep.  Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:    * Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.  Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Incorrect submission (verdict: Wrong Answer):
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)

n = int(input())
T = [[] for _ in range(n+1)]
for _ in range(n-1):
  a, b = map(int, input().split())
  T[a].append(b)
  T[b].append(a)
k = int(input())
VP = tuple(tuple(map(int, input().split())) for _ in range(k))
P = [-1]*(n+1)
A = [-1]*(n+1)
r = VP[0][0]
for v, p in VP:
  P[v] = p
C = [None]*(n+1)
def dfs1(v, par=-1):
  if P[v] == -1:
    l = -float("inf")
    r = float("inf")
  else:
    l = P[v]
    r = P[v]
  for nv in T[v]:
    if nv == par:
      continue
    nl, nr = dfs1(nv, v)
    l = max(l, nl-1)
    r = min(r, nr+1)
  if l > r:
    print("No")
    exit()
  P[v] = (l, r)
  return P[v]
dfs1(r)
def dfs2(v, ini, par=-1):
  l, r = P[v]
  if l <= ini-1 <= r:
    A[v] = ini-1
  else:
    A[v] = ini+1
  for nv in T[v]:
    if nv == par:
      continue
    dfs2(nv, A[v], v)
dfs2(r, P[r][0])
print("Yes")
print(*A[1:], sep="\n")

Fixed, accepted submission:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)

INF = 10**10
n = int(input())
T = [[] for _ in range(n+1)]
for _ in range(n-1):
  a, b = map(int, input().split())
  T[a].append(b)
  T[b].append(a)
k = int(input())
VP = tuple(tuple(map(int, input().split())) for _ in range(k))
P = [-1]*(n+1)
A = [-1]*(n+1)
r, c = VP[0]
for v, p in VP:
  P[v] = p
def dfs1(v, par=-1, color=c%2):
  if P[v] == -1:
    l = -INF
    r = INF
  else:
    l = P[v]
    r = P[v]
  for nv in T[v]:
    if nv == par:
      continue
    nl, nr = dfs1(nv, v, color^1)
    l = max(l, nl-1)
    r = min(r, nr+1)
  if l&1 ^ color:
    l += 1
  if r&1 ^ color:
    r -= 1
  if l > r:
    print("No")
    exit()
  P[v] = (l, r)
  return P[v]
dfs1(r)
def dfs2(v, ini, par=-1):
  l, r = P[v]
  if l == r:
    A[v] = l
  else:
    if l <= ini-1 <= r:
      A[v] = ini-1
    else:
      A[v] = ini+1
  for nv in T[v]:
    if nv == par:
      continue
    dfs2(nv, A[v], v)
dfs2(r, P[r][0])
print("Yes")
print(*A[1:], sep="\n")
Problem:
Snuke has a favorite restaurant.  The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.  So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.

Incorrect submission (verdict: Wrong Answer):
from math import factorial
print(factorial(int(input()))%(10**9+7))

Fixed, accepted submission:
x=int(input());print(x*800-(x//15)*200)
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):
# C
ABCD = input()
A = ABCD[0]
B = ABCD[1]
C = ABCD[2]
D = ABCD[3]


for i in range(8):
    res = A
    if i in [0, 2, 4, 6]:
        res = res + '+' + B
    else:
        res = res + '-' + B
    if i in [0, 1, 4, 5]:
        res = res + '+' + C
    else:
        res = res + '-' + C
    if i in [0, 1, 2, 3]:
        res = res + '+' + D
    else:
        res = res + '-' + D
        
    if eval(res) == 7:
        print(res)
        break



Fixed, accepted submission:
# C
ABCD = input()
A = ABCD[0]
B = ABCD[1]
C = ABCD[2]
D = ABCD[3]


for i in range(8):
    res = A
    if i in [0, 2, 4, 6]:
        res = res + '+' + B
    else:
        res = res + '-' + B
    if i in [0, 1, 4, 5]:
        res = res + '+' + C
    else:
        res = res + '-' + C
    if i in [0, 1, 2, 3]:
        res = res + '+' + D
    else:
        res = res + '-' + D
        
    if eval(res) == 7:
        print(res+'=7')
        break
Problem:
There is a hotel with the following accommodation fee:    * X yen (the currency of Japan) per night, for the first K nights   * Y yen per night, for the (K+1)-th and subsequent nights  Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

Incorrect submission (verdict: Wrong Answer):
N = [int(input()) for i in range(4)]
if N[0] >= N[1]:
  for i in range(N[0]):
    print(str(i+1) + '泊目は' + str(N[2]) + '円です')
else:
  for i in range(N[0]):
    print(str(i+1) + '泊目は' + str(N[2]) + '円です')
  for i in range(N[1] - N[0]):
    print(str(N[0] + i+1) + '泊目は' + str(N[3]) + '円です')

Fixed, accepted submission:
N = [int(input()) for i in range(4)]
if N[0] <= N[1]:
    print(N[2] * N[0])
else:
    print(N[2] * N[1] + N[3] * (N[0] - N[1]))
Problem:
You have final scores of an examination for n students. Calculate standard deviation of the scores s1, s2 ... sn.  The variance α2 is defined by  α2 = (∑n _i_ =1(s _i_ \- m)2)/n  where m is an average of si. The standard deviation of the scores is the square root of their variance.

Incorrect submission (verdict: Wrong Answer):
import math
while True:
    n=int(input())
    a=[int(x) for x in input().split()]
    if a==[0]:
        break
    else:
        m=sum(a)/n
        print("{0:.5f}".format(math.sqrt(sum([(x-m)**2/n for x in a]))))
 

Fixed, accepted submission:
import math
while True:
    n = int(input())
    if n==0: break
    s = list(map(int,input().split()))
    m = sum(s)/float(n)
    print("{0:.8f}".format(math.sqrt(sum([(x-m)**2 for x in s])/n)))
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):
l = list(map(int, input().split()))
if l.count(5) == 2 and l.count(7) == 1:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
l = sorted(list(map(int, input().split())))

if l.count(5) == 2 and l.count(7) == 1:
    print("YES")
else:
    print("NO")
Problem:
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.  Given an integer N, determine whether it is a Harshad number.

Incorrect submission (verdict: Time Limit Exceeded):
def f(n):
  ans = 0
  while n > 0:
    ans += n % 10
    n // 10
  return ans

n = int(input())
if n % f(n) == 0:
  print('Yes')
else:
  print('No')

Fixed, accepted submission:
n = input()
ni = int(n)
f = 0
for ns in n:
  f += int(ns)
if ni % f == 0:
  print('Yes')
else:
  print('No')