TnT/CodeNet4Repair huggingface dataset:
Problem:
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`.  You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

Incorrect submission (verdict: Wrong Answer):
s = input()
print(s[0],len(s)-2,s[-1])

Fixed, accepted submission:
s = input()
print("{}{}{}".format(s[0],len(s)-2,s[-1]))
Problem:
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.

Incorrect submission (verdict: Wrong Answer):
a=str(input())
print(a.swapcase)

Fixed, accepted submission:
a=str(input())
print(a.swapcase())
Problem:
There are N positive integers written on a blackboard: A_1, ..., A_N.  Snuke can perform the following operation when all integers on the blackboard are even:    * Replace each integer X on the blackboard by X divided by 2.  Find the maximum possible number of operations that Snuke can perform.

Incorrect submission (verdict: Time Limit Exceeded):
N = int(input())
A = [int(i) for i in input().split()]

odd = False
count = 0

while True:
    for i in A:
        if i%2 != 0:
            odd = True

    if odd:
        break

    for i in A:
        i = int(i/2)

    count += 1

print(count)


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

cnt =0
flag = 0
while True:
  a = list(map(lambda x: x/2, a))
  for i in range(n):
    print
    if not a[i].is_integer():
      flag = 1
      break
  if flag == 1:
    break
  cnt += 1

print(cnt)
Problem:
Write a program which finds a pattern $p$ in a ring shaped text $s$.     

Incorrect submission (verdict: Wrong Answer):
import sys
hoge = list()
for line in sys.stdin:
    hoge.append(line)
hoge[0] = hoge[0] + hoge[0]
if hoge[0].find(hoge[1]) == -1:
    print ('No')
else:
    print ('Yes')


Fixed, accepted submission:
hoge = list()
for _ in range(2):
    hoge.append(input())
hoge[0] = hoge[0] + hoge[0]
if hoge[0].find(hoge[1]) == -1:
    print ('No')
else:
    print ('Yes')
Problem:
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.   According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.   Determine if he is correct.

Incorrect submission (verdict: Wrong Answer):
c = [list(map(int, input().split())) for i in range(3)]



for a1 in range(101):
    b1 = c[0][0] - a1
    b2 = c[1][0] - a1
    b3 = c[2][0] - a1
    a2 = c[0][1] - b1
    if a2 != (c[1][1] - b2) and a2 != (c[2][1] - b3):
        print("No")
        exit()
    a3 = c[2][0] - b1
    if a3 != (c[2][1] - b2) and a3 != (c[2][2] - b3):
        print("No")
        exit()
    if a1 < 0 or a2 < 0 or a3 < 0 or b1 < 0 or b2 < 0 or b3 < 0:
        print("No")
        exit()
print("Yes")

Fixed, accepted submission:
c = [list(map(int, input().split())) for i in range(3)]



for a1 in range(101):
    flag = True
    b1 = c[0][0] - a1
    b2 = c[1][0] - a1
    b3 = c[2][0] - a1
    a2 = c[0][1] - b1
    if a2 != (c[1][1] - b2) or a2 != (c[2][1] - b3):
        flag = False
    a3 = c[0][2] - b1
    if a3 != (c[1][2] - b2) or a3 != (c[2][2] - b3):
        flag = False

    if a1 < 0 or a2 < 0 or a3 < 0 or b1 < 0 or b2 < 0 or b3 < 0:
        flag = False

    if flag:
        print("Yes")
        exit()
print("No")
Problem:
Write a program which reads a sequence and prints it in the reverse order.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
a = []
a = list(map(int, input().split()))
print(N)
for i in range(1, len(a)+1):
    print(a[-i], end =" ")

Fixed, accepted submission:
N = int(input())
a = input().split(" ")
print(" ".join(reversed(a)))
Problem:
_Kode Festival_ is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".)  This year, 2^N stones participated. The hardness of the i-th stone is A_i.  In the contest, stones are thrown at each other in a knockout tournament.  When two stones with hardness X and Y are thrown at each other, the following will happen:    * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y.    * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same.    * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X.  The 2^N stones will fight in a knockout tournament as follows:    1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ...    2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ...    3. And so forth, until there is only one stone remaining.  Determine the eventual hardness of the last stone remaining.

Incorrect submission (verdict: Wrong Answer):
3
1
3
2
4
6
8
100
104


Fixed, accepted submission:
def battle(first, second):
    if first > second:
        return first - second
    elif first < second:
        return second - first
    else:
        return first

n = 1 << int(input())
a = []
for i in range(n):
    a.append(int(input()))
while True:
    if n // 2 == 0:
        break
    new = []
    for i in range(n // 2):
        new.append(battle(a[i*2], a[i*2+1]))
    a = new
    n //= 2
print(a[0])
Problem:
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand.  How many grams of sand will the upper bulb contains after t seconds?

Incorrect submission (verdict: Wrong Answer):
def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))

x,t =iim()
print(min(0,x-t))

Fixed, accepted submission:
def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))

x,t =iim()
print(max(0,x-t))
Problem:
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.  Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.    * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.   * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.  A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.  Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.

Incorrect submission (verdict: Wrong Answer):
N, A, B = map(int, input().split())
S = input()
a = b = 0
for s in S:
  if s == 'a' and a+b < A+B:
    print('Yes')
    a += 1
  elif s == 'b' and b < B:
    print('Yes')
    b += 1
  else:
    print('No')

Fixed, accepted submission:
N, A, B = map(int, input().split())
S = input()
a = b = 0
for s in S:
  if s == 'a' and a+b < A+B:
    print('Yes')
    a += 1
  elif s == 'b' and a+b < A+B and b < B:
    print('Yes')
    b += 1
  else:
    print('No')
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())
s = [input() for _ in range(n)]
#print(s)

print('AC × ' + str(s.count('AC')))
print('WA × ' + str(s.count('WA')))
print('TLE × ' + str(s.count('TLE')))
print('RE × ' + str(s.count('RE')))


Fixed, accepted submission:
n = int(input())
s = [input() for _ in range(n)]
#print(s)

print('AC x ' + str(s.count('AC')))
print('WA x ' + str(s.count('WA')))
print('TLE x ' + str(s.count('TLE')))
print('RE x ' + str(s.count('RE')))
Problem:
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

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

x = int(input())

print(x*(x+1)/2)

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

x = int(input())

def answer(x):
    for i in range(x):
        if(i*(i+1)//2 < x):
            continue
        else:
            return i

result = answer(x)

if(result == None):
    print(x)
else:
    print(answer(x))
Problem:
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.  Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.    * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.   * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.  A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.  Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.

Incorrect submission (verdict: Wrong Answer):
N, A, B = list(map(int, input().split()))
S = input()
passed_num = 0
passed_foreigners_num = 0
for i in S:
    if passed_num < A + B:
        if i == 'a':
            print('yes')
            passed_num += 1 
        elif i == 'b' and passed_foreigners_num < B:
            print(i)
            print('yes')
            passed_num += 1 
            passed_foreigners_num += 1
        else:
            print('No')
    else:
        print('No')

Fixed, accepted submission:
N, A, B = list(map(int, input().split()))
S = input()
passed_num = 0
passed_foreigners_num = 0
for i in S:
    if passed_num < A + B:
        if i == 'a':
            print('Yes')
            passed_num += 1 
        elif i == 'b' and passed_foreigners_num < B:
            print('Yes')
            passed_num += 1 
            passed_foreigners_num += 1
        else:
            print('No')
    else:
        print('No')
Problem:
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).  A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.  At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
A = [[int(i) for i in reversed(input().split())] for n in range(N)]
B = [[int(i) for i in input().split()] for n in range(N)]
A.sort(); A.reverse(); B.sort()
C = 0
i = 0
for b in B:
  found = False
  while not found:
    a = A[i]
    if a[1] < b[0] and a[0] < b[1]:
      found = True
      C += 1
      A.pop(i)
      i = 0
    elif i+1 == len(A):
      found = True
      i = 0
    else:
      i += 1
    print(a,b,found,C)
print(C)

Fixed, accepted submission:
N = int(input())
A = [[int(i) for i in reversed(input().split())] for n in range(N)]
B = [[int(i) for i in input().split()] for n in range(N)]
A.sort(); A.reverse(); B.sort()
C = 0
i = 0
for b in B:
  found = False
  while not found:
    a = A[i]
    if a[1] < b[0] and a[0] < b[1]:
      found = True
      C += 1
      A.pop(i)
      i = 0
    elif i+1 == len(A):
      found = True
      i = 0
    else:
      i += 1
print(C)
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):
import collections
n = int(input())
s = [input() for _ in range(n)]
s = collections.Counter(s)
print("AC x " + str(s["AC"]))
print("WA x " + str(s["WA"]))
print("ATLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"]))


Fixed, accepted submission:
import collections
n = int(input())
s = [input() for _ in range(n)]
s = collections.Counter(s)
print("AC x " + str(s["AC"]))
print("WA x " + str(s["WA"]))
print("TLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"]))
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 = input()
print("YES" if n[0]=='9' or n[1]=='9' else "NO")

Fixed, accepted submission:
n = input()
print("Yes" if n[0]=='9' or n[1]=='9' else "No")
Problem:
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes:    * Red boxes, each containing R red balls   * Green boxes, each containing G green balls   * Blue boxes, each containing B blue balls  Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?

Incorrect submission (verdict: Wrong Answer):
R,G,B,N = map(int,input().split())
ans = 0
RGB = sorted([R,G,B],reverse=True)
print(RGB)
R,G,B = RGB[0],RGB[1],RGB[2]
for r in range(-(-N//R)+1):
    for g in range(((N-R*r)//G)+1):
        for b in range((N-(R*r+G*g)//B)+1):
            if(R*r + G*g + B*b == N):
                ans += 1
print(ans)


Fixed, accepted submission:
R,G,B,N = map(int,input().split())
ans = 0
for r in range(N//R+1):
    for g in range((N-R*r)//G+1):
        if (N-R*r-G*g)%B == 0:
            ans += 1
print(ans)
Problem:
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`.  Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
S = input()
ans = 2*10**5+1
tmp = 0

S_list = list(S)
S_list.insert(0,"+")
S_list.insert(N+1,"+")

for i in reversed(range(1,N+1)):
    if (S_list[i] == "."):
        for j in range(0,i):
            if (S_list[j] == "#"):
                tmp += 1
        for j in range(i+1,len(S_list)):
            if (S_list[j] == "."):
                tmp += 1
        print(tmp)
        if tmp < ans :
            ans = tmp
            tmp = 0

print(ans)


Fixed, accepted submission:
N = int(input())
S = input()
lS = len(S)
l = 0
bw = [0, 0]
for i in range(lS):
    if (S[i] == "."):
        bw[1] += 1

ans = sum(bw)
for i in range(lS):
    if (S[i] == "#"):
        bw[0] += 1
    if (S[i] == "."):
        bw[1] -= 1

    if (ans > sum(bw)):
        ans = sum(bw)

print(ans)
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())

ans=1000-n%1000

if ans==1000:
    ans=0
print(ans)
Problem:
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.  The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.  Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.  Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

Incorrect submission (verdict: Wrong Answer):
if __name__ == "__main__":
    N = int(input())
    a = [int(input()) for _ in range(N)]

    index = 0
    count = 0
    number = 1
    flags = [0 for _ in range(N)]
    flags[index] = 1
    while(number != 2):
        number = a[index]
        index = number - 1
        count += 1
        print(flags)
        if flags[index] == 1:
            break
        else:
            flags[index] = 1

    if number == 2:
        print(count)
    else:
        print(-1)


Fixed, accepted submission:
if __name__ == "__main__":
    N = int(input())
    a = [int(input()) for _ in range(N)]

    index = 0
    count = 0
    number = 1
    flags = [0 for _ in range(N)]
    flags[index] = 1
    while(number != 2):
        number = a[index]
        index = number - 1
        count += 1
        if flags[index] == 1:
            break
        else:
            flags[index] = 1

    if number == 2:
        print(count)
    else:
        print(-1)
Problem:
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?

Incorrect submission (verdict: Wrong Answer):
s=input()
print(s[0])
print(str(len(s)-2))
print(s[-1])


Fixed, accepted submission:
a,b=map(int,input().split())
print((a-1)*(b-1))
Problem:
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i.  Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different.  Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.

Incorrect submission (verdict: Wrong Answer):
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  n=I()
  edges=[[]*n for _ in range(n)]

  for _ in range(n-1):
    a,b=LI()
    a-=1
    b-=1
    edges[a].append(b)
    edges[b].append(a)

  colors=[0]*n
  colors[0]=1
  q=collections.deque()
  q.append((0,1))

  while q:
    a,c=q.popleft()

    color=1
    for b in edges[a]:
      if color==c:
        color+=1

      if colors[b]!=0:
        continue

      colors[b]=color
      q.append((b,color))

      color+=1

  for x in colors:
    print(x)

main()
# print(main())


Fixed, accepted submission:
import math,itertools,fractions,heapq,collections,bisect,sys,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  n=I()
  G=[[]*n for _ in range(n)]
  vp=[]

  for _ in range(n-1):
    a,b=LI()
    a-=1
    b-=1
    G[a].append(b)
    G[b].append(a)
    vp.append((a,b))

  k=0
  d={}
  cs=[0]*n

  used=[False]*n
  used[0]=True
  q=collections.deque()
  q.append(0)
  while q:
    v=q.popleft()
    k=max(k,len(G[v]))
    cur=1
    for u in G[v]:
      if used[u]:
        continue
      if cur==cs[v]:
        cur+=1
      cs[u]=d[(u,v)]=d[(v,u)]=cur
      cur+=1
      used[u]=True
      q.append(u)

  print(k)
  for a,b in vp:
    print(d[(a,b)])

main()
# print(main())
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):
N = int(input())

n = 0

for i in range(N):
  n =+ i+1
print(n)

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

n = 0

for i in range(N):
  n += i+1
print(n)
Problem:
In this problem, we use the 24-hour clock.  Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?

Incorrect submission (verdict: Wrong Answer):
H, M, h, m, K = map(int, input().split())
print((60*h + m - 60*H - M)//K)

Fixed, accepted submission:
H, M, h, m, K = map(int, input().split())
print(60*h + m - 60*H - M - K)
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):
input()
DATA=input()
R=DATA.count("R")
B=DATA.count("B")
if R< B:
    print("R")
elif B>R:
    print("B")
else:
    print("No")

Fixed, accepted submission:
input()
DATA=input()
R=DATA.count("R")
B=DATA.count("B")
if R> B:
    print("Yes")
else:
    print("No")
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())

if n % 2 == 0:
  print(n/2-1)
else:
  print((n-1)/2)

Fixed, accepted submission:
n = int(input())
if n%2 == 0:
  print(int(n/2-1))
else:
  print(int((n-1)/2))