TnT/CodeNet4Repair huggingface dataset:
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

Incorrect submission (verdict: Wrong Answer):
S,T=open(0);l,m=len(S),len(T)
for i in range(l-m):
 if all(c in"?"+d for c,d in zip(S[-i-1-m:],T)):S=S.replace("?","a");print(S[:-i-1-m]+T+S[l-i:]);quit()
print("UNRESTORABLE")

Fixed, accepted submission:
S,T=input(),input();l,m=len(S),len(T)
for i in range(l-m+1):
 if all(c in"?"+d for c,d in zip(S[-i-m:],T)):S=S.replace("?","a");print(S[:-i-m]+T+S[l-i:]);quit()
print("UNRESTORABLE")
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, b, h = int(input()), int(input()), int(input())
print((a + b) * h / 2)


Fixed, accepted submission:
a, b, h = int(input()), int(input()), int(input())
print((a + b) * h // 2)
Problem:
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.  He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.  To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.  How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Incorrect submission (verdict: Wrong Answer):





from bisect import bisect_left, bisect_right
import sys
read = sys.stdin.readline


def read_ints():
    return list(map(int, read().split()))


def read_a_int():
    return int(read())


N = read_a_int()
A = read_ints()
B = read_ints()
C = read_ints()

A.sort()
B.sort()
C.sort()

print(A)
print(B)
print(C)

ans = 0
for b in B:
    ans += bisect_left(A, b) * (N - bisect_right(C, b))
    
print(ans)


Fixed, accepted submission:





from bisect import bisect_left, bisect_right
import sys
read = sys.stdin.readline


def read_ints():
    return list(map(int, read().split()))


def read_a_int():
    return int(read())


N = read_a_int()
A = read_ints()
B = read_ints()
C = read_ints()

A.sort()
C.sort()

ans = 0
for b in B:
    ans += bisect_left(A, b) * (N - bisect_right(C, b))
    
print(ans)
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(10**8)
def main():
    n, a, b, c, d = map(int, input().split())
    memo = {}
    def dfs(x):
        if x == 0:
            return 0
        elif x == 1:
            return d
        elif x in memo:
            return memo[x]
        else:
            ret = min(
                d * x,
                d * (x - x//2*2) + a + dfs(x//2),
                d * ((x+1)//2*2-x) + a + dfs((x+1)//2),
                d * (x - x//3*2) + b + dfs(x//3),
                d * ((x+2)//3*3 - x) + b + dfs((x+2)//3),
                d * (x - x//5*5) + c + dfs(x//5),
                d * ((x+4)//5*5 - x) + c + dfs((x+4)//5)
            )
            memo[x] = ret
            return ret
    print(dfs(n))

t = int(input())
for i in range(t):
    main()

Fixed, accepted submission:
import sys
sys.setrecursionlimit(10**8)
def main():
    n, a, b, c, d = map(int, input().split())
    memo = {}
    def dfs(x):
        if x == 0:
            return 0
        elif x == 1:
            return d
        elif x in memo:
            return memo[x]
        else:
            ret = min(
                d * x,
                d * (x - x//2*2) + a + dfs(x//2),
                d * ((x+1)//2*2 - x) + a + dfs((x+1)//2),
                d * (x - x//3*3) + b + dfs(x//3),
                d * ((x+2)//3*3 - x) + b + dfs((x+2)//3),
                d * (x - x//5*5) + c + dfs(x//5),
                d * ((x+4)//5*5 - x) + c + dfs((x+4)//5)
            )
            memo[x] = ret
            return ret
    print(dfs(n))

t = int(input())
for i in range(t):
    main()
Problem:
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times:    * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n.  What is the largest possible sum of the integers written on the blackboard after K operations?

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int,input().split())
k = int(input())
note = max(a,b,c)
for i in range(k):
	note *= 2
print(note)

Fixed, accepted submission:
a = [int(i) for i in input().split()]
k = int(input())
a.sort()
note = a[2]
for i in range(k):
	note *= 2
print(a[0] + a[1] + note)
Problem:
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`.  You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Incorrect submission (verdict: Wrong Answer):
import sys
def no():
    print("No")
    sys.exit()
a,b=map(int,input().split())
s=input()
num=[str(i) for i in range(1,10)]
if len(s)==a+b+1:
    for i in range(a):
        if s[i] not in num: no()
    if s[a]!="-": no()
    for i in range(a+1,a+b+1):
        if s[i] not in num: no()   

else: no()

print("Yes")

Fixed, accepted submission:
import sys
def no():
    print("No")
    sys.exit()

a,b=map(int,input().split())
s=input()
num=[str(i) for i in range(10)]

if len(s)==a+b+1:
    for i in range(a):
        if s[i] not in num: no()
    if s[a]!="-": no()
    for i in range(a+1,a+b+1):
        if s[i] not in num: no()   

else: no()

print("Yes")
Problem:
Let us consider rectangles whose height, _h_ , and width, _w_ , are both integers. We call such rectangles _integral rectangles_. In this problem, we consider only wide integral rectangles, i.e., those with _w_ > _h_.  We define the following ordering of wide integral rectangles. Given two wide integral rectangles,    1. The one shorter in its diagonal line is smaller, and    2. If the two have diagonal lines with the same length, the one shorter in its height is smaller.     Given a wide integral rectangle, find the smallest wide integral rectangle bigger than the given one.

Incorrect submission (verdict: Wrong Answer):
def intRec(h,w):
    diag=h**2+w**2
    ans=[]
    temp=1e8
    for H in range(1,151):
        for W in range(w,151):
            if (H**2+W**2)>diag and (H**2+W**2)<temp:
                temp=H**2+W**2
                ans=[H,W]
    return ans

if __name__=="__main__":
    while True:
        h,w=map(int,input().split())
        if h==0:
            break
        ans=intRec(h,w)
        print(ans[0],ans[1])


Fixed, accepted submission:
def intRec(h,w):
    diag=h**2+w**2
    minD=1e8
    ans=[]
    for H in range(1,151):
        for W in range(1,151):
            if H<W:
                Diag=H**2+W**2
                if diag==Diag and H>h:
                    return [H,W]
                elif diag<Diag and Diag<minD:
                    minD=Diag
                    ans=[H,W]
    return ans

if __name__=="__main__":
    while True:
        h,w=map(int,input().split())
        if h==0:
            break
        ans=intRec(h,w)
        print(ans[0],ans[1])
Problem:
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems.  The score allocated to the i-th (1≦i≦N) problem is i points.  Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve.  As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him.  Determine the set of problems that should be solved.

Incorrect submission (verdict: Wrong Answer):
# 2019/08/09
from itertools import accumulate
from bisect import bisect_left

n=int(input())
a=[0]+list(accumulate(list(range(1,n+1))))


def bsct(a,key):
    l=0
    r=len(a)
    mid=(l+r)//2
    while l<=r:
        if a[mid]==key:
            return mid
        elif a[mid]>key:
            r=mid-1
        else:
            l=mid+1
        mid=(l+r)//2
    return 0

res=bsct(a,n)
if res:
    print(res)
    exit()

idx=bisect_left(a,n)
rem=a[idx]-n

ans=list(range(1,idx+1))
del ans[rem]
print(*ans,sep='\n')

Fixed, accepted submission:
# 2019/08/09
from bisect import bisect_left

n=int(input())
acc=[0]
for i in range(1,n+1):
    acc.append(i+acc[i-1])
    if acc[i]==n:
        print(*range(1,i+1),sep='\n')
        exit()
    elif acc[i]>n:
        idx=i
        break

rem=acc[idx]-n

res=0
for i in range(1,idx+1):
    if i!=rem:
        print(i)
Problem:
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

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

import sys
input = sys.stdin.readline

def io_generator():
	return input()

#+++++++++++++++++++

def bbs(v):
	s = [int(ci) for ci in list(str(v))]
	return sum(s)
	
def main(io):
	n, a, b = map(int, io().split())
	return sum([x for x in range(n) if a<= bbs(x) and bbs(x) <= b])

#++++++++++++++++++++

if __name__ == "__main__":
	io= lambda : io_generator()
	print (main(io))

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

import sys
input = sys.stdin.readline

def io_generator():
	return input()

#+++++++++++++++++++

def bbs(v):
	s = [int(ci) for ci in list(str(v))]
	return sum(s)
	
def main(io):
	n, a, b = map(int, io().split())
	return sum([x for x in range(1,n+1) if a<= bbs(x) and bbs(x) <= b])

#++++++++++++++++++++

if __name__ == "__main__":
	io= lambda : io_generator()
	print (main(io))
Problem:
AtCoDeer the deer and his friend TopCoDeer is playing a game. The game consists of N turns. In each turn, each player plays one of the two _gestures_ , _Rock_ and _Paper_ , as in Rock-paper-scissors, under the following condition:  (※) After each turn, (the number of times the player has played Paper)≦(the number of times the player has played Rock).  Each player's score is calculated by (the number of turns where the player wins) - (the number of turns where the player loses), where the outcome of each turn is determined by the rules of Rock-paper-scissors.  _(For those who are not familiar with Rock-paper-scissors: If one player plays Rock and the other plays Paper, the latter player will win and the former player will lose. If both players play the same gesture, the round is a tie and neither player will win nor lose.)_  With his supernatural power, AtCoDeer was able to foresee the gesture that TopCoDeer will play in each of the N turns, before the game starts. Plan AtCoDeer's gesture in each turn to maximize AtCoDeer's score.  The gesture that TopCoDeer will play in each turn is given by a string s. If the i-th (1≦i≦N) character in s is `g`, TopCoDeer will play Rock in the i-th turn. Similarly, if the i-th (1≦i≦N) character of s in `p`, TopCoDeer will play Paper in the i-th turn.

Incorrect submission (verdict: Wrong Answer):
s=input()
g=s.count("g")
p=len(s)-g
a=len(s)//2
if p>g:
  g=0
  p=p-a-(len(s)-a-g)
  print(g+p)
elif p==g:
  print(0)
else:
  g=len(s)-a-p
  p=0
  print(g+p)

Fixed, accepted submission:
s=input()
p=s.count("p")
print(len(s)//2-p)
Problem:
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

Incorrect submission (verdict: Wrong Answer):
N, K = map(int, input().split())
if (N / 2) + 1 >= K:
    print('Yes')
else:
    print('No')

Fixed, accepted submission:
N, K = map(int, input().split())
if (N / 2) + 1 > K:
    
    print('YES')
else:
    print('NO')
Problem:
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.  Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence.  Here, the distance between two points s and t on a number line is represented by |s-t|.

Incorrect submission (verdict: Wrong Answer):
x,a,b=map(int,input().split())
if abs(x-a)<abs(x-b):
    print("B")
elif abs(x-a)>abs(x-b):
    print("A")

Fixed, accepted submission:
x,a,b=map(int,input().split())
if abs(x-a)<abs(x-b):
    print("A")
else:
    print("B")
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):
n = int(input())

ans = n % 1000

print(ans)

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

if n % 1000 == 0:
    print(0)
else:
    print(1000 - (n % 1000))
Problem:
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total?  Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

Incorrect submission (verdict: Wrong Answer):
coin_500 = int(input())
coin_100 = int(input())
coin_50 = int(input())
total = int(input())

def coin_check(coin_total, total):
    if coin_total==total:
        return 1
    else:
        return 0

count = 0
for i in range(coin_500 + 1):
    coin_total = i*500
    count+=coin_check(coin_total, total)
    for j in range(coin_100 + 1):
        coin_total = i*500 + j*100
        count+=coin_check(coin_total, total)
        for k in range(coin_50 + 1):
            coin_total = i*500 + j*100 + k*50
            count+=coin_check(coin_total, total)
print(count)

Fixed, accepted submission:
coin_500 = int(input())
coin_100 = int(input())
coin_50 = int(input())
total = int(input())

def coin_check(coin_total, total):
    if coin_total==total:
        return 1
    else:
        return 0

count = 0
for i in range(coin_500 + 1):
    for j in range(coin_100 + 1):
        for k in range(coin_50 + 1):
            coin_total = i*500 + j*100 + k*50
            count+=coin_check(coin_total, total)
print(count)
Problem:
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.  The problem has N test cases.  For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively.  See the Output section for the output format.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
result_dict = {'AC':0,'WA':0,'TLE':0,'RE':0}
for i in range(n):
    s = input()
    result_dict[s] += 1
 
print('AC'+' × '+str(result_dict['AC']))
print('WA'+' × '+str(result_dict['WA']))
print('TLE'+' × '+str(result_dict['TLE']))
print('RE'+' × '+str(result_dict['RE']))

Fixed, accepted submission:
n = int(input())
result_dict = {'AC':0,'WA':0,'TLE':0,'RE':0}
for i in range(n):
    s = input()
    result_dict[s] += 1
 
print('AC'+' x '+str(result_dict['AC']))
print('WA'+' x '+str(result_dict['WA']))
print('TLE'+' x '+str(result_dict['TLE']))
print('RE'+' x '+str(result_dict['RE']))
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):
sub = int(input())
print((1+sub) * sub / 2)

Fixed, accepted submission:
subin = int(input())
res = int((1+subin) * subin / 2)
print(res)
Problem:
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.  Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.  Now, the following Q operations will be performed:    * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.  Find the value of the counter on each vertex after all operations.

Incorrect submission (verdict: Wrong Answer):
import math 
import bisect
import heapq
from operator import mul
from collections import deque

n, q = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n-1)]
px = [list(map(int, input().split())) for _ in range(q)]



def dfs(start):
    if start == 1:
        return [True] * n
    task = deque([])
    seen = [False] * n

    task.appendleft(start)
    seen[start-1] = True

    while task:
        task.popleft()
        for i in range(len(ab)):
            if ab[i][0] == start and seen[ab[i][1]-1] == False:
                next = ab[i][1]
                seen[next-1] = True
                task.appendleft(next)


    return seen


ans = [0]*n

for i in range(q):
    for j in range(n):
        ans[j] += px[i][1] if dfs(px[i][0])[j] else 0
print(ans)



Fixed, accepted submission:
import math 
import bisect
import heapq
from operator import mul
from collections import deque

n, q = map(int, input().split())
g=[[] for _ in range(n)]

for i in range(n-1):
    a,b=map(int,input().split())
    g[a-1].append(b-1)
    g[b - 1].append(a - 1)
c = [0] * n

for j in range(q):
    p, x = map(int, input().split())
    c[p - 1] += x

seen = [0] * n
seen[0] = 1
next = deque([0])


while next:
    v = next.popleft()
    cost = c[v]

    for nv in g[v]:

        if seen[nv]:
            continue
        seen[nv] = 1
        next.append(nv)
        c[nv] += cost

print(*c)
Problem:
How many multiples of d are there among the integers between L and R (inclusive)?

Incorrect submission (verdict: Wrong Answer):
L,R,d = map(int,input().split())

print(R//d - L//d + 1)


Fixed, accepted submission:
L,R,d = map(int,input().split())

ll = L//d
if L % d != 0 :
    ll += 1
rr = R//d

print(rr-ll+1)
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):
a, b = input().split()
ab = int(a+b)
sqrt_ab = ab**(1/2)
if int(sqrt_ab)**2 == ab:
    print('Yes')
else:
    print('No')
print(ab, sqrt_ab)

Fixed, accepted submission:
a, b = input().split()
ab = int(a+b)
sqrt_ab = ab**(1/2)
if int(sqrt_ab)**2 == ab:
    print('Yes')
else:
    print('No')
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(("No","Yes")[x-a<=b])

Fixed, accepted submission:
a,b,x = map(int,input().split())
if x - a < 0:
    print("NO")
else:
    print(("NO","YES")[x-a<=b])
Problem:
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied:    * Each lowercase letter of the English alphabet occurs even number of times in w.  You are given the string w. Determine if w is beautiful.

Incorrect submission (verdict: Wrong Answer):
w=input()
l=[]
c=0
for i in w:
  l.append(i)
l.sort()
for k in range(len(l)-1):
  if l[k]==l[k+1]:
    c+=1
  elif c%2==1:
    c=0
    continue
  else:
    print('NO')
    exit()
print('YES')

Fixed, accepted submission:
w=input()
l=[]
c=0
for i in w:
  l.append(i)
l.sort()
if len(l)<2:
  print('No')
  exit()
for k in range(len(l)-1):
  if l[k]==l[k+1]:
    c+=1
  elif c%2==1:
    c=0
    continue
  else:
    print('No')
    exit()
print('Yes')
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:
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):

import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
from collections import Counter, deque
from collections import defaultdict
from itertools import combinations, permutations, accumulate, groupby, product
from bisect import bisect_left,bisect_right
from heapq import heapify, heappop, heappush
from math import floor, ceil,pi,factorial
from operator import itemgetter
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
def LI2(): return [int(input()) for i in range(n)]
def MXI(): return [[LI()]for i in range(n)]
def SI(): return input().rstrip()
def printns(x): print('\n'.join(x))
def printni(x): print('\n'.join(list(map(str,x))))
inf = 10**17
mod = 10**9 + 7

n=I()
s=SI()
i=0
state=1
count=0
l=[];r=[]
for i in range(n):
    if state==1:
        if s[i]=="(":
            count+=1
        elif s[i]==")":
            l.append(count)
            state=-1
            count=1
    elif state==-1:
        if s[i]==")":
            count+=1
        elif s[i]=="(":
            r.append(count)
            state=1
            count=1
if s[-1]==")":
    r.append(count)
else:
    l.append(count)
    r.append(0)
'''print(l)
print(r)'''
minus,plus=0,0
for i in range(len(l)):
    if l[i]-r[i]<0:
        minus+=r[i]-l[i]
    else:
        plus+=l[i]-r[i]
for i in range(minus):
    print("(",end="")
print(s,end="")
for j in range(plus):
    print(")",end="")

Fixed, accepted submission:
import sys

stdin = sys.stdin

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

n = ni()
s = ns().strip()
h = 0
minh = 0
for c in s:
    if c == "(":
        h += 1
    else:
        h -= 1
    minh = min(minh, h)
print("("*(-minh) + s + ")"*(h-minh))