TnT/CodeNet4Repair huggingface dataset:
Problem:
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L.  She will concatenate all of the strings in some order, to produce a long string.  Among all strings that she can produce in this way, find the lexicographically smallest one.  Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds:    * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i.   * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

Incorrect submission (verdict: Wrong Answer):
N,L = list(map(int,input().split(" ")))
S = [input() for i in range(N)]
S.sort()
print(S)

Fixed, accepted submission:
N,L = list(map(int,input().split(" ")))
S = [input() for i in range(N)]
S.sort()
print("".join(S))
Problem:
There are N towns located in a line, conveniently numbered 1 through N. Takahashi the merchant is going on a travel from town 1 to town N, buying and selling apples.  Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows:    * _Move_ : When at town i (i < N), move to town i + 1.   * _Merchandise_ : Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for A_i yen (the currency of Japan) at town i (1 ≦ i ≦ N), where A_i are distinct integers. Also, you can assume that he has an infinite supply of money.  For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.)  During the travel, Takahashi will perform actions so that the _profit_ of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel.  Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change A_i into another arbitrary non-negative integer A_i' for any town i, any number of times. The cost of performing this operation is |A_i - A_i'|. After performing this operation, different towns may have equal values of A_i.  Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen.

Incorrect submission (verdict: Wrong Answer):
import sys
from bisect import bisect_left as bi_l

n, t, *a = map(int, sys.stdin.read().split())

def main():
  cand = []
  mi = a[0]
  for x in a[1:]:
    cand.append(x - mi)
    mi = min(mi, x)
  cand.sort() 
  print(cand)
  ans = n - bi_l(cand, cand[-1]) - 1
  print(ans)
  
if __name__ == '__main__':
  main()

Fixed, accepted submission:
import sys
from bisect import bisect_left as bi_l

n, t, *a = map(int, sys.stdin.read().split())

def main():
  cand = []
  mi = a[0]
  for x in a[1:]:
    cand.append(x - mi)
    mi = min(mi, x)
  cand.sort() 
  ans = n - bi_l(cand, cand[-1]) - 1
  print(ans)
  
if __name__ == '__main__':
  main()
Problem:
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.  All strings that can be obtained in this way can be evaluated as formulas.  Evaluate all possible formulas, and print the sum of the results.

Incorrect submission (verdict: Wrong Answer):
S = input()
combination = 2 ** (len(S) - 1)

ans = 0
for i in range(combination):
    
    bin_str = bin(i)[2:]

    
    bin_str = '0' * (len(S) - len(bin_str) - 1) + bin_str

    formula = ''
    for i, s_char in enumerate(S):
        
        plus = ''
        if len(S) - 1 != i and bin_str[i] == '1':
            plus = '+'

        formula += s_char + plus

    
    ans += eval(formula)

Fixed, accepted submission:
S = input()
combination = 2 ** (len(S) - 1)

ans = 0
for i in range(combination):
    
    bin_str = bin(i)[2:]

    
    bin_str = '0' * (len(S) - len(bin_str) - 1) + bin_str

    formula = ''
    for i, s_char in enumerate(S):
        
        plus = ''
        if len(S) - 1 != i and bin_str[i] == '1':
            plus = '+'

        formula += s_char + plus

    
    ans += eval(formula)

print(ans)
Problem:
E869120 has A 1-yen coins and infinitely many 500-yen coins.   Determine if he can pay exactly N yen using only these coins.

Incorrect submission (verdict: Wrong Answer):

def main():
    N = int(input())
    a = int(input())

    r = N%500

    if a >= r:
        print ('yes')
    else:
        print ('No')


if __name__ == "__main__":
    # global stime
    # stime = time.clock()
    main()


Fixed, accepted submission:

def main():
    N = int(input())
    a = int(input())

    r = N%500

    if r == 0:
        print ('Yes')
    elif a >= r:
        print ('Yes')
    else:
        print ('No')


if __name__ == "__main__":
    # global stime
    # stime = time.clock()
    main()
Problem:
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i.  Snuke can perform the following operation zero or more times:    * Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.  After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).  What is the minimum number of roads he must build to achieve the goal?

Incorrect submission (verdict: Wrong Answer):
class union_find():
    
    def __init__(self,n):
        self.n=n
        self.root=[-1]*(n+1)
        self.rank=[0]*(n+1)

    def find_root(self,x):
        if self.root[x]<0:
            return x
        else:
            self.root[x]=self.find_root(self.root[x])
            return self.root[x]

    def unite(self,x,y):
        x=self.find_root(x)
        y=self.find_root(y)
        

        if x==y:
            return 
        elif self.rank[x]>self.rank[y]:
            self.root[x]+=self.root[y]
            self.root[y]=x
        else:
            self.root[y]+=self.root[x]
            self.root[x]=y

            if self.rank[x]==self.rank[y]:
                self.rank[y]+=1


N,M = map(int,input().split())
g = union_find(N)
group = 0
for i in range(M):
  A,B= map(int,input().split())
  g.unite(A,B)

print(g.root)

for j in g.root:
  if j < 0:
    group += 1

print(group - 2)
  

Fixed, accepted submission:
class union_find():
    
    def __init__(self,n):
        self.n=n
        self.root=[-1]*(n+1)
        self.rank=[0]*(n+1)

    def find_root(self,x):
        if self.root[x]<0:
            return x
        else:
            self.root[x]=self.find_root(self.root[x])
            return self.root[x]

    def unite(self,x,y):
        x=self.find_root(x)
        y=self.find_root(y)
        

        if x==y:
            return 
        elif self.rank[x]>self.rank[y]:
            self.root[x]+=self.root[y]
            self.root[y]=x
        else:
            self.root[y]+=self.root[x]
            self.root[x]=y

            if self.rank[x]==self.rank[y]:
                self.rank[y]+=1


N,M = map(int,input().split())
g = union_find(N)
group = 0
for i in range(M):
  A,B= map(int,input().split())
  g.unite(A,B)

for j in g.root:
  if j < 0:
    group += 1

print(group - 2)
Problem:
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information:  If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest.  Given this information, determine whether TopCoDeer is honest.

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

print("H" if 'a' == 'b' else "D")

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

print("H" if a == b else "D")
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):
n = int(input())
b = list(map(int, input().split()))

c = []
while 0 < len(b):
    candidate = []
    for i, b_i in enumerate(b):
        if (i + 1) == b_i:
            candidate.append(i)
    if 0 == len(candidate):
        print(-1)
        exit(0)
    c.append(b[candidate[-1]])
    b.pop(candidate[-1])

if 0 != len(b):
    print(-1)
    exit(0)

c.reverse()
print(c)


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

result = []
for i in range(len(b)):
    for j in reversed(list(range(len(b)))):
        if b[j] == j + 1:
            result.append(b.pop(j))
            break
if len(b) != 0:
    print(-1)
    exit(0)

for result_i in reversed(result):
    print(result_i)
Problem:
You are given an H × W grid.   The squares in the grid are described by H strings, S_1,...,S_H.   The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W).   `.` stands for an empty square, and `#` stands for a square containing a bomb.  Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square.   (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.)   He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square.  Print the strings after the process.

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

def check_mass(i,j):
  c=0
  if i>0:
    if j>0:
      if mass[i-1][j-1]=="#":
        c+=1
    if mass[i-1][j]=="#":
        c+=1
    if j<W-1:
      if mass[i-1][j+1]=="#":
        c+=1
  if True:
      if mass[i][j-1]=="#":
        c+=1
      if mass[i][j]=="#":
          c+=1
      if j<W-1:
        if mass[i][j+1]=="#":
          c+=1
  if i<H-1:
      if mass[i+1][j-1]=="#":
        c+=1
      if mass[i+1][j]=="#":
          c+=1
      if j<W-1:
        if mass[i+1][j+1]=="#":
          c+=1
  return c

H,W= list(map(int, input().split()))
mass= []
for i in range(H):
  s = list(input()) 
  mass.append(s)

print(mass)

for i in range(H):
  for j in range(W):
    if mass[i][j] == ".":
      c=check_mass(i,j)
      mass[i][j] = str(c)

for i in range(H):
  for j in range(W):
    print(mass[i][j], end="")
  print()


Fixed, accepted submission:
import numpy as np

def check_mass(i,j):
  c=0
  if i>0:
    if j>0:
      if mass[i-1][j-1]=="#":
        c+=1
    if mass[i-1][j]=="#":
        c+=1
    if j<W-1:
      if mass[i-1][j+1]=="#":
        c+=1
  if True:
      if j>0:
        if mass[i][j-1]=="#":
          c+=1
      if mass[i][j]=="#":
          c+=1
      if j<W-1:
        if mass[i][j+1]=="#":
          c+=1
  if i<H-1:
      if j>0:
        if mass[i+1][j-1]=="#":
          c+=1
      if mass[i+1][j]=="#":
          c+=1
      if j<W-1:
        if mass[i+1][j+1]=="#":
          c+=1
  return c

H,W= list(map(int, input().split()))
mass= []
for i in range(H):
  s = list(input()) 
  mass.append(s)

# print(mass)

for i in range(H):
  for j in range(W):
    if mass[i][j] == ".":
      c=check_mass(i,j)
      mass[i][j] = str(c)

for i in range(H):
  for j in range(W):
    print(mass[i][j], end="")
  print()
Problem:
There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other.  Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro:    * Choose an element x in A, and remove exactly x stones from the pile.  A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner.

Incorrect submission (verdict: Wrong Answer):
n, k = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))

dp = [0 for i in range(k+1)]
for i in range(len(dp)):
    for j in arr:
        if j+i < k+1:
            dp[j+i] = (dp[i]+1)%2

ans = "First" if dp[-1] else "Second"
print(ans)



Fixed, accepted submission:
n, k = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))

dp = [0 for i in range(k+1)]

for i in range(len(dp)):
    if dp[i] == 0:
        for j in arr:
            if j+i < k+1:
                dp[j+i] = 1

ans = "First" if dp[-1] else "Second"
print(ans)
Problem:
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

Incorrect submission (verdict: Wrong Answer):
import sys

stdin = sys.stdin

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

a,b = na()
import math
print(math.ceil((a+b / 2)))


Fixed, accepted submission:
import sys

stdin = sys.stdin

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

a,b = na()
import math
print(math.ceil(((a+b) / 2)))
Problem:
There are N people numbered 1 to N. Each person wears a red hat or a blue hat.  You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`.  Determine if there are more people wearing a red hat than people wearing a blue hat.

Incorrect submission (verdict: Wrong Answer):
s=list(input())
x=s.count('R')
y=s.count('B')
if x>>y:
    print('Yes')
else:
    print('No')


Fixed, accepted submission:
a=input()
s=list(input())
x=s.count('R')
y=s.count('B')
if x>y:
    print('Yes')
else:
    print('No')
Problem:
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.

Incorrect submission (verdict: Wrong Answer):
s = input()
if s in "aiueo":
    print("Vowel")
else:
    print("Consonant")

Fixed, accepted submission:
s = input()
if s in "aiueo":
    print("vowel")
else:
    print("consonant")
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 = []
if b - a < k:
    k = b - a
for i in range(k):
    ans.append(a+i)
    ans.append(b-i)
ans = list(set(sorted(ans)))
for i in ans:
    print(i)

Fixed, accepted submission:
a, b, k = map(int, input().split())
if 2*k >= b-a+1:
    for i in range(a, b+1):
        print(i)
else:
    for i in range(a, a+k):
        print(i)
    for j in range(b-k+1, b+1):
        print(j)
Problem:
Snuke loves working out. He is now exercising N times.  Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i.  Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

Incorrect submission (verdict: Wrong Answer):
s=int(input())
ans=1
for i in range(1,s):
  ans*=i
  
print(ans)

Fixed, accepted submission:
s=int(input())
ans=1
for i in range(1,s+1):
  ans=ans%(10**9+7)*i
  
  
print(ans%(10**9+7))
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):
A,B = map(int,input().split())
S = input()
L = [0,1,2,3,4,5,6,7,8,9]
for i in (range(A) and range(A+1,A+B+1)):
  if (S[i] in L):
    if (S[A] == '-'):
      print('Yes')
  else:
    print('No')

Fixed, accepted submission:
A,B = map(int,input().split())
S = list(input())
flag = True
L = ['1','2','3','4','5','6','7','8','9','0']
if (S[A] == '-'):
  del S[A]
  for i in range(len(S)):
    if (S[i] not in L):
      flag = False
      break
else:
  flag = False
if flag:
  print('Yes')
else:
  print('No')
Problem:
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.    An example of a trapezoid  Find the area of this trapezoid.

Incorrect submission (verdict: Wrong Answer):
a = int(input())
b = int(input())
h = int(input())
print( (a+b)*h/2 )

Fixed, accepted submission:
a = int(input())
b = int(input())
h = int(input())
s = int((a+b)*h/2)
print(s)
Problem:
You are given three strings A, B and C. Check whether they form a _word chain_.  More formally, determine whether both of the following are true:    * The last character in A and the initial character in B are the same.   * The last character in B and the initial character in C are the same.  If both are true, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
a,b,c = input().split()
print(('No','Yes')[a[-1]==b[0] and b[-1]==c[0]])


Fixed, accepted submission:
a,b,c = input().split()
print(('NO','YES')[a[-1]==b[0] and b[-1]==c[0]])
Problem:
Write a program which reads a sequence and prints it in the reverse order.

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

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

numbers.reverse()
print(numbers)

Fixed, accepted submission:
input()

numbers = input().split()

numbers.reverse()
print(" ".join(numbers))
Problem:
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N.  There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i.  Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.  Help him.

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

a = [[0 for i in range(2)] for j in range(M)]
b = []

for l in range(M):
  a[l][0] , a[l][1] = map(int,input().split())

for k in range(M):
  if a[k][0] == a[0][1]:
    b.append(k)

for m in b:
  if a[m][1] == N:
    print ('possible')
  else:
    print('impossible')


Fixed, accepted submission:
import sys 

N,M  = map(int,input().split())
     
a = [[0 for i in range(2)] for j in range(M)]
b = [] 
c = []

for i in range(M):
  a[i][0] , a[i][1] = map(int,input().split())
  first, second = a[i][0],a[i][1]
  if first == 1:
    b.append(second)
  if second == N:
    c.append(first)

b_set = set(b)
c_set = set(c)

if list(b_set & c_set) is None or len(list(b_set & c_set)):
  res = ('POSSIBLE')
else:
  res = ('IMPOSSIBLE')

print(res)
Problem:
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.  Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Incorrect submission (verdict: Wrong Answer):
N, P = [int(i) for i in input().split()]


div = {}
x = P
y = 2
while y*y <= x:
    while x % y == 0:
        try:
            div[y] += 1
        except KeyError:
            div[y] = 1
        x //= y
    y += 1
if x > 1:
    div[x] = 1

res = 1
for k, v in div.items():
    res *= max(1, (v//4)) * k
print(res)

Fixed, accepted submission:
N, P = [int(i) for i in input().split()]


div = {}
x = P
y = 2
while y*y <= x:
    while x % y == 0:
        try:
            div[y] += 1
        except KeyError:
            div[y] = 1
        x //= y
    y += 1
if x > 1:
    div[x] = 1

res = 1
for k, v in div.items():
    res *= k**(v//N)
print(res)
Problem:
Snuke loves puzzles.  Today, he is working on a puzzle using `S`\- and `c`-shaped pieces. In this puzzle, you can combine two `c`-shaped pieces into one `S`-shaped piece, as shown in the figure below:    Snuke decided to create as many `Scc` groups as possible by putting together one `S`-shaped piece and two `c`-shaped pieces.  Find the maximum number of `Scc` groups that can be created when Snuke has N `S`-shaped pieces and M `c`-shaped pieces.

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split())
if 2*n >= m:
    ans = m//2
else:
    ans = n - (2*n - m)//4 -1
print(ans)

Fixed, accepted submission:
n, m = map(int, input().split())
if 2*n >= m:
    ans = m//2
else:
    ans = n + (m - 2*n)//4
print(ans)
Problem:
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters.  Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s.  Given the name of the contest, print the abbreviation of the name.

Incorrect submission (verdict: Wrong Answer):
a,b,c=map(str,input().split())
print('A'+b+'C')

Fixed, accepted submission:
a,b,c=map(str,input().split())
print('A'+b[0].upper()+'C')
Problem:
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.  Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut:    * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).   * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.  Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a = [int(i) for i in input().split()]

length = sum(a)
if length %2 ==0:
    check = 0
    half = length//2
    idx = 0
    for i in range(len(a)):
        check+=a[i]
        if check > half:
            idx = i
            break
    if check - half > half - (check-a[idx]):
        check -=a[idx]
    
    ans = (half-check)*2
else:
    check = 0
    half = length//2
    idx = 0
    for i in range(len(a)):
        check+=a[i]
        if check > half:
            idx = i
            break
    if check - half > half - (check-a[idx]):
        check -=a[idx]
    
    ans = abs((half-check))*2
print(ans)
    
                
            

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

length = sum(a)
if length %2 ==0:
    check = 0
    half = length//2
    idx = 0
    for i in range(len(a)):
        check+=a[i]
        if check > half:
            idx = i
            break
    if check - half > half - (check-a[idx]):
        check -=a[idx]
    
    ans = abs((half-check)*2)
else:
    check = 0
    half = length//2
    idx = 0
    for i in range(len(a)):
        check+=a[i]
        if check > half:
            idx = i
            break
    check -=a[idx]
    afte = sum(a[idx:])
    ans = half-check+afte-half
    
    check +=a[idx]
    afte = sum(a[idx+1:])
    ans = min(ans,abs(check-half)+abs(afte-half))
    
print(ans)