TnT/CodeNet4Repair huggingface dataset:
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):
import sys
N, Y = map(int, input().split())

num = [min(N, Y//10000), min(N, Y//5000), min(N, Y//1000)]

for i in reversed(range(0, num[0]+1)):
    for j in reversed(range(0, N-i+1)):
        if 10000*i + 5000*j > Y:
            continue
        for k in reversed(range(0, N-i-j+1)):
            if 10000*i + 5000*j + 1000*k == Y:
                print('%d %d %d' % (i, j, k))
                sys.exit()
print('-1 -1 -1')


Fixed, accepted submission:
import sys
N, Y = map(int, input().split())

num = min(N, Y//10000)
if 10000*N < Y:
    print('-1 -1 -1')
    sys.exit()

for i in reversed(range(0, num+1)):
    for j in reversed(range(0, N-i+1)):
        if 10000*i + 5000*j > Y:
            continue
        if 10000*i + 5000*j + 1000*(N-i-j) == Y:
            print('%d %d %d' % (i, j, N-i-j))
            sys.exit()
print('-1 -1 -1')
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):
c=str(input())
if c in "aiueo":
  print("YES")
else:
  print("NO")

Fixed, accepted submission:
c=str(input())
if c in "aiueo":
  print("vowel")
else:
  print("consonant")
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())
print(n%1000)

Fixed, accepted submission:
n=int(input())
a=n%1000
if a==0:
  print(0)
else:
  print(1000-n%1000)
Problem:
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

Incorrect submission (verdict: Wrong Answer):
from math import factorial
n = int(input())
lst = list(map(int,input().split()))
lst.sort()
num = lst[-1]
ans = 0
for i in range(n):
    x = factorial(num) / factorial(lst[i]) / factorial(num-lst[i])
    if (x > ans):
        ans = x
    else:
        print(lst[i-1],num)
        exit()
    

Fixed, accepted submission:
n = int(input())
A = list(map(int,input().split()))
A.sort()
y = max(A)
x = A[0]
for a in A:
    if abs(y/2-a) < abs(y/2 - x):
        x = a
print(y, x)
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):
a=int(input())
print(int(a*(a-1)/2))

Fixed, accepted submission:
a=int(input())
print(int(a*(a+1)/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):
N = int(input())
A = list(map(int, input().split()))
#A = [int(x) for x in input().split()]
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
A = [-float('inf')] + A + [float('inf')]
B.sort()
C.sort()
C = [-float('inf')] + C + [float('inf')]

total = 0

for i in range(N):
    low = 0
    high = N + 1
    while high - low > 1:
        mid1 = (low + high) // 2
        if A[mid1] < B[i]:
            low = mid1
        else:
            high = mid1
    mid1 = low
    low = 0
    high = N + 1
    while high - low > 1:
        mid2 = (low + high) // 2
        if C[mid2] <= B[i]:
            low = mid2
        else:
            high = mid2
    mid2 = high
    print(mid1, mid2)
    total += mid1 * (N + 1 - mid2)

print(total)

Fixed, accepted submission:
N = int(input())
A = list(map(int, input().split()))
#A = [int(x) for x in input().split()]
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
A = [-float('inf')] + A + [float('inf')]
B.sort()
C.sort()
C = [-float('inf')] + C + [float('inf')]

total = 0

for i in range(N):
    low = 0
    high = N + 1
    while high - low > 1:
        mid1 = (low + high) // 2
        if A[mid1] < B[i]:
            low = mid1
        else:
            high = mid1
    mid1 = low
    low = 0
    high = N + 1
    while high - low > 1:
        mid2 = (low + high) // 2
        if C[mid2] <= B[i]:
            low = mid2
        else:
            high = mid2
    mid2 = high
    total += mid1 * (N + 1 - mid2)

print(total)
Problem:
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

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

a.sort()

if n == 2:
	print(str(a[1])+" "+str(a[0]))
else:
	maximum = a[-1]
	a.pop()
	half = maximum/2
	a.append(half)
	a.sort()

	position = a.index(half)

	large = a[position+1]
	small = a[position-1]

	print(position)
	print(large)
	print(small)

	if large - (half) < (half) - small:
		print(str(maximum)+" "+str(large))
	else:
		print(str(maximum)+" "+str(small))

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

a.sort()

if n == 2:
	print(str(a[1])+" "+str(a[0]))
else:
	maximum = a[-1]
	a.pop()
	half = maximum/2
	a.append(half)
	a.sort()

	position = a.index(half)

	if position+1 >= len(a):
		print(str(maximum)+" "+str(int(a[position-1])))
		exit()
	else:
		large = int(a[position+1])
	if position-1 <= 0:
		print(str(maximum)+" "+str(int(a[position+1])))
		exit()
	else:
		small = int(a[position-1])

	if large - (half) < (half) - small:
		print(str(maximum)+" "+str(large))
	else:
		print(str(maximum)+" "+str(small))
Problem:
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.  You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.

Incorrect submission (verdict: Wrong Answer):
o=input()
e=input()
ans=""
for i in range(len(e)):
    ans+=o[i//2]
    ans+=e[i//2]
if len(o)!=len(e):
    ans+=o[len(o)-1]
print(ans)


Fixed, accepted submission:
o=input()
e=input()
ans=""
for i in range(len(e)):
    ans+=o[i]
    ans+=e[i]
if len(o)!=len(e):
    ans+=o[len(o)-1]
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):
s= list(map(str,input().split()))
print('A' + ''+ s[0] + '' +'C')

Fixed, accepted submission:
a,s,c=input().split()
print(a[0]+s[0]+c[0])
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):
# -*- coding: utf-8 -*-
"""
Created on Sat Apr 25 18:20:35 2020
"""

import sys
import numpy as np

sys.setrecursionlimit(10 ** 9)
#def input():
#    return sys.stdin.readline()[:-1]
mod = 10**9+7

#N = int(input())
N, K = map(int,input().split())
A = np.array(list(map(int,input().split())))

dp = np.arange(K + A[-1]+1) >= 0
for i in range(K, -1, -1):
    if np.all(dp[A+i]):
        dp[i] = False
#    print(i, dp[i],A+i,dp[A+i])
#print(dp)

if dp[0]:
    ans = 'Frist'
else:
    ans = 'Second'
print(ans)

Fixed, accepted submission:
# -*- coding: utf-8 -*-
"""
Created on Sat Apr 25 18:20:35 2020
"""

import sys
#import numpy as np

sys.setrecursionlimit(10 ** 9)
#def input():
#    return sys.stdin.readline()[:-1]
mod = 10**9+7

#N = int(input())
N, K = map(int,input().split())
A = list(map(int,input().split()))

dp = [i <= K for i in range(K + A[-1]+1)]
for i in range(K, -1, -1):
    for a in A:
        if dp[a+i]:
            dp[i] = False
            break

#    print(i, dp[i],A+i,dp[A+i])
#print(dp)

if dp[0]:
    ans = 'Second'
else:
    ans = 'First'
print(ans)
Problem:
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once.  Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

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

if s.startswith('k') and s.startswith('eyence'):
  print('YES')
elif s.startswith('ke') and s.startswith('yence'):
  print('YES')
elif s.startswith('key') and s.startswith('ence'):
  print('YES')
elif s.startswith('keye') and s.startswith('nce'):
  print('YES')
elif s.startswith('keyen') and s.startswith('ce'):
  print('YES')
elif s.startswith('keyenc') and s.startswith('e'):
  print('YES')
else:
  print('NO')

Fixed, accepted submission:
s = input()

if s.startswith('k') and s.endswith('eyence'):
  print('YES')
elif s.startswith('ke') and s.endswith('yence'):
  print('YES')
elif s.startswith('key') and s.endswith('ence'):
  print('YES')
elif s.startswith('keye') and s.endswith('nce'):
  print('YES')
elif s.startswith('keyen') and s.endswith('ce'):
  print('YES')
elif s.startswith('keyenc') and s.endswith('e'):
  print('YES')
else:
  print('NO')
Problem:
Write a program which computes the digit number of sum of two integers a and b.

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

Fixed, accepted submission:
while True:
	try:
		a=input().split()
	except:
		break
	b=int(a[0])+int(a[1])
	print(len(str(b)))
Problem:
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi.  How many ways are there to choose K consecutive integers?

Incorrect submission (verdict: Wrong Answer):
a,b = map(int,input().split())
print(b-a + 1)

Fixed, accepted submission:
a,b = map(int,input().split())
print(a-b + 1)
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):
#15dpro
n=int(input())
#s=
#print(n)
for i in range(1,n-1):
 #   print(i)
    if i%3==0:
        print(' '+str(i),end='')
    else:
        x=i
  #      print(x)
        while x>0 and x%10!=3:
      #      print(x)
            x//=10
        if x%10==3:
            print(' '+str(i),end='')

Fixed, accepted submission:
#15dpro
n=int(input())
s=''
#print(n)
for i in range(1,n+1):
 #   print(i)
    if i%3==0:
        s+=' '+str(i)
    else:
        x=i
  #      print(x)
        while x>0 and x%10!=3:
      #      print(x)
            x//=10
        if x%10==3:
            s+=' '+str(i)
print(s)
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):
a, b, c = map(int, input().split())

count=[a,b,c]
d =max(count)
v=0
count.remove(d)
for i in range(len(count)):
    v += count[i]

if v ==d :
    print('YES')

else:
    print('NO')

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

count=[a,b,c]
d =max(count)
v=0
count.remove(d)
for i in range(len(count)):
    v += count[i]

if v ==d :
    print('Yes')

else:
    print('No')
Problem:
We will call a string x _good_ if it satisfies the following condition:    * Condition: x can be represented as a concatenation of two copies of another string y of length at least 1.  For example, `aa` and `bubobubo` are good; an empty string, `a`, `abcabcabc` and `abba` are not good.  Eagle and Owl created a puzzle on good strings. Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem.    * 1 ≤ |s| ≤ 200   * Each character of s is one of the 100 characters represented by the integers 1 through 100.   * Among the 2^{|s|} subsequences of s, exactly N are good strings.

Incorrect submission (verdict: Wrong Answer):
def calc(n, k):
  if n==1:
    return str(k) + " " + str(k)
  if n%2==0:
    return calc(n-1, k+1) + " " + str(k) + " " + str(k)
  return str(k) + " " + calc((n-1)//2, k+1) + " " + str(k)

print(calc(int(input()), 1))

Fixed, accepted submission:
fact = [1]
for i in range(1, 101):
    fact.append(fact[-1]*i)
    
def C(a, b):
    return fact[a]//(fact[b]*fact[a-b])

def sC(a, b):
    ret = 0
    for i in range(min(b, a-b) + 1):
        ret += C(b, i) * C(a - b, i)
    return ret

def calc(n, k):
    if n<=0:
        return []
    i = 0
    while (1<<(i+1)) -1 <= n:
        i += 1
        
    ret = [str(k)] * (i+1)
    rem = n-((1<<i) - 1)
    kk = k + 1
    i += 1
    j = i // 2
    
    while j > 0:
        while True:
            tmp = sC(i, j)
            if tmp <= rem:
                ret.insert(j, str(kk))
                ret.append(str(kk))
                kk += 1
                rem -= tmp

            else:
                break
                
        j -= 1
        
    return ret + calc(rem, kk)

N = int(input())

X = calc(N, 1)
print(len(X))
print(" ".join(X))
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()
if a[-1] == b[0] and b[-1] == c[0]:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
a,b,c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
    print("YES")
else:
    print("NO")
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):
import math

a,b,c=map(int,input().split())
l=[a,b,c]
sorted(l)


print("Yes" if l[0]+l[1]==l[2] else "No")


Fixed, accepted submission:
import math

a,b,c=map(int,input().split())
l=[a,b,c]
l.sort()


print("Yes" if l[0]+l[1]==l[2] else "No")
Problem:
AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i. Because he loves big numbers, he calls a subset of the cards _good_ when the sum of the numbers written on the cards in the subset, is K or greater.  Then, for each card i, he judges whether it is _unnecessary_ or not, as follows:    * If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary.   * Otherwise, card i is NOT unnecessary.  Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.

Incorrect submission (verdict: Wrong Answer):
n,k=map(int,input().split())
l=[int(i) for i in input().split()]
#sum of sub >=k 
l.sort()
s=sum(l)
if s<k:
    print(n)
else: 
    cnt=0 
    for i in l:
        if s-i<k:
            cnt+=1 
    print(cnt)

Fixed, accepted submission:
N, K = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
ans = N
psum = 0
for i in range(N-1, -1, -1) :
    if psum+a[i] < K :
        psum += a[i]
    else :
        ans = min(ans, i)
print(ans)

'''
notes:
1-->we include elems that are less than k  
2-->if some x is necessary all elems grtr than x are also necessary 
Base for some x to be necessary:: 
     atleast one of the sums in range [sm-x,sm) must be possible 
     excluding x  
now at any instance after traversing x elements 
we make sure that we have max posible sm <k 
now 
if x+sm <k 
    we cannot include this anyway 
otherwise
    we can make a set with {sm,x} now x is important 
BOOYAH (:
'''
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
import numpy as np
import math
import sys
def sinput(): return sys.stdin.readline()
def iinput(): return int(sinput())
def imap(): return map(int, sinput().split())
def fmap(): return map(float, sinput().split())
def iarr(): return list(imap())
def farr(): return list(fmap())
def sarr(): return sinput().split()

s = list(input())
s.reverse()
v = []
v.append(list("dream"))
v.append(list("erase"))
v.append(list("eraser"))
v.append(list("dreamer"))
for i in range(4):
    v[i].reverse()

now = 0
while(1):
    check = False
    for i in range(4):
        if now+len(v[i]) > len(s): continue
        tmp = ''.join(s[now:now+len(v[i])])
        x = ''.join(v[i])
        if tmp == x:
            print(tmp, x)
            check = True
            now += len(v[i])
        if check: break
    if not check: break
    if now == len(s): break

if check: print("YES")
else: print("NO")

Fixed, accepted submission:
import numpy as np
import math
import sys
def sinput(): return sys.stdin.readline()
def iinput(): return int(sinput())
def imap(): return map(int, sinput().split())
def fmap(): return map(float, sinput().split())
def iarr(): return list(imap())
def farr(): return list(fmap())
def sarr(): return sinput().split()

s = list(input())
s.reverse()
v = []
v.append(list("dream"))
v.append(list("erase"))
v.append(list("eraser"))
v.append(list("dreamer"))
for i in range(4):
    v[i].reverse()

now = 0
while(1):
    check = False
    for i in range(4):
        if now+len(v[i]) > len(s): continue
        tmp = ''.join(s[now:now+len(v[i])])
        x = ''.join(v[i])
        if tmp == x:
            check = True
            now += len(v[i])
        if check: break
    if not check: break
    if now == len(s): break

if check: print("YES")
else: print("NO")
Problem:
We have N cards. A number a_i is written on the i-th card.   Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first.   The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

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

a.sort()

alice = sum(a[0::2])
bob = sum(a[1::2])

print(alice - bob)


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

a.sort(reverse=True)

alice = sum(a[0::2])
bob = sum(a[1::2])

print(alice - bob)
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):
s = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,s):
    ch = input()
    if ch == "AC":
        ac += 1
    elif ch == "WA":
        wa += 1
    elif ch == "TLE":
        tle +=1
    elif ch == "RE":
        re += 1

print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))

Fixed, accepted submission:
s = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(1,s+1):
    ch = input()
    if ch == "AC":
        ac += 1
    elif ch == "WA":
        wa += 1
    elif ch == "TLE":
        tle +=1
    elif ch == "RE":
        re += 1

print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
Problem:
You are parking at a parking lot. You can choose from the following two fee plans:    * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours.   * Plan 2: The fee will be B yen, regardless of the duration.  Find the minimum fee when you park for N hours.

Incorrect submission (verdict: Wrong Answer):
N, A, B = map(int, input().split())
print(min(N-A, B))

Fixed, accepted submission:
N, A, B = map(int, input().split())
print(min(N*A, B))