TnT/CodeNet4Repair huggingface dataset:
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())
  X = [int(_) for _ in input().split()]
  Y = sorted(X)
  mid = (Y[(N-1)//2] + Y[N//2]) / 2
  for i in range(N):
    print(Y[N//2] if X[i] <= mid else print(Y[(N-1)//2]))
  return

if __name__ == '__main__':
  main()
  

Fixed, accepted submission:
def main():
  N = int(input())
  X = [int(_) for _ in input().split()]
  Y = sorted(X)
  mid = (Y[(N-1)//2] + Y[N//2]) / 2
  for i in range(N):
    print(Y[N//2] if X[i] <= mid else Y[(N-1)//2])
  return

if __name__ == '__main__':
  main()
Problem:
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.

Incorrect submission (verdict: Wrong Answer):
S=input().split()
print(S[0][0]+S[1][0]+S[2][0])

Fixed, accepted submission:
S=input().split()
print(str.upper(S[0][0]+S[1][0]+S[2][0]))
Problem:
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.   The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.  An edge whose removal disconnects the graph is called a _bridge_.   Find the number of the edges that are bridges among the M edges.

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split())
edges = [list(map(int, input().split())) for i in range(m)]
count = 0
print(n,m,edges)

for x in edges:
    l = list(range(n))
    print(l)
    for y in edges:
        if y != x:
            l = [l[y[0]-1] if l[i] == l[y[1]-1] else l[i] for i in range(n)]
            print(l)
    if len(set(l)) != 1:
        count += 1

print(count)

Fixed, accepted submission:
n, m = map(int, input().split())
edges = [list(map(int, input().split())) for i in range(m)]
count = 0
 
for x in edges:
    l = list(range(n))
    for y in edges:
        if y != x:
            l = [l[y[0]-1] if l[i] == l[y[1]-1] else l[i] for i in range(n)]
    if len(set(l)) != 1:
        count += 1
 
print(count)
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):
def main():
    N = int(input())
    ans = 0
    if N < 1000:
        ans = 1
    else:
        ans = int(N / 1000)
        if N % 1000 > 0:
            ans += 1

    print(f'{ans}')


if __name__ == '__main__':
    main()

Fixed, accepted submission:
def main():
    N = int(input())
    remainder = N % 1000
    quotient = int(N / 1000)
    ans = 0
    if N < 1000:
        ans = 1000 - N
    elif remainder == 0:
        ans = 0
    elif remainder > 0:
        ans = 1000 - remainder

    print(f'{ans}')


if __name__ == '__main__':
    main()
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()
ans = 0
for i in range(n):
    if s[i] == 'I':
        ans += 1
    else:
        ans -= 1
print(ans)
    

Fixed, accepted submission:
n = int(input())
s = input()
ans = []
if s[0] == 'I':
    ans.append(1)
else:
    ans.append(-1)
for i in range(1,n):
    if s[i] == 'I':
        ans.append(ans[i-1] +1)
    else:
        ans.append(ans[i-1] -1)
if max(ans) < 0:
    print('0')
else:
    print(max(ans))
Problem:
Snuke has an empty sequence a.  He will perform N operations on this sequence.  In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).  You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Incorrect submission (verdict: Wrong Answer):
import sys

# import re
import math
import collections

import bisect
# import itertools
import fractions
# import functools
import copy
import heapq
import decimal
# import statistics
import queue

sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7

ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))


# ===CODE===


def main():
    n = ni()
    a = na()

    for i, ai in enumerate(a):
        if ai > i + 1:
            print(-1)
            exit(0)
        else:
            print(ai)


if __name__ == '__main__':
    main()


Fixed, accepted submission:
import sys

# import re
import math
import collections

import bisect
# import itertools
import fractions
# import functools
import copy
import heapq
import decimal
# import statistics
import queue

sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7

ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))


# ===CODE===


def main():
    n = ni()
    a = na()

    ans = []
    while len(a) > 0:
        flg = True
        for i in range(len(a) - 1, -1, -1):
            if i + 1 == a[i]:
                ans.append(i + 1)
                del a[i]
                flg = False
                break
        if flg:
            print(-1)
            exit(0)

    for i in range(n-1,-1,-1) :
        print(ans[i])


if __name__ == '__main__':
    main()
Problem:
You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen.  Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .

Incorrect submission (verdict: Wrong Answer):
from collections import defaultdict
import math




if __name__ == '__main__' :
    n = int(input())
    R = [int(input()) for i in range(n)]
    tmp = [max(R[i+1::])  - R[i] for i in range(n-1)]


Fixed, accepted submission:
from collections import defaultdict
import math




if __name__ == '__main__' :
    n = int(input())
    R = [int(input()) for i in range(n)]
    maxv = R[1]-R[0]
    minv = min(R[0],R[1])
    for j in range(2,n):
        maxv = max(maxv,R[j]-minv)
        minv = min(minv,R[j])
    print(maxv)
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
input()
D = map(int, input().split())
input()
T = map(int, input().split())

CD = Counter(D)
CT = Counter(T)
for x in CD:
  c = CT.get(x)
  if not c is None and CD[x] <= c:
    pass
  else:
    break
else:
  print("YES")
  sys.exit()
print("NO")

    

Fixed, accepted submission:
import sys
from collections import Counter
input()
D = map(int, input().split())
input()
T = map(int, input().split())

CD = Counter(D)
CT = Counter(T)
for x in CT:
  c = CD.get(x)
  if not c is None and CT[x] <= c:
    pass
  else:
    break
else:
  print("YES")
  sys.exit()
print("NO")
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):
x = int(input())
while x<=0:
  x -=1000
print(abs(x))

Fixed, accepted submission:
x = int(input())
while x > 0:
    x -= 1000
print(abs(x))
Problem:
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops.  Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto.  Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements.                 void call(int n){   int i = 1;  CHECK_NUM:   int x = i;   if ( x % 3 == 0 ){     cout << " " << i;     goto END_CHECK_NUM;   }  INCLUDE3:   if ( x % 10 == 3 ){     cout << " " << i;     goto END_CHECK_NUM;   }   x /= 10;   if ( x ) goto INCLUDE3;  END_CHECK_NUM:   if ( ++i <= n ) goto CHECK_NUM;    cout << endl; }

Incorrect submission (verdict: Wrong Answer):
n = int(input())
for x in range(3, n+1, 3):
    print(x, end=' ')



Fixed, accepted submission:
n = int(input())
result = ''
for x in range(3, n+1):
    if x % 3 == 0 or '3' in str(x):
        result += (' '+str(x))
print(result)
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):
n,y=map(int, input().split())
for i in range(y//10000+1):
    for j in range(y//5000+1):
        if 10000*i + 5000*j +1000*(n-i-j) == y:
            print(i,j,n-i-j)
            break
    else:
        continue
    break

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

ans=[-1,-1,-1]
for i in range(n+1):
    for j in range(n+1-i):
        if y-i*10000-j*5000==1000*(n-i-j):
            ans=[i,j,n-i-j]
            
print(*ans)
Problem:
We have a grid with N rows and M columns of squares. Initially, all the squares are white.  There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted.  Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Incorrect submission (verdict: Wrong Answer):
import sys

n,m,k = map(int, input().split())

for a in range(n):
	if n-2*a == 0:
		continue
	b = ((k-a*m))/(n-2*a)
	if (b >= 0) and (b <= m) and (b%1 == 0):
		print("YES")
		sys.exit()

print("NO")

Fixed, accepted submission:
import sys

n,m,k = map(int, input().split())

for a in range(n):
	if n-2*a == 0:
		continue
	b = ((k-a*m))/(n-2*a)
	if (b >= 0) and (b <= m) and (b%1 == 0):
		print("Yes")
		sys.exit()

print("No")
Problem:
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?

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

import math

if math.sqrt(a)+math.sqrt(b)>math.sqrt(c):
  print('Yes')
else:
  print('No')

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

import math

if (a+b)>=c:
  print('No')
elif 4*a*b<c**2-2*c*(a+b)+(a+b)**2:
  print('Yes')
else:
  print('No')
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())

if a%3 == 0 or b %3 == 0 or (a + b)%3 == 0:
    print("Yes")
else:
    print("No")

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

if a%3 == 0 or b %3 == 0 or (a + b)%3 == 0:
    print("Possible")
else:
    print("Impossible")
Problem:
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements?  Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:    * r_1 + c_1 = r_2 + c_2   * r_1 - c_1 = r_2 - c_2  For example, in the following figure, the bishop can move to any of the red squares in one move:  

Incorrect submission (verdict: Wrong Answer):
H, W = map(int, input().split())

x = H * W

if x % 2 == 0:
    print(x/2)
else:
    print(x//2 + 1)


Fixed, accepted submission:
H, W = map(int, input().split())

x = H * W

if H == 1 or W == 1:
    print(1)
else:
    if x % 2 == 0:
        print(int(x/2))
    else:
        print(int(x//2 + 1))
Problem:
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces.  Find the number of chocolate pieces prepared at the beginning of the camp.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
D,X=map(int,input().split())
print(1+sum((D-1)//int(input())for _ in[0]*N)+X)

Fixed, accepted submission:
N=int(input())
D,X=map(int,input().split())
print(sum(1+(D-1)//int(input())for _ in[0]*N)+X)
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):
N = int(input())
X = [int(i) for i in input().split()]
Y = sorted(X)
a = Y[-1 + N // 2]
b = Y[N // 2]
print(a, b)
for Xi in X:
    if Xi <= a:
        print(b)
    else:
        print(a)


Fixed, accepted submission:
N = int(input())
X = [int(i) for i in input().split()]
Y = sorted(X)
a = Y[-1 + N // 2]
b = Y[N // 2]
for Xi in X:
    if Xi <= a:
        print(b)
    else:
        print(a)
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
def main():
    list = [int(x) for x in input().split()]

    sum = list[0] + list[1] + list[2]
    max = list[0]
    if max <= list[1]:
        max = list[1]
    if max <= list[2]:
        max = list[2]

    if max * 2 == sum:
        print("YES")
    else:
        print("NO")

if __name__ == "__main__":
    main()


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

    if a == b + c or  b == c + a or c == a + b:
        print("Yes")
    else:
        print("No")

if __name__ == "__main__":
    main()
Problem:
You start with the number 0 and you want to reach the number N.  You can change the number, paying a certain amount of coins, with the following operations:    * Multiply the number by 2, paying A coins.   * Multiply the number by 3, paying B coins.   * Multiply the number by 5, paying C coins.   * Increase or decrease the number by 1, paying D coins.  You can perform these operations in arbitrary order and an arbitrary number of times.  What is the minimum number of coins you need to reach N?  **You have to solve T testcases.**

Incorrect submission (verdict: Wrong Answer):
import sys
sys.setrecursionlimit(2147483647)
from functools import lru_cache

@lru_cache(maxsize=None)
def solve(n, a, b, c, d):
    if n == 0:
        return 0
    if n == 1:
        return d
    res = 10**18
    res = min(res, d * n)
    res = min(res, d * (n - n // 2 * 2) + a + solve(n // 2, a, b, c, d))
    res = min(res, d * (n - n // 3 * 3) + b + solve(n // 3, a, b, c, d))
    res = min(res, d * (n - n // 5 * 5) + c + solve(n // 5, a, b, c, d))
    res = min(res, d * (n - (n + 2 - 1) // 2 * 2 + a + solve((n + 2 - 1) // 2, a, b, c, d)))
    res = min(res, d * (n - (n + 3 - 1) // 3 * 3 + b + solve((n + 3 - 1) // 3, a, b, c, d)))
    res = min(res, d * (n - (n + 5 - 1) // 5 * 5 + c + solve((n + 5 - 1) // 5, a, b, c, d)))
    return res

T = int(input())
for _ in range(T):
    N, A, B, C, D = map(int, input().split())
    print(solve(N, A, B, C, D))

Fixed, accepted submission:
import sys
sys.setrecursionlimit(2147483647)
from functools import lru_cache

@lru_cache(maxsize=None)
def solve(n, a, b, c, d):
    if n == 0:
        return 0
    if n == 1:
        return d
    res = 10**18
    res = min(res, d * n)
    res = min(res, d * (n - n // 2 * 2) + a + solve(n // 2, a, b, c, d))
    res = min(res, d * (n - n // 3 * 3) + b + solve(n // 3, a, b, c, d))
    res = min(res, d * (n - n // 5 * 5) + c + solve(n // 5, a, b, c, d))
    res = min(res, d * ((n + 2 - 1) // 2 * 2 - n) + a + solve((n + 2 - 1) // 2, a, b, c, d))
    res = min(res, d * ((n + 3 - 1) // 3 * 3 - n) + b + solve((n + 3 - 1) // 3, a, b, c, d))
    res = min(res, d * ((n + 5 - 1) // 5 * 5 - n) + c + solve((n + 5 - 1) // 5, a, b, c, d))
    return res

T = int(input())
for _ in range(T):
    N, A, B, C, D = map(int, input().split())
    print(solve(N, A, B, C, D))
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,u = map(int,input().split())
t = int(input())
if v >= u:
  print('NO')
  quit()
dis = (u-v) * t
f = abs(a-b)
if f <= dis:
  print('YES')
else:
  print('NO')

Fixed, accepted submission:
a,u = map(int,input().split())
b,v = map(int,input().split())
t = int(input())
if v >= u:
  print('NO')
  quit()
dis = (u-v) * t
f = abs(a-b)
if f <= dis:
  print('YES')
else:
  print('NO')
Problem:
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
s = [2]
q = 1

for i in range(n):
	if i<3:
		continue
	t = []
	for j in range(len(s)):
		t.append((i)%s[j])
	if 0 in t:
		q = 3
	else:
		s.append(i)

x = 1

for i in range(len(s)):
	m = 0
	for j in range(10):
		m += n//(s[i]**(j+1))
	x *= m+1
print(x%(10**9+7))

Fixed, accepted submission:
n = int(input())
s = [2]
q = 1

for i in range(n+1):
	if i<3:
		continue
	t = []
	for j in range(len(s)):
		t.append((i)%s[j])
	if 0 in t:
		q = 3
	else:
		s.append(i)

x = 1

for i in range(len(s)):
	m = 0
	for j in range(10):
		m += n//(s[i]**(j+1))
	x *= m+1
print(x%(10**9+7))
Problem:
Compute A \times B, truncate its fractional part, and print the result as an integer.

Incorrect submission (verdict: Wrong Answer):
inputs = input().split(' ')
a,b = int(inputs[0]), float(inputs[1])
int(a*b)

Fixed, accepted submission:
import math
from decimal import Decimal

inputs = input().split(' ')
a,b = int(inputs[0]), Decimal(inputs[1])
print(math.floor(a*b))
Problem:
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_.   Here, a correct bracket sequence is defined as follows:    * `()` is a correct bracket sequence.   * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence.   * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence.   * Every correct bracket sequence can be derived from the rules above.  Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

Incorrect submission (verdict: Wrong Answer):
def resolve():
    s = input()
    tmp = s
    while "()" in tmp:
        tmp.replace("()", "")
    l = tmp.count("(")
    r = tmp.count(")")
    ans = "(" * r + s + ")" * l
    print(ans)


if __name__ == "__main__":
    resolve()


Fixed, accepted submission:
def resolve():
    n = int(input())
    s = input()
    l, r = 0, 0
    for c in s:
        if c == "(":
            l += 1
        elif l > 0 and c == ")":
            l -= 1
    for c in s[::-1]:
        if c == ")":
            r += 1
        elif r > 0 and c == "(":
            r -= 1
    ans = "(" * r + s + ")" * l
    print(ans)


if __name__ == "__main__":
    resolve()
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())
if(n<=64):
  print(64)
elif(n<=32):
  print(32)
elif(n<=16):
  print(16)
elif(n<=8):
  print(8)
elif(n<=4):
  print(4)
elif(n<=2):
  print(2)
else:
  print(1)

Fixed, accepted submission:
n=int(input())
if(n>=64):
  print(64)
elif(n>=32):
  print(32)
elif(n>=16):
  print(16)
elif(n>=8):
  print(8)
elif(n>=4):
  print(4)
elif(n>=2):
  print(2)
else:
  print(1)
Problem:
How many multiples of d are there among the integers between L and R (inclusive)?

Incorrect submission (verdict: Wrong Answer):
c=0
a,b,d=map(int,input().split())
for i in range(a,b):
  if i%d==0: c+=1
print(c)

Fixed, accepted submission:
c=0
a,b,d=map(int,input().split())
for i in range(a,b+1):
  if i%d==0: c+=1
print(c)
Problem:
Print all the integers that satisfies the following in ascending order:    * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

Incorrect submission (verdict: Wrong Answer):
a,b,k=map(int,input().split())
ans=[]
for i in range(0,k):
  ans.append(a+i)
  ans.append(b-i)
ans=list(set(ans))
ans.sort()
print(ans)

Fixed, accepted submission:
a,b,k=map(int,input().split())
ans=[]
i=0
while i<k and a+i<=b-i:
  ans.append(a+i)
  ans.append(b-i)
  i=i+1
ans=list(set(ans))
ans.sort()
for i in range(0,len(ans)):
  print(ans[i])