TnT/CodeNet4Repair huggingface dataset:
Problem:
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter.  Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition:    * Every row and column in A' can be read as a palindrome.  Determine whether he can create a matrix satisfying the condition.

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

h, w = map(int,input().split())
a = [list(input()) for _ in range(h)]

c=Counter(sum(a,[]))

if h%2==0 and w%2==0:
    g1=0
    g2=0
    g4=(h//2)*(w//2)
elif h%2==1 and w%2==0:
    g1=0
    g2=w//2
    g4=(h//2)*(w//2)
elif h%2==0 and w%2==1:
    g1=0
    g2=h//2
    g4=(h//2)*(w//2)
else:
    g1=1
    g2=(h//2)+(w//2)
    g4=(h//2)*(w//2)


print(c)

g1_cnt=0
g2_cnt=0
g4_cnt=0

for key in c:
    if c[key]%4==1:
        g1_cnt+=1
        g4_cnt+=c[key]//4
    elif c[key]%4==2:
        g2_cnt+=1
        g4_cnt+=c[key]//4
    elif c[key]%4==2:
        g1_cnt+=1
        g2_cnt+=1
        g4_cnt+=c[key]//4
    else:
        g4_cnt+=c[key]//4

# print(g1,g2,g4)

# print(g1_cnt,g2_cnt,g4_cnt)

print('Yes' if g1==g1_cnt and g2==g2_cnt and g4==g4_cnt else 'No')


Fixed, accepted submission:
from collections import Counter
import math

h, w = map(int,input().split())
a = [list(input()) for _ in range(h)]

c=Counter(sum(a,[]))

if h%2==0 and w%2==0:
    g1=0
    g2=0
    g4=(h//2)*(w//2)
elif h%2==1 and w%2==0:
    g1=0
    g2=w//2
    g4=(h//2)*(w//2)
elif h%2==0 and w%2==1:
    g1=0
    g2=h//2
    g4=(h//2)*(w//2)
else:
    g1=1
    g2=(h//2)+(w//2)
    g4=(h//2)*(w//2)

g1_cnt=0
g2_cnt=0
g4_cnt=0

for key in c:
    if c[key]%4==1:
        g1_cnt+=1
        g4_cnt+=c[key]//4
    elif c[key]%4==2:
        g2_cnt+=1
        g4_cnt+=c[key]//4
    elif c[key]%4==2:
        g1_cnt+=1
        g2_cnt+=1
        g4_cnt+=c[key]//4
    else:
        g4_cnt+=c[key]//4

# print(g1,g2,g4)

# print(g1_cnt,g2_cnt,g4_cnt)

flag=False
if g1==g1_cnt:
    if g2==g2_cnt and g4==g4_cnt:
        flag=True
    elif g2_cnt<g4_cnt and g2>g4 and g2*2+g4*4==g2_cnt*2+g4_cnt*4:
        flag=True

print('Yes' if flag else 'No')
Problem:
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
S = input()
print(max(len(set[:i]&set[i:])) for i in range(N))

Fixed, accepted submission:
N = int(input())
S = input()
print(max(len(set(S[:i])&set(S[i:])) for i in range(N)))
Problem:
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.  Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.

Incorrect submission (verdict: Wrong Answer):
s=input()
ret=1
for i in range(97, 97+26):
  x=s.count(chr(i))
  ret*=x+1
print(ret-1)

Fixed, accepted submission:
n=input()
s=input()
ret=1
for i in range(97, 97+26):
  x=s.count(chr(i))
  ret*=x+1
  ret%=1000000007
print((ret-1)%1000000007)
Problem:
Write a program which reads three integers, and prints them in ascending order.

Incorrect submission (verdict: Wrong Answer):
a,b,c = list(map(int,input().split()))
if a < b:
	if a < c:
		if b < c:
			print(a < b < c)
		else:
			print("a < c < b")
	else:
		print("c < a < b")
elif a > b:
	if b < c:
		if a < c:
			print("b < a < c")
		else:
			print("b < c < a")
	else:
		print("c < b < a")

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

num.sort()

print(num[0], num[1], num[2])
Problem:
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops.  Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto.  Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements.                 void call(int n){   int i = 1;  CHECK_NUM:   int x = i;   if ( x % 3 == 0 ){     cout << " " << i;     goto END_CHECK_NUM;   }  INCLUDE3:   if ( x % 10 == 3 ){     cout << " " << i;     goto END_CHECK_NUM;   }   x /= 10;   if ( x ) goto INCLUDE3;  END_CHECK_NUM:   if ( ++i <= n ) goto CHECK_NUM;    cout << endl; }

Incorrect submission (verdict: Wrong Answer):
n = int(input())
for i in range(1, n + 1):
    x = i
    if x % 3 == 0:
        print(" ", i, sep="")
    else:
        while x > 0:
            if x % 10 == 3:
                print(" ", i, sep="")
                break
            x //= 10
print()

Fixed, accepted submission:
n = int(input())
for i in range(1, n + 1):
    if i % 3 == 0 or "3" in str(i):
        print(" ", i, sep="", end="")
print()
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):
import sys
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline()

A, B, C, X = [int(input()) for i in range(4)]

cnt = 0
for a in range(A):
	for b in range(B):
		for c in range(C):
			if 500*a + 100*b + 50*c == X:
				cnt += 1

print(cnt)

Fixed, accepted submission:
import sys
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline()

A, B, C, X = [int(input()) for i in range(4)]

cnt = 0
for a in range(A+1):
	for b in range(B+1):
		for c in range(C+1):
			if 500*a + 100*b + 50*c == X:
				cnt += 1

print(cnt)
Problem:
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions:    * The graph is simple and connected.   * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S.  It can be proved that at least one such graph exists under the constraints of this problem.

Incorrect submission (verdict: Wrong Answer):
n = int(input())

if n%2==0:
    lis=[[i,n+1-i] for i in range(1,int(n/2)+1)]
else:
    lis=[[i,n-i] for i in range(1,int((n-1)/2)+1)] + [[n]]

connect_lis=[]

for i in lis:
    for j in i:
        for k in lis:
            if k!=i:
                for p in k:
                    if j < p:
                        connect_lis.append([j,p])
        

for i in connect_lis:
    print(i[0],'',i[1])

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

if n%2==0:
    lis=[[i,n+1-i] for i in range(1,int(n/2)+1)]
else:
    lis=[[i,n-i] for i in range(1,int((n-1)/2)+1)] + [[n]]

connect_lis=[]

for i in lis:
    for j in i:
        for k in lis:
            if k!=i:
                for p in k:
                    if j < p:
                        connect_lis.append([j,p])
        
print(len(connect_lis))
for i in connect_lis:
    print(i[0],i[1])
Problem:
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.

Incorrect submission (verdict: Wrong Answer):
s=int(input())
h=s/3600
s=s%3600
m=s/60
s=s%60
print(str(h),":",str(m),":",str(s))

Fixed, accepted submission:
import math

s=int(input())
h=math.floor(s/3600)
s=math.floor(s%3600)
m=math.floor(s/60)
s=math.floor(s%60)
print(str(h)+":"+str(m)+":"+str(s))
Problem:
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

Incorrect submission (verdict: Wrong Answer):
n1,n2,n3,n4 = map(int,input().split())

if n1==1 or n2==1 or n3==1 or n4==1:
  if n1==7 or n2==7 or n3==7 or n4==7:
    if n1==9 or n2==9 or n3==9 or n4==9:
      if n1==4 or n2==4 or n3==4 or n4==4:
        print("Yes")
      else:
        print("No")
    else:
      print("No")
  else:
    print("No")
else:
  print("No")

Fixed, accepted submission:
n1,n2,n3,n4 = map(int,input().split())
 
if n1==1 or n2==1 or n3==1 or n4==1:
  if n1==7 or n2==7 or n3==7 or n4==7:
    if n1==9 or n2==9 or n3==9 or n4==9:
      if n1==4 or n2==4 or n3==4 or n4==4:
        print("YES")
      else:
        print("NO")
    else:
      print("NO")
  else:
    print("NO")
else:
  print("NO")
Problem:
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers).  The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.

Incorrect submission (verdict: Wrong Answer):
import itertools
n = int(input())
c = list(itertools.product(['S', 'H', 'C', 'D'], range(1,14)))
for _ in range(n):
    s, num = input().split()
    c.remove((s, int(num)))
for i in range(4*13 - n):
    print(*c, sep = '\n')

Fixed, accepted submission:
import itertools
n = int(input())
c = list(itertools.product(['S', 'H', 'C', 'D'], range(1,14)))
for _ in range(n):
    s, num = input().split()
    c.remove((s, int(num)))
for i in range(4*13 - n):
    print(c[i][0], c[i][1])
Problem:
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key.  To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:    * The `0` key: a letter `0` will be inserted to the right of the string.   * The `1` key: a letter `1` will be inserted to the right of the string.   * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.  Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

Incorrect submission (verdict: Wrong Answer):
S = input()
str = []
print(len(S))
result = ""
flg = 0
for i in range(len(S)):
    if flg == 1:
        flg  = 0
        continue
    if S[len(S)-1 - i] == 'b':
        flg = 1
    else:
        str.append(S[len(S) - 1 - i])
str.reverse()
for i in str:
    result += i
print(result)


Fixed, accepted submission:
S = input()
str = []
result = ""
flg = 0
for i in range(len(S)):
    if S[i] == 'B':
        result = result[:-1]
    else:
        result += S[i]
print(result)
Problem:
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

Incorrect submission (verdict: Wrong Answer):
N, M = map(int,input().split())
a = []
b = []
for i in range(M):
    a.append(list(map(int, input().split())))
print (a)
a=[flatten for inner in a for flatten in inner]


for i in range(N):
    b.append((a.count(i+1)))
print (b)


Fixed, accepted submission:
N, M = map(int,input().split())
a = []
b = []
for i in range(M):
    a.append(list(map(int, input().split())))

a=[flatten for inner in a for flatten in inner]
 
for i in range(N):
    b.append((a.count(i+1)))


c = list(map(str, b))

answer = '\n'.join(c)
print(answer)
Problem:
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.  According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

Incorrect submission (verdict: Wrong Answer):

n, y = map(int, input().split())

y = y // 1000
print(n, y)


def find_by_for(v, n):
    """
        the tuple by example
        ((0, 0, 0), (10, 5, 1), 100, 10)
    """
    queue = []
    queue.append(((0, 0, 0), (1, 5, 10), v, n))
    while queue:
        xs, fv, v, n = queue.pop()
        if v == 0 and n == 0:
            return "%d %d %d"%(xs[2], xs[1], xs[0])
        elif not fv or v <= 0 or n <= 0:
            pass
        else:
            queue.append((xs, fv[:-1], v, n))
            z = list(xs) 
            idx = len(fv) -1
            z[idx] +=1
            queue.append((tuple(z), fv, v - fv[idx], n -1))
    return "-1 -1 -1"

print(find_by_for(y, n))
    


Fixed, accepted submission:
def find_min(v):
    x = v % 10
    return (x % 5, x//5, v // 10)

def find_max(v):
    return (v, 0, 0)


def conv10to5(xs):
    return (xs[0], xs[1] +2, xs[2] - 1)

def undoconv10to5(xs):
    return (xs[0], xs[1] -2, xs[2] + 1)


def conv5to1(xs):
    return (xs[0]+5, xs[1] -1, xs[2])


def next(xs):
    if xs[2] == 0:
        if xs[1] == 0:
            return None
        else:
            return conv5to1(xs)
    else:
        if xs[1] < 7:
            return conv10to5(xs)
        else:
            return conv5to1(undoconv10to5(undoconv10to5(undoconv10to5(xs))))


def find(v, n):
    xs = find_min(v)
    m = sum(xs)
    if m > n or n > v:
        return (-1, -1, -1)

    while xs:
        if sum(xs) == n:
            return xs
        xs = next(xs)
    return (-1, -1, -1)



if __name__ == "__main__":
    n, y = map(int, input().split())
    y = y // 1000
    print("{2} {1} {0}".format(*find(y, n)))
Problem:
Joisino is planning to open a shop in a shopping street.  Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.  There are already N stores in the street, numbered 1 through N.  You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.  Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.  Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

Incorrect submission (verdict: Wrong Answer):

def profit(xs, n, F, P):
    return sum([P[i][sum(map(lambda x: x[0]*x[1], zip(xs, F[i])))] for i in range(n)])


def vector(k):
    return [1 if k & 1 << i else 0 for i in range(10)]


def foo(n, F, P):
    return max([profit(vector(i), n, F, P) for pat in range(1, 1024)])


if __name__ == "__main__":
    n = int(input())
    F = {}
    P = {}
    for i in range(n):
        F[i] = list(map(int, input().split()))
    for i in range(n):
        P[i] = list(map(int, input().split()))
        
    print(foo(n, F, P))

Fixed, accepted submission:

def profit(xs, n, F, P):
    return sum([P[i][sum(map(lambda x: x[0]*x[1], zip(xs, F[i])))] for i in range(n)])


def vector(k):
    return [1 if k & 1 << i else 0 for i in range(10)]


def foo(n, F, P):
    return max([profit(vector(pat), n, F, P) for pat in range(1, 1024)])


if __name__ == "__main__":
    n = int(input())
    F = {}
    P = {}
    for i in range(n):
        F[i] = list(map(int, input().split()))
    for i in range(n):
        P[i] = list(map(int, input().split()))
        
    print(foo(n, F, P))
Problem:
_n_ _q_   _name 1 time1_   _name 2 time2_   ...   _name n timen_    In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space.  In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.

Incorrect submission (verdict: Wrong Answer):
n, q = map(int, input().split())

p = []
time = 0

for i in range(n):
	p.append(input().split())
	p[i][1] = int(p[i][1])

print("{0} {1}".format(p[0][0], time))

while(len(p) > 0):
	if (p[0][1]<= q):
		time += p[0][1]
		print("{0} {1}".format(p[0][0], time))
		p.pop(0)
	else:
		time += q
		p[0][1] -= q
		p.append(p[0])
		p.pop(0)

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

p = []
time = 0

for i in range(n):
	p.append(input().split())
	p[i][1] = int(p[i][1])

while(len(p) > 0):
	if (p[0][1]<= q):
		time += p[0][1]
		print("{0} {1}".format(p[0][0], time))
		p.pop(0)
	else:
		time += q
		p[0][1] -= q
		p.append(p[0])
		p.pop(0)
Problem:
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i.  How many squares i satisfy both of the following conditions?    * The assigned number, i, is odd.   * The written integer is odd.

Incorrect submission (verdict: Wrong Answer):
lst=list(map(int,input().split()))
cnt=0
for i,j in enumerate(lst,1):
    if i%2!=0 and j%2!=0:
        cnt+=1
print(cnt)

Fixed, accepted submission:
n=int(input())
lst=list(map(int,input().split()))
cnt=0
for i,j in enumerate(lst,1):
    if i%2==0:
        pass
    else:
        if j%2==0:
            pass
        else:
            cnt+=1
print(cnt)
Problem:
Write a program which reads three integers, and prints them in ascending order.

Incorrect submission (verdict: Wrong Answer):
print(list(sorted(map(int,input().split()))))

Fixed, accepted submission:
print(*list(sorted(map(int,input().split()))))
Problem:
Write a program which solve a simultaneous equation:      ax + by = c   dx + ey = f      The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.

Incorrect submission (verdict: Time Limit Exceeded):
import itertools
import operator

dataset = []
while True:
    try:
        dataset.append(input())
    except EOFError:
        break

for item in dataset:
    a, b, c, d, e, f = [int(i) for i in item.split()]
    for i in range(-1000, 1001):
        for j in range(-1000, 1001):
            if a * i + b * j == c and d * i + e * j == f:
                print("{:.3f} {:.3f}".format(i, j))

Fixed, accepted submission:
import sys
dataset = sys.stdin.readlines()

for item in dataset:
    a, b, c, d, e, f = list(map(int, item.split()))

    y = (c * d - f * a)/(b * d - e * a)
    x = (c * e - b * f) / (a * e - b * d)
    if x == 0:
        x = 0
    print("{:.3f} {:.3f}".format(x, y))
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):
import math
a,b=map(str,input().split())
d=math.sqrt(int(a+b))


Fixed, accepted submission:
import math
a,b=input().split()
d=int(a+b)
if math.sqrt(d)%1==0:
    print("Yes")
else:
    print("No")
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())
for i in range(a,a+k):
  print(i) 
for i in range(b-k,b):
  print(i)

Fixed, accepted submission:
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:k])|set(r[-k:])):print(i)
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())
ans=[0]*4
for i in range(n):
  s=input()
  if s=='AC':
    ans[0]+=1
  elif s=='WA':
    ans[1]+=1
  elif s=='TLE':
    ans[2]+=1
  else:
    ans[3]+=1
for i in range(4):
  print("AC x "+str(ans[i]))

Fixed, accepted submission:
n=int(input())
ans=[0]*4
for i in range(n):
  s=input()
  if s=='AC':
    ans[0]+=1
  elif s=='WA':
    ans[1]+=1
  elif s=='TLE':
    ans[2]+=1
  else:
    ans[3]+=1
print("AC x "+str(ans[0]))
print("WA x "+str(ans[1]))
print("TLE x "+str(ans[2]))
print("RE x "+str(ans[3]))
Problem:
Takahashi loves numbers divisible by 2.  You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.  Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.  For example,    * 6 can be divided by 2 once: 6 -> 3.   * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.   * 3 can be divided by 2 zero times.

Incorrect submission (verdict: Wrong Answer):

n = int(input())
count = 0

while 2 ** count <= 100:
    count += 1
print(count)


Fixed, accepted submission:

n = int(input())
count = 0

while 2 ** count <= n:
    count += 1
print(2 ** (count-1))
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())
x=n//2
if x<k:
    print("NO")
else:
    print("YES")


Fixed, accepted submission:
n,k=map(int,input().split())
x=-(-n//2)
if x<k:
    print("NO")
else:
    print("YES")