TnT/CodeNet4Repair huggingface dataset:
Problem:
A railroad running from west to east in Atcoder Kingdom is now complete.  There are N stations on the railroad, numbered 1 through N from west to east.  Tomorrow, the opening ceremony of the railroad will take place.  On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.  The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.  Here, it is guaranteed that F_i divides S_i.  That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.  For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
CSF = []
for n in range(N-1):
  CSF.append([int(csf) for csf in input().split()])
print(CSF)

for i in range(N-1):
  total_time = 0
  for j in range(i, N-1):
    c12, s1, f1 = CSF[j]
    if s1 >= total_time:
      total_time = s1 + c12
    else:
      how_many_arrival_train = (total_time - s1) // f1
      how_long_waiting_time = (total_time - s1) % f1
      if how_long_waiting_time == 0:
        total_time = s1 + f1 * how_many_arrival_train + c12
      else:
        total_time = s1 + f1 * (how_many_arrival_train + 1) + c12
  print(total_time)
print(0)

Fixed, accepted submission:
N = int(input())
CSF = []
for n in range(N-1):
  CSF.append([int(csf) for csf in input().split()])

for i in range(N-1):
  total_time = 0
  for j in range(i, N-1):
    c12, s1, f1 = CSF[j]
    if s1 >= total_time:
      total_time = s1 + c12
    else:
      how_many_arrival_train = (total_time - s1) // f1
      how_long_waiting_time = (total_time - s1) % f1
      if how_long_waiting_time == 0:
        total_time = s1 + f1 * how_many_arrival_train + c12
      else:
        total_time = s1 + f1 * (how_many_arrival_train + 1) + c12
  print(total_time)
print(0)
Problem:
Write a program which reads three integers, and prints them in ascending order.

Incorrect submission (verdict: Wrong Answer):
Arr = []
Arr = input().split()
for i in range(2):
    Arr[i] = int(Arr[i])

Fixed, accepted submission:
Arr = []
Arr = input().split()
for i in range(3):
    Arr[i] = int(Arr[i])
for j in range(0, 2, 1):
    for i in range(0 , 2-j , 1):
        if(Arr[i] > Arr[i+1]):
            buf = Arr[i]
            Arr[i] = Arr[i+1]
            Arr[i+1]=buf
print("{} {} {}".format(Arr[0],Arr[1],Arr[2]))
Problem:
We held two competitions: Coding Contest and Robot Maneuver.  In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen.  DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.

Incorrect submission (verdict: Wrong Answer):
from sys import stdin
x,y = map(int, stdin.readline().rstrip().split())
z = {1:300000, 2:200000,3:100000}
ans = 0
if x == 1 and y == 1:
    print(1000000)
else:
    if x < 4:
        ans += z[x]
    elif y < 4:
        ans += z[y]
print(ans)


Fixed, accepted submission:
from sys import stdin
x,y = map(int, stdin.readline().rstrip().split())

if x == 1 and y == 1:
    print(1000000)
    exit()
else:
    pass
ans = 0
if x == 1:
    ans += 300000
elif x == 2:
    ans += 200000
elif x == 3:
    ans += 100000
else:
    pass
if y == 1:
    ans += 300000
elif y == 2:
    ans += 200000
elif y == 3:
    ans += 100000
else:
    pass

print(ans)
Problem:
We have a grid with N rows and M columns of squares. Initially, all the squares are white.  There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted.  Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Incorrect submission (verdict: Wrong Answer):
if __name__ == "__main__":
    N, M, K = map(int, input().split())

    for k in range(1, N + 1):
        for l in range(1, M + 1):
            b = k * (M - l) + (N-k) * l
            if b == K:
                print("Yes")
    
    print("No")


Fixed, accepted submission:
if __name__ == "__main__":
    N, M, K = map(int, input().split())

    for k in range(0, N + 1):
        for l in range(0, M + 1):
            b = k * (M - l) + (N - k) * l
            if b == K:
                print("Yes")
                exit()

    print("No")
Problem:
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.  The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.  How many different PIN codes can he set this way?  Both the lucky number and the PIN code may begin with a 0.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a=input()
ans=0
for i in range(1000):
    i=str(i).zfill(3)
    x=a.find(i[0])
    y=a[x+1:].find(i[1])
    z=a[y+1:].find(i[2])
    if x!=-1 and y!=-1 and z!=-1:
        ans+=1
print(ans)

Fixed, accepted submission:
n=int(input())
a=input()
ans=0

import collections as co
d=co.defaultdict(list)
for x,i in enumerate(a):
    d[i].append(x)

se=set()
for i in d.keys():
    se.add(i)

for pin in range(1000):
    pin=str(pin)
    pin=pin.zfill(3)
    now=-1
    cnt=0
    flag=True
    
    for num in pin:
        if num not in se:
            flag=False
    
        for i in d[num]:
            if i>now:
                now=i
                cnt+=1
                break

    if cnt==3 and flag==True:
        ans+=1
print(ans)
Problem:
AtCoDeer has three cards, one red, one green and one blue.   An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card.   We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer.   Is this integer a multiple of 4?

Incorrect submission (verdict: Wrong Answer):
print(['No','Yes'][int(''.join(list(input().split())))%4==0])

Fixed, accepted submission:
print('NYOE S'[int(''.join(input().split()))%4==0::2])
Problem:
Nathan O. Davis くんは，あるゲームを攻略中であり，非常にレアなアイテムを手に入れようと四苦八苦していた．このレアなアイテムは，カジノのスロットマシーンで特殊な絵柄を一列に揃えたときに手に入れることができる．スロットマシーンには _N_ 個のリールがあり，ボタンを一回押すと現在回転している最も左側のリールが停止する．そのため，このレアアイテムを手に入れるためには _N_ 回ボタンを押す必要がある．また，特殊な絵柄で停止させるためにはあるタイミングでボタンを押す必要がある．ところが，このボタンを押すタイミングは非常にシビアであるため，なかなか上手くいかずに困っていた．そこで，Nathan くんはメモリビューアを利用して，ゲーム中におけるメモリの値を確認しながらゲームを解析することにした．  Nathan くんの解析の結果，その特殊な絵柄でリールを停止させるためには，ボタンが押された時に，メモリの 007E0D1F 番地に書き込まれた「乱数」が特定の値になっている必要があることを突き止めた．乱数の値は 1 フレーム毎に線形合同法によって変化しており，またボタンが押されたかどうかの判定は 1 フレーム毎に 1 回行われていることも分かった．ここで，線形合同法とは擬似乱数を生成するための方法の一つであり，以下の式によって値を決定する．  _x'_ = ( _A_ × _x_ \+ _B_ ) mod _C_  ここで， _x_ は現在の乱数の値， _x'_ は次の乱数の値， _A_ , _B_ , _C_ は何らかの定数である．また， _y_ mod _z_ は _y_ を _z_ で割ったときの余りを表す．  例えば，2 個のリールを持つスロットマシーンで _A_ = 5， _B_ = 7， _C_ = 11 で，最初の「乱数」の値が 10 であったとする．そして，特殊な絵柄でリールを止めるための条件は，左側のリールが 2，右側のリールが 4 であったとする．このとき，1 フレーム目における乱数の値は (5 × 10 + 7) mod 11 = 2 であるため，1 フレーム目にボタンを押せば左側のリールは特殊な絵柄で停止する．続く 2 フレーム目での乱数の値は (5 × 2 + 7) mod 11 = 6 であるから，2 フレーム目にボタンを押しても右側のリールは特殊な絵柄では停止しない．続く 3 フレーム目では乱数の値が (5 × 6 + 7 ) mod 11 = 4 になるので，3 フレーム目にボタンを押せば右側のリールは特殊な絵柄で停止する．よって，最短 3 フレームで全てのリールを特殊な絵柄で停止されることができ，レアなアイテムを手に入れることができる．  あなたの仕事は，最短で何フレーム目に全てのリールを特殊な絵柄で停止させることができるかを求めるプログラムを書いて，Nathan くんがレアなアイテムを手に入れられるよう助けてあげることである．

Incorrect submission (verdict: Wrong Answer):
class LCG:
    def __init__(self, a, b, c, x):
        self.a = a
        self.b = b
        self.c = c
        self.x = x
        self.count = 0

    def get_next(self):
        self.x = (self.a * self.x + self.b) % self.c
        self.count += 1
        return self.x

    def get_count(self):
        return self.count

if __name__ == "__main__":
    while True:
        n, a, b, c, x = map(int, input().split())
        if n == 0: break
        ran = LCG(a, b, c, x)
        reel = map(int, input().split())
        for r in reel:
            while x != r:
                x = ran.get_next()
                if ran.get_count() > 10000:
                    print(-1)
                    break
            else:
                continue
            break
        else:
            print(ran.get_count())

Fixed, accepted submission:
class LCG:
    def __init__(self, a, b, c, x):
        self.a = a
        self.b = b
        self.c = c
        self.x = x
        self.count = 0

    def get_next(self):
        self.x = (self.a * self.x + self.b) % self.c
        self.count += 1
        return self.x

    def get_count(self):
        return self.count

if __name__ == "__main__":
    while True:
        n, a, b, c, x = map(int, input().split())
        if n == 0: break
        ran = LCG(a, b, c, x)
        reel = map(int, input().split())
        try:
            for r in reel:
                while x != r:
                    x = ran.get_next()
                    if ran.get_count() > 10000:
                        raise Exception
                else:
                    x = ran.get_next() 
            print(ran.get_count() - 1)
        except Exception:
            print(-1)
Problem:
A railroad running from west to east in Atcoder Kingdom is now complete.  There are N stations on the railroad, numbered 1 through N from west to east.  Tomorrow, the opening ceremony of the railroad will take place.  On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.  The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.  Here, it is guaranteed that F_i divides S_i.  That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.  For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Incorrect submission (verdict: Wrong Answer):
import sys
input = sys.stdin.readline
import numpy as np

 
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

n = I()
mylist = tuple(tuple(MI()) for i in range(n-1))

def main(j):
    time = 0
    for i in range(j,n-1):
        c, s, f = mylist[i]
        if time < s:
            time = s + c
        else:
            time += (f - (time - s) % f) + c
    print(time)
    
    
for i in range(n-1):
    main(i)
print(0)

Fixed, accepted submission:
import sys
input = sys.stdin.readline
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

n = I()
mylist = tuple(tuple(MI()) for i in range(n-1))

def main(j):
    time = 0
    for i in range(j,n-1):
        c, s, f = mylist[i]
        if time < s:
            time = s
        elif (time - s) % f != 0:
            time += (f - (time - s) % f) 
        time += c
    print(time)
    
def main2():    
    for i in range(n-1):
        main(i)
    print(0)
main2()
Problem:
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.

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

Fixed, accepted submission:
exec("print("+input()+")")
Problem:
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi:    * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person.  Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.

Incorrect submission (verdict: Wrong Answer):
A,B,K=map(int,input().split())
for i in range(K):
  if A%2==1:
    A-=1
  else:
    pass
  A/=2
  B+=A
  if B%2==1:
    B-=1
  else:
    pass
  B/=2
  A+=B
print(A,B)  

Fixed, accepted submission:
A,B,K=map(int,input().split())
while True:
  A-=A%2
  A,B=A//2,B+A//2
  K-=1
  if K==0:
    break
  B-=B%2
  B,A=B//2,A+B//2
  K-=1
  if K==0:
    break
print(A,B)
Problem:
Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated.

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

S = [list(map(int, input().split())) for _ in range(D)]

contest = [0 for _ in range(26)]
ans = 0

for i in range(D):
    t = int(input())
    contest[t-1] += i+1
    ans += S[i][t-1]
    for j, con in enumerate(contest):
        if con:
            ans -= C[j]*(i+1-con)
        else:
            ans -= C[j]*(i+1)
    print(ans)

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

S = [list(map(int, input().split())) for _ in range(D)]

contest = [0 for _ in range(26)]
ans = 0

for i in range(D):
    t = int(input())
    contest[t-1] = i+1
    ans += S[i][t-1]
    for j, con in enumerate(contest):
        if con:
            ans -= C[j]*(i+1-con)
        else:
            ans -= C[j]*(i+1)
    print(ans)
Problem:
Mr.X, who the handle name is T, looked at the list which written N handle names, S_1, S_2, ..., S_N.   But he couldn't see some parts of the list. Invisible part is denoted `?`.      Please calculate all possible index of the handle name of Mr.X when you sort N+1 handle names (S_1, S_2, ..., S_N and T) in lexicographical order.   Note: If there are pair of people with same handle name, either one may come first.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
S=[input() for i in range(N)]
T=input()
L=1
H=N+1
for i in range(N):
    A=S[i].replace("?","a")
    Z=S[i].replace("?","z")
    if T<A:
        H-=1
    if Z<T:
        L+=1
st=""
for i in range(L,H+1):
    st+=str(i)+" "
print(st)


Fixed, accepted submission:
N=int(input())
S=[input() for i in range(N)]
T=input()
L=1
H=N+1
for i in range(N):
    A=S[i].replace("?","a")
    Z=S[i].replace("?","z")
    if T<A:
        H-=1
    if Z<T:
        L+=1
st=""
for i in range(L,H+1):
    st+=str(i)
    if i==H:
        continue
    st+=" "
print(st)
Problem:
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.  

Incorrect submission (verdict: Wrong Answer):
x, y = [int(_) for _ in input().split()]
c1 = [1,3,5,7,8,10,12]
c2 = [4,6,9,11]
c3 = [2]
for t in [c1, c2, c3]:
    if (x in t) and (y in t):
        print("Yes")
print("No")

Fixed, accepted submission:
x, y = [int(_) for _ in input().split()]
c1 = [1,3,5,7,8,10,12]
c2 = [4,6,9,11]
c3 = [2]
cnt = 0
for t in [c1, c2, c3]:
    if (x in t) and (y in t):
        cnt = 1
if cnt == 1:
    print("Yes")
else:       
    print("No")
Problem:
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.  Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Incorrect submission (verdict: Wrong Answer):
a,b,c,x,y = map(int,input().split())
p = 0

if x > y:
    p += (x - y) * a
    x = y
if x < y:
    p += (y - x) * b
    y = x
if x % 2 == 1:
    p += x + y
    x -= 1
    y -= 1
if x > 0:
    n = x // 2 * c
    m = x * a + y * b
    p += min(n,m)

print(p)

Fixed, accepted submission:
a,b,c,x,y = map(int,input().split())
c *= 2
ans = 10 ** 16
for i in range(max(x,y)+1):
    m = 0
    m += (a*(max(0,x-i)))+(b*(max(0,y-i)))+(c*i)
    ans = min(ans,m)
print(ans)
Problem:
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.  Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Incorrect submission (verdict: Wrong Answer):
class Atcoder:


    def Contest0901(self):
        lines = []
        for i in range(3):
            line = input()
            if line:
                lines.append(line)
            else:
                break
        print(lines)
        return lines[0][0] + lines[1][1] + lines[2][2]

if __name__ == '__main__':
    atcoder = Atcoder()
    output = atcoder.Contest0901()
    print(output)

Fixed, accepted submission:
class Atcoder:


    def Contest0901(self):
        lines = []
        for i in range(3):
            line = input()
            if line:
                lines.append(line)
            else:
                break
        return lines[0][0] + lines[1][1] + lines[2][2]


if __name__ == '__main__':
    atcoder = Atcoder()
    output = atcoder.Contest0901()
    print(output)
Problem:
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.

Incorrect submission (verdict: Wrong Answer):
n, k = map(int, input().split())
a = [list(map(int, input().split())) for i in range(n)]
i = 0

while True:
  if (k - a[i][1]) > 0:
    k -= a[i][1]
    ++i
  else:
    print(a[i][0])
    break


Fixed, accepted submission:
n, k = map(int, input().split())
a = sorted([(list(map(int, input().split()))) for i in range(n)])
i = 0

while True:
  if (k - a[i][1]) > 0:
    k -= a[i][1]
    i += 1
  else:
    print(a[i][0])
    break
Problem:
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.  If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
for i in range(n):
    t, x, y = map(int, input().split())
    if x + y < t or (x + y + t) % 2:
        print("No")
        exit()
print("Yes")

Fixed, accepted submission:
n = int(input())
for i in range(n):
    t, x, y = map(int, input().split())
    if x + y > t or (x + y + t) % 2:
        print("No")
        exit()
print("Yes")
Problem:
You are given a string S.  Takahashi can insert the character `A` at any position in this string any number of times.  Can he change S into `AKIHABARA`?

Incorrect submission (verdict: Wrong Answer):
need = ["K","I","H","B","R"]
noneed = ["C","D","E","F","G","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
s = "KIHBRC"

def judge(s: str) -> bool:
    pos = [0]*5
    for i,c in enumerate(need):
        if s.count(c) != 1:
            return False
        else:
            pos[i] = s.index(c)
            
    for c in noneed:
        if s.count(c) != 0:
            return False

    
    if pos[0] > 2:
        return False
    
    if pos[1] - pos[0] > 1:
        return False
    
    if pos[2] - pos[1] > 1:
        return False
    
    if pos[3] - pos[2] > 2:
        return False
    
    if pos[4] - pos[3] > 2:
        return False
    
    if len(s) - pos[4] > 2:
        return False
    
    return True


print(judge(s))

Fixed, accepted submission:
import re

s = input()

if re.match("A?KIHA?BA?RA?$", s):
    print("YES")
else:
    print("NO")
Problem:
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.  If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
t=0
x=0
y=0
for i in range(N):
    t_next,x_next,y_next=map(int,input().split())
    zikan=t_next - t
    kyori=abs(x_next-x)+abs(y_next-y)
    if kyori>zikan or (zikan-kyori)%2==1:
        print('NO')
        exit(0)
print('YES')

Fixed, accepted submission:
N=int(input())
t=0
x=0
y=0
for i in range(N):
    t_next,x_next,y_next=map(int,input().split())
    zikan=t_next - t
    kyori=abs(x_next-x)+abs(y_next-y)
    if kyori>zikan or (zikan-kyori)%2==1:
        print('No')
        exit(0)
    t,x,y=t_next,x_next,y_next
print('Yes')
Problem:
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:    * 1 \leq x,y,z   * x^2 + y^2 + z^2 + xy + yz + zx = n  Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Incorrect submission (verdict: Time Limit Exceeded):
from collections import defaultdict

D = defaultdict(int)

def f(x,y,z):
    return x**2+y**2+z**2+x*y+y*z+z*x
    return (x+y+z)**2+(x*y+y*z+z*x)

N=int(input())
for i in range(1,N+1):
    ans=0
    for x in range(1,101):
        for y in range(1,101):
            for z in range(1,101):
                ret = f(x,y,z)
                if ret==N:
                    ans+=1
    print(ans)


Fixed, accepted submission:
from collections import defaultdict

D = defaultdict(int)

def f(x,y,z):
    return x**2+y**2+z**2+x*y+y*z+z*x
    return (x+y+z)**2+(x*y+y*z+z*x)

N=int(input())
for x in range(1,101):
    for y in range(1,101):
        for z in range(1,101):
            ret = f(x,y,z)
            if ret>N:
                break
            D[f(x,y,z)]+=1

for i in range(1,N+1):
    print(D[i])
Problem:
Write a program which reads two integers a and b, and calculates the following values:    * a ÷ b: d (in integer)   * remainder of a ÷ b: r (in integer)   * a ÷ b: f (in real number)

Incorrect submission (verdict: Wrong Answer):
(a, b) = [int(i) for i in input().split()]
print(a//b,a%b,a/b)

Fixed, accepted submission:
(a, b) = [int(i) for i in input().split()]
d = a // b #d = int(a/b)
r = a%b
f = a /b
print('%s %s %.5f' % (d, r, f))