TnT/CodeNet4Repair huggingface dataset:
Problem:
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi:    * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person.  Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.

Incorrect submission (verdict: Wrong Answer):
*AB, K = map(int, input().split())
turn = 0
for _ in range(K):
    AB[turn] //= 2
    AB[turn ^ 1] += AB[turn]
print(*AB)

Fixed, accepted submission:
*AB, K = map(int, input().split())
for i in range(K):
    AB[i % 2] //= 2
    AB[i % 2 ^ 1] += AB[i % 2]
print(*AB)
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())
D = sorted([int(input()) for _ in range(N)])
i,prev = 0,D[0]
while i != (len(D)-1):
    if D[i] == prev: 
        D.pop(i)
    else:
        i +=1
        prev = D[i]
print(len(D))

Fixed, accepted submission:
N = int(input())
D = sorted([int(input()) for _ in range(N)])
if len(D) > 1:
    i,prev = 0,None
    while i < (len(D)-1):
        if D[i] == D[i+1]: 
            D.pop(i+1)
        else:
            prev = D[i]
            i +=1
print(len(D))
Problem:
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

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

odd = s[1::2]

print(odd)

Fixed, accepted submission:
s = input()

odd = s[0::2]

print(odd)
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):
print(int(input()) % 1000)

Fixed, accepted submission:
x = 1000 - int(input())%1000
print(0 if x == 1000 else x)
Problem:
Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II.    * insert  $k$: Insert a node containing $k$ as key into $T$.   * find $k$: Report whether $T$ has a node containing $k$.    * delete $k$: Delete a node containing $k$.   * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.  The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases:    1. If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$).   2. If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent.   3. If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.

Incorrect submission (verdict: Wrong Answer):

Fixed, accepted submission:
import sys

class Node:
    __slots__ = ['key', 'left', 'right']
    def __init__(self, key):
        self.key = key
        self.left = self.right = None

root = None

def insert(key):
    global root
    x, y = root, None
    while x: x, y = x.left if key < x.key else x.right, x
    if y is None: root = Node(key)
    elif key < y.key: y.left = Node(key)
    else: y.right = Node(key)

def find(target):
    result = root
    while result and target != result.key:
        result = result.left if target < result.key else result.right
    return result is None

def delete(target):
    def remove_node(p, c, a):
        if p.left == c: p.left = a
        else: p.right = a
    p, c = None, root
    while c.key != target: p, c = c, c.left if target < c.key else c.right
    if c.left is None:
        remove_node(p, c, c.right)
    elif c.right is None:
        remove_node(p, c, c.left)
    elif c.right.left is None:
        c.right.left = c.left
        remove_node(p, c, c.right)
    else:
        g = c.right
        while g.left.left: g = g.left
        c.key = g.left.key
        g.left = g.left.right

def inorder(node):
    return inorder(node.left) + f' {node.key}' + inorder(node.right) if node else ''
def preorder(node):
    return f' {node.key}' + preorder(node.left) + preorder(node.right) if node else ''

input()
for e in sys.stdin:
    if e[0] == 'i': insert(int(e[7:]))
    elif e[0] == 'd': delete(int(e[7:]))
    elif e[0] == 'f': print(['yes','no'][find(int(e[5:]))])
    else: print(inorder(root)); print(preorder(root))
Problem:
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

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

print(N / 2 + 1)

Fixed, accepted submission:
from math import floor

N = int(input())

if N % 2 == 0:
  print(round(N / 2 - 1))
else:
  print(floor(N / 2))
Problem:
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i.  Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.

Incorrect submission (verdict: Wrong Answer):
N= int(input())
Alist = list(map(int, input().split())) 

if N % 2 ==0:
    for i in range(0,N//2):
        if Alist.count(2*i+1) != 2:
            print(0)
    print(2**(N//2) % (10**9+7))

elif N % 2 ==1:
    for i in range(0,(N+1)//2):
        if Alist.count(2*i) != 2:
            print(0)
    print(2**((N-1)//2) % (10**9+7))

Fixed, accepted submission:
from collections import Counter

N= int(input())
Alist = list(map(int, input().split())) 

count = Counter(Alist)
res = 1
if N % 2 ==0:
    for i in range(1,N,2):
        if count[i] != 2:
            print(0)
            exit()
else:
    if count[0] != 1:
        print(0)
        exit()
    else:
        for i in range(2,N,2):
            if count[i] != 2:
                print(0)
                exit()
print(2**(N//2) % (10**9+7))
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

        print(' '.join(map(str, A)))

if __name__ == "__main__":
    n = int(input())
    a = [int(e) for e in input().split()]
    insertionSort(a, n)

Fixed, accepted submission:
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

        print(' '.join(map(str, A)))

if __name__ == "__main__":
    n = int(input())
    a = [int(e) for e in input().split()]
    print(' '.join(map(str, a)))
    insertionSort(a, n)
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 = [int(i) for i in input().split()]
b = [0] * n

if (n % 2 == 0) :
  b[n // 2] =a[0]
  b[0] = a[n - 1]
  for i in range(n//2 - 1) :
    b[(n // 2) - i - 1] = a[2 * i + 1]
    b[(n // 2) + i + 1] = a[2 * (i+1)]
  for i in range(n) :
    print(b[i], end = '')
  print()

if (n % 2 == 1) :
  b[n // 2] =a[0]
  for i in range(n//2) :
    b[(n // 2) + i + 1] = a[2 * i + 1]
    b[(n // 2) - i - 1] = a[2 * (i+1)]
  for i in range(n) :
    print(b[i], end = '')
  print()

Fixed, accepted submission:
n = int(input())
a = [int(i) for i in input().split()]
b = [0] * n

if (n % 2 == 0) :
  b[n // 2] =a[0]
  b[0] = a[n - 1]
  for i in range(n//2 - 1) :
    b[(n // 2) - i - 1] = a[2 * i + 1]
    b[(n // 2) + i + 1] = a[2 * (i+1)]
  for i in range(n-1) :
    print(b[i], end = ' ')
  print(b[n-1])

if (n % 2 == 1) :
  b[n // 2] =a[0]
  for i in range(n//2) :
    b[(n // 2) + i + 1] = a[2 * i + 1]
    b[(n // 2) - i - 1] = a[2 * (i+1)]
  for i in range(n-1) :
    print(b[i], end = ' ')
  print(b[n-1])
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())
c = 1
if a > b:
    c = -1

if (c == 1 and a + c*v*t >= b + c*w*t) or (c == -1 and a + c*v*t <= b + c*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 a <= b and a + v*t >= b + w*t:
    print("YES")
elif a > b and a - v*t <= b - w*t:
    print("YES")
else:
    print("NO")
Problem:
It is September 9 in Japan now.  You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

Incorrect submission (verdict: Wrong Answer):
N = str(input())

if N in '9':
  print("Yes")
  
else:
  print('N0')
  

Fixed, accepted submission:
N = str(input())
li = [N[0], N[1]]

if '9' in li :
  print("Yes")
  
else:
  print('No')
Problem:
You have written N problems to hold programming contests. The i-th problem will have a score of P_i points if used in a contest.  With these problems, you would like to hold as many contests as possible under the following condition:    * A contest has three problems. The first problem has a score not greater than A points, the second has a score between A + 1 and B points (inclusive), and the third has a score not less than B + 1 points.  The same problem should not be used in multiple contests. At most how many contests can be held?

Incorrect submission (verdict: Wrong Answer):
import numpy as np

N = int(input())

A, B = [int(i) for i in input().split()]

P = [int(i) for i in input().split()]



def Three_Categorize(P,A,B):
    count_1 = 0
    count_2 = 0
    count_3 = 0
    for i in P:
        if i <= A:
            count_1 +=1
        elif (i > A) & (i < B):
            count_2 +=1
        elif i > B:
            count_3 +=1
    
    return [count_1, count_2, count_3]
            
    
count = Three_Categorize(P,A,B)
ans=1
for i in count:
    ans = ans*i
    
print(ans)

Fixed, accepted submission:
import numpy as np

N = int(input())

A, B = [int(i) for i in input().split()]

P = [int(i) for i in input().split()]



def Three_Categorize(P,A,B):
    count_1 = 0
    count_2 = 0
    count_3 = 0
    for i in P:
        if i <= A:
            count_1 +=1
        elif (i > A) & (i <= B):
            count_2 +=1
        elif i > B:
            count_3 +=1
    
    return [count_1, count_2, count_3]
            
    
count = Three_Categorize(P,A,B)
ans=min(count)
print(ans)
Problem:
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i.  Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

Incorrect submission (verdict: Wrong Answer):
print(sum(map(int,input().split())))

Fixed, accepted submission:
print(input().count("1"))
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):
def resolve():
  N = int(input())
  X = input()
  base_1_count = X.count("1")

  for i in range(N):
    tar_X = list(X)
    first_1_count = base_1_count
    if tar_X[i] == "1":
      first_1_count -= 1
      tar_X[i] = "0"
    else:
      first_1_count += 1
      tar_X[i] = "1"
    target_int = int(''.join(tar_X), 2)
    count = 0
    num_of_1 = first_1_count
    while target_int != 0:
      target_int = target_int % num_of_1
      num_of_1 = bin(target_int).count("1")
      print(bin(target_int))
      count += 1
    print(count)

if __name__ == "__main__":
  resolve()


Fixed, accepted submission:
def resolve():
  N = int(input())
  X = input()

  if N == 1:
    if X.count("1"):
      print(0)
    else:
      print(1)
    return True

  base_1_count = X.count("1")
  if base_1_count == 0:
    for _ in range(N):
      print(1)
    return True

  X_int = int(X, 2)
  X_int_p = X_int%(base_1_count + 1)

  
  if base_1_count == 1:
    for i in range(N):
      if X[i] == "1":
        print(0)
      else:
        if i == N - 1:
          Xi = X_int_p + 1
        else:
          Xi = X_int_p
        count = 1
        while Xi != 0:
          Xi %= bin(Xi).count("1")
          count += 1
        print(count)
    return True

  X_int_m = X_int%(base_1_count - 1)
  for i in range(N):
    
    if X[i] == "1":
      temp_1_count = base_1_count-1
      pow_2 = pow(2, N-i-1, base_1_count-1)
      if X_int_m <= pow_2:
        Xi = X_int_m - pow_2 + base_1_count-1
      else:
        Xi = X_int_m - pow_2
    else:
      temp_1_count = base_1_count+1
      Xi = X_int_p + pow(2, N-i-1, base_1_count+1)
      while Xi >= base_1_count+1:
        Xi %= base_1_count+1

    # print("1_c={0}".format(temp_1_count))
    if temp_1_count == Xi:
      print(1)
      continue

    count = 1
    while Xi != 0:
      num_of_1 = bin(Xi).count("1")
      Xi %= num_of_1
      count += 1
    print(count)

if __name__ == "__main__":
    resolve()
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())

max=max(a,b)

if n>a+b:
  min=0
  
else:
  min=min(a,b)
  
  
print(int(max),int(min))
  

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

max=min(a,b)

if n>a+b:
  min=0
  
else:
  
  min=a+b-n
  
  
print(int(max),int(min))
Problem:
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}.  Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

Incorrect submission (verdict: Wrong Answer):
s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
  print("Yes")
else:
  print("No")

Fixed, accepted submission:
s = input()
t = input()
if s[0] == t[2] and s[1] == t[1] and s[2] == t[0]:
  print("YES")
else:
  print("NO")
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):
x = []
while True:
	if 0 in x:
		break
	x.append(int(input()))
x.pop()
for i, v in enumerate(x):
	print(('case %d: %d') % (i+1, v))

Fixed, accepted submission:
a = []
while True:
	n = input()
	if n == "0":
		break
	a.append(n)

for i in range(len(a)):
	print("Case " + str(i + 1) + ": " + a[i])
Problem:
Joisino is planning to record N TV programs with recorders.  The TV can receive C channels numbered 1 through C.  The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i.  Here, there will never be more than one program that are broadcast on the same channel at the same time.  When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T).  Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.

Incorrect submission (verdict: Wrong Answer):
import sys
import numpy as np
input = sys.stdin.readline


n, c = map(int, input().split())
tv_guide = np.zeros(10**5+1, dtype=np.int)

for _ in range(n):
    s, t, c = map(lambda x: int(x)-1, input().split())
    tv_guide[s:t+1] += 1

print(tv_guide.max())


Fixed, accepted submission:
import sys
input = sys.stdin.readline


n, c = map(int, input().split())
tv_guide = []
for _ in range(n):
    start, end, channel = map(int, input().split())
    tv_guide.append([start, 1, channel-1])
    tv_guide.append([end, 0, channel-1])

tv_guide.sort(key=lambda x:(x[0], -x[1]))

channel_count = [0] * c
channel_set = set()
ans = 0
for time, on, channel in tv_guide:
    if on:
        channel_set.add(channel)
        channel_count[channel] += 1

    else:
        channel_count[channel] -= 1
        if channel_count[channel] == 0:
            channel_set.discard(channel)

    candidate = len(channel_set)
    if ans < candidate:
        ans = candidate

print(ans)
Problem:
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.  Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.

Incorrect submission (verdict: Wrong Answer):
for _ in[0]*int(input()):
 X,Y=' '+input(),' '+input()
 m,n=len(X),len(Y)
 c=[[0]*n for _ in[0]*m]
 print(c)
 for i in range(1,m):
  for j in range(1,n):c[i][j]=(max(c[i-1][j],c[i][j-1]),1+c[i-1][j-1])[X[i]==Y[j]]
 print(c[m-1][n-1])


Fixed, accepted submission:
def m():
 e=input
 a=''
 for _ in[0]*int(e()):
  X,z=e(),[]
  for y in e():
   s=i=0
   for k in z:
    t=X.find(y,s)+1
    if t<1:break
    if t<k:z[i]=t
    s=k;i+=1
   else:
    t=X.find(y,s)+1
    if t:z+=[t]
  a+=f'\n{len(z)}'
 print(a[1:])
if'__main__'==__name__:m()
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):
print("テスト", end="")
print("テスト")

Fixed, accepted submission:
n = int(input())
n %= 1000
n = 1000 - n
if n == 1000:
  n = 0
print(n)
Problem:
You are given an integer sequence of length N, a_1,a_2,...,a_N.  For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing.  After these operations, you select an integer X and count the number of i such that a_i=X.  Maximize this count by making optimal choices.

Incorrect submission (verdict: Wrong Answer):
from collections import Counter
import sys
n = int(input())
a = list(map(int, input().split()))
count_a = list(Counter(a).items())
count_a = sorted(count_a, reverse=True)
result = 1
c = 0
for x, y in count_a:
    if y>=2:
        result *= x
        c += 1
        if y>=4 and c == 1:
            result *= x
            c += 1
        if c == 2:
            print(result)
            sys.exit()
print(0)

Fixed, accepted submission:
from collections import Counter
n = int(input())
a = list(map(int, input().split()))
set_a = list(set(a))

count_a = Counter(a)

result = []
for i in set_a:
    result.append(count_a[i]+count_a[i-1]+count_a[i+1])
print(max(result))
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):
from sys import stdin
from itertools import combinations
from math import factorial
import numpy as np

a,b = [int(x) for x in stdin.readline().rstrip().split()]

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

Fixed, accepted submission:
from sys import stdin
from itertools import combinations
from math import factorial
import numpy as np

a,b = [int(x) for x in stdin.readline().rstrip().split()]

if a*b%2 == 1:
    print("Odd")
else:
    print("Even")
Problem:
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

Incorrect submission (verdict: Wrong Answer):
# B - 1 21


a,b = input().split()

c = int(a + b)

i = 0

while i != -1 and c != i:
    if (i * i) == c:
        print("Yes")
        exit
    i = i + 1

print("No")


Fixed, accepted submission:
# B - 1 21


import sys

a,b = input().split()

c = int(a + b)

i = 0

while i <= 10000:
    if (i * i) == c:
        print("Yes")
        sys.exit()

    i = i + 1

print("No")