TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n. It is known that each of the n integers 1,...,n appears at least once in this sequence.  For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.

Incorrect submission (verdict: Wrong Answer):
from collections import Counter
n = int(input())
a = list(map(int, input().split()))
mod = 10 ** 9 + 7
class nCr():
    def __init__(self, n):
        self.n = n
        self.fa = [1] * (self.n + 1)
        self.fi = [1] * (self.n + 1)
        for i in range(1, self.n + 1):
            self.fa[i] = self.fa[i - 1] * i % mod
            self.fi[i] = pow(self.fa[i], mod - 2, mod)
    def comb(self, n, r):
        if n < r:return 0
        if n < 0 or r < 0:return 0
        return self.fa[n] * self.fi[r] % mod * self.fi[n - r] % mod
comb = nCr(2 * n)
c = Counter(a)
num = c.most_common()[0][0]
l, r = a.index(num) + 1, n - list(reversed(a)).index(num) + 1
print(l, r)
for i in range(1, n + 2):
    print(comb.comb(n + 1, i) - comb.comb(n - (r - l) - 1, i - 1))

Fixed, accepted submission:
from collections import Counter
n = int(input())
a = list(map(int, input().split()))
mod = 10 ** 9 + 7
class nCr():
    def __init__(self, n):
        self.n = n
        self.fa = [1] * (self.n + 1)
        self.fi = [1] * (self.n + 1)
        for i in range(1, self.n + 1):
            self.fa[i] = self.fa[i - 1] * i % mod
            self.fi[i] = pow(self.fa[i], mod - 2, mod)
    def comb(self, n, r):
        if n < r:return 0
        if n < 0 or r < 0:return 0
        return self.fa[n] * self.fi[r] % mod * self.fi[n - r] % mod
comb = nCr(2 * n)
c = Counter(a)
num = c.most_common()[0][0]
l, r = a.index(num), n - list(reversed(a)).index(num)
for i in range(1, n + 2):
    print((comb.comb(n + 1, i) - comb.comb(n - (r - l), i - 1)) % mod)
Problem:
You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?

Incorrect submission (verdict: Wrong Answer):
a,b,c,d = map(int,input())

if a==2:
  a=1
if b==2:
  b=1
if c==2:
  c=1
if d==2:
  d=1
  
print(a+b+c+d)

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

if a==2:
  e=1
else:
    e=0
    
if b==2:
  f=1
else:
    f=0
  
if c==2:
  g=1
else:
    g=0
    
if d==2:
  h=1
else:
    h=0
  
print(e+f+g+h)
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
X = list(input())

for i in range(2**3):
    W = ["+"]*3
    for j in range(3):
        if (i >> j) & 1:
            W[j] = "-"
    formula = ""
    for k,l in zip(X,W+[""]):
        formula += k+l
    if eval(formula) == 7:
        print(formula)
        break


Fixed, accepted submission:
X = list(input())

for i in range(2**3):
    W = ["+"]*3
    for j in range(3):
        if (i >> j) & 1:
            W[j] = "-"
    formula = ""
    for k,l in zip(X,W+[""]):
        formula += k+l
    if eval(formula) == 7:
        print(formula+"=7")
        break
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())
numList = [x for x in range(a, b+1)]

ansList = numList[:k] + numList[-k:]
for num in list(set(ansList)):
  print(num)

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

if (b - a) / 2 >= k:
  for i in range(a, a+k):
    print(i)

  for i in range(b-k+1, b+1):
    print(i)

else:
  for i in range(a, b+1):
    print(i)
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

Incorrect submission (verdict: Wrong Answer):
S = list(input())
T = list(input())

s_reversed = list(reversed(S))
T_reversed = list(reversed(T))
F = 0

for i in range(len(S)-len(T)+1):
    s = s_reversed[i:i+len(T)]
    flag = 0
    for j in range(len(T)):
        if s[j] == T_reversed[j] or s[j] == '?':
            continue
        else:
            flag += 1
            break
    if flag:
        F += 1
        for j in range(len(T)):
            s_reversed[i+j-1] = T_reversed[j]
        break

if F:
    print('UNRESTORABLE')
else:
    R = ''.join(list(reversed(s_reversed)))
    print(R.replace('?','a'))





Fixed, accepted submission:
S = list(input())
T = list(input())

S.reverse()
T.reverse()
flag = False

for i in range(len(S)-len(T)+1):
    for j in range(len(T)):
        if S[j+i] == T[j] or S[j+i] == '?':
            pass
        else:
            break

        if j == len(T)-1:
            flag = True
            for k in range(len(T)):
                if S[k+i] == '?':
                    S[k+i] = T[k]
            break
    if flag:
        break

if flag:
    for i in range(len(S)):
        if S[i] == '?':
            S[i] = 'a'
    S.reverse()
    print(''.join(S))

else:
    print('UNRESTORABLE')
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(2*len(o) -1):
    if i%2==0:
        ans.append(o[i//2])
    else:
        ans.append(e[i//2])

print(*ans, sep='')

Fixed, accepted submission:
o=input()
e=input()
ans=[]
cnt=0

for i in range(2*len(e)):
    if  i%2==0:
        ans.append(o[i//2])
    else:
        ans.append(e[i//2])

if len(o)!=len(e):
    ans.append(o[-1])

print(*ans, sep='')
Problem:
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.  In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white.  Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Incorrect submission (verdict: Wrong Answer):
S=input()
t=len(S)
ans=0
for i in range(t-1):
 if S[i-1]!=S[i]:
  ans+=1
 else:
  ans+=0
print(ans)
 

Fixed, accepted submission:
S=input()
t=len(S)
ans=0
for i in range(t-1):
 if S[i]!=S[i+1]:
  ans+=1
 else:
  ans+=0
print(ans)
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())
result = 0
i = 1
while result < n:
    result = i * 3
    print(" {0}".format(result), end="")
    i += 1

Fixed, accepted submission:
n = int(input())
for i in range(3, n+1):
    if i % 3 == 0:
        print(" " + str(i), end = "")
    elif "3" in str(i):
        print(" " + str(i), end = "")
print()
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: Time Limit Exceeded):
X = map(int,input().split())
#X=7
P = {0}
i=1
while True:
    Q = {0}
    for p in P:
        Q.add(p+i)
        Q.add(p-i)
    P = P.union(Q)
    if X in P:
        print(i)
        break
    i = i+1


Fixed, accepted submission:
X = int(input())
a = 0
while X >0:
    a += 1
    X -= a
print (a)
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):
r, g, b = map(int, input().split())

if (10*g + b)%4 == 0:
  print('Yes')
else:
  print('No')

Fixed, accepted submission:
print('YES'*(int(input()[::2])%4==0)or'NO')
Problem:
There is a grid with H rows and W columns, where each square is painted black or white.  You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.  Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:    * There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Incorrect submission (verdict: Wrong Answer):
h,w=map(int,input().split())

maze=[list(input()) for _ in range(h)]

group=[[0]*w for _ in range(h)]

group_no=0
color_cnt=[]

for y in range(h):
    for x in range(w):
        if group[y][x]==0:
            stack=[(y,x)]
            group_no+=1

            color_cnt_tmp=[0,0]

            while len(stack)>0:
                tmp=stack.pop()
                group[tmp[0]][tmp[1]]=group_no
                print(tmp)
                if maze[tmp[0]][tmp[1]]=="#":
                    color_cnt_tmp[0]+=1
                else:
                    color_cnt_tmp[1]+=1

                for (ny,nx) in [(0,1),(1,0),(-1,0),(0,-1)]:
                    my=ny+tmp[0]
                    mx=nx+tmp[1]
                    if 0<=my<h and 0<=mx<w:
                        if group[my][mx]==0:
                            if maze[tmp[0]][tmp[1]]!=maze[my][mx]:
                                stack.append((my,mx))
                                group[my][mx]=group_no
            color_cnt.append(color_cnt_tmp)

ans=0
for (b,w) in color_cnt:
    ans+=b*w

print(ans)



Fixed, accepted submission:
h,w=map(int,input().split())

maze=[list(input()) for _ in range(h)]

group=[[0]*w for _ in range(h)]

group_no=0
color_cnt=[]

for y in range(h):
    for x in range(w):
        if group[y][x]==0:
            stack=[(y,x)]
            group_no+=1

            color_cnt_tmp=[0,0]

            while len(stack)>0:
                tmp=stack.pop()
                group[tmp[0]][tmp[1]]=group_no
                #print(tmp)
                if maze[tmp[0]][tmp[1]]=="#":
                    color_cnt_tmp[0]+=1
                else:
                    color_cnt_tmp[1]+=1

                for (ny,nx) in [(0,1),(1,0),(-1,0),(0,-1)]:
                    my=ny+tmp[0]
                    mx=nx+tmp[1]
                    if 0<=my<h and 0<=mx<w:
                        if group[my][mx]==0:
                            if maze[tmp[0]][tmp[1]]!=maze[my][mx]:
                                stack.append((my,mx))
                                group[my][mx]=group_no
            color_cnt.append(color_cnt_tmp)

ans=0
for (b,w) in color_cnt:
    ans+=b*w

print(ans)
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_1, m_1, h_2, m_2, k = map(int, input().split())
print(h_1*60+m_1-h_2*60-m_2-k)

Fixed, accepted submission:
h_1, m_1, h_2, m_2, k = map(int, input().split())
print(-h_1*60-m_1+h_2*60+m_2-k)
Problem:
In this problem, we only consider strings consisting of lowercase English letters.  Strings s and t are said to be **isomorphic** when the following conditions are satisfied:    * |s| = |t| holds.   * For every pair i, j, one of the following holds:     * s_i = s_j and t_i = t_j.     * s_i \neq s_j and t_i \neq t_j.  For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not.  A string s is said to be in **normal form** when the following condition is satisfied:    * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison.  For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`.  You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.

Incorrect submission (verdict: Wrong Answer):
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**9)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

al = ['a','b','c','d','e','f','g','h','i','j']

n = ni()

import numpy as np

for i in range(2**(n-1)):
  k = [0]+list(map(int,format(i, 'b')))
  p = np.cumsum(k)
  ans = [str(al[j]) for j in p]
  print(' '.join(ans))

Fixed, accepted submission:
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**9)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

al = ['a','b','c','d','e','f','g','h','i','j']

import copy
n = ni()

def f(a,b,m):
  if a == n:
    print(''.join([al[i] for i in b]))
  else:
    for i in range(m+2):
      c = copy.copy(b)
      c.append(i)
      f(a+1,c,max(m,i))

f(1,[0],0)
Problem:
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.

Incorrect submission (verdict: Wrong Answer):
print(int(int(input())**0.5))

Fixed, accepted submission:
print(int(int(input())**0.5)**2)
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a,b = map(int,input().split())
if a*b%2==0:
        print("Odd")
else:
        print("Even")

Fixed, accepted submission:
a,b = map(int,input().split())
c = a*b
print("Odd" if c%2==1 else "Even")
Problem:
N persons are standing in a row. The height of the i-th person from the front is A_i.  We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person:  Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool.  Find the minimum total height of the stools needed to meet this goal.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
h = list(map(int, input().split()))
left_max = h[0]
steps = 0
for i in range(1, n):
  if(left_max > h[i]):
    steps += h[i]-left_max
    left_max = steps + h[i]
print(steps)

Fixed, accepted submission:
n = int(input())
h = list(map(int, input().split()))
left_max = h[0]
steps = 0
for i in range(1, n):
  if(left_max > h[i]):
    steps += left_max-h[i]
  else:
    left_max = h[i]
print(steps)
Problem:
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied:    * Each lowercase letter of the English alphabet occurs even number of times in w.  You are given the string w. Determine if w is beautiful.

Incorrect submission (verdict: Wrong Answer):
import collections

w = input()
count_dict = collections.Counter(w)
for k, v in count_dict.items():
    if v % 2 != 0:
        print('No')
        break
    print('yes')


Fixed, accepted submission:
import collections

w = input()
flg = 0
count_dict = collections.Counter(w)
for k, v in count_dict.items():
    if v % 2 != 0:
        flg = 1
        break
if flg == 1:
    print('No')
else:
    print('Yes')
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):
A = int(input()) # 500
B = int(input()) # 100
C = int(input()) # 50
X = int(input()) # target

ans = 0

for i in range(A+1):
  amount = 500* (i - 1)
  for j in range(B+1):
    amount += 100*(j - 1)
    for h in range(C+1):
      amount += 50*(h - 1)
      if amount == X:
        ans += 1
        
        
print(ans)

Fixed, accepted submission:
A = int(input()) # 500
B = int(input()) # 100
C = int(input()) # 50
X = int(input()) # target
 
ans = 0
 
for i in range(A+1):
  for j in range(B+1):
    for h in range(C+1):
      amount = 500*i + 100 * j + 50 * h
      if amount == X:
        ans += 1

print(ans)
Problem:
タナカ氏が HW アールの果樹園を残して亡くなりました。果樹園は東西南北方向に H × W の区画に分けられ、区画ごとにリンゴ、カキ、ミカンが植えられています。タナカ氏はこんな遺言を残していました。  果樹園は区画単位でできるだけ多くの血縁者に分けること。ただし、ある区画の東西南北どれかの方向にとなりあう区画に同じ種類の果物が植えられていた場合は、区画の境界が分からないのでそれらは 1 つの大きな区画として扱うこと。  例えば次のような 3 × 10 の区画であれば ('リ'はリンゴ、'カ'はカキ、'ミ'はミカンを表す)      同じ樹がある区画の間の境界を消すと次のようになり、      結局 10 個の区画、つまり 10 人で分けられることになります。  雪が降って区画の境界が見えなくなる前に分配を終えなくてはなりません。あなたの仕事は果樹園の地図をもとに分配する区画の数を決めることです。  果樹園の地図を読み込み、分配を受けられる血縁者の人数を出力するプログラムを作成してください。

Incorrect submission (verdict: Wrong Answer):
import sys
sys.setrecursionlimit(100000)

fields = []
i = 0
H, W = 0, 0
field = []
for line in sys.stdin:
	line = line.rstrip()
	if line[0] not in ["#", "@", "*"]:
		line = line.split(" ")
		tmpH = H
		tmpW = W
		H = int(line[0])
		W = int(line[1])
		if (len(field) != 0):
			fields.append([field, tmpH, tmpW])
			field = []
		if (H == 0 and W == 0):
			break

	else:
		field.append(list(line))
	i += 1

def dfs(x, y, fruit):
	field[x][y] = "0"
	for dx in [-1, 0, 1]:
		if dx == -1: width = [0]
		if dx == 0: width = [-1, 0, 1]
		if dx == 1: width = [0]
		for dy in width:
			nx = x+dx
			ny = y+dy
			inField = (0 <= nx) and (0 <= ny) and (nx < H) and (ny < W)
			if inField and field[nx][ny] == fruit:
				dfs(nx, ny, fruit)
print(fields)
count = 0
for field_info in fields:
	field = field_info[0]
	H = field_info[1]
	W = field_info[2]
	for x in range(H):
		for y in range(W):
			if (field[x][y] != "0"):
				dfs(x, y, field[x][y])
				count += 1
	print(count)
	count = 0

Fixed, accepted submission:
import sys
sys.setrecursionlimit(100000)

def load_fields():
	fields = []
	i = 0
	H, W = 0, 0
	field = []
	for line in sys.stdin:
		line = line.rstrip()
		if line[0] not in ["#", "@", "*"]:
			line = line.split(" ")
			H, W = int(line[0]), int(line[1])
			if (len(field) != 0):
				fields.append(field)
				field = []
			if (H == 0 and W == 0):
				break

		else:
			field.append(list(line))
		i += 1
	return fields

def dfs(x, y, fruit):
	field[x][y] = "0"
	for dx in [-1, 0, 1]:
		if dx == -1: width = [0]
		if dx == 0: width = [-1, 0, 1]
		if dx == 1: width = [0]
		for dy in width:
			nx = x+dx
			ny = y+dy
			inField = (0 <= nx < H) and (0 <= ny < W)
			if inField and (field[nx][ny] == fruit):
				dfs(nx, ny, fruit)

fields = load_fields()
count = 0
for field in fields:
	H = len(field)
	W = len(field[0])
	for x in range(H):
		for y in range(W):
			if (field[x][y] != "0"):
				dfs(x, y, field[x][y])
				count += 1
	print(count)
	count = 0
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):
print("vowel" if "aiueo".find(input())>0 else "consonant")

Fixed, accepted submission:
print("vowel" if "aiueo".find(input())>=0 else "consonant")
Problem:
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.

Incorrect submission (verdict: Wrong Answer):
from sys import stdin

A, B = stdin.readline().rstrip().split()


a = int(A)
b = int(B)
print(type(A))
print(max(a+b, a-b, a*b))

Fixed, accepted submission:
from sys import stdin

A, B = stdin.readline().rstrip().split()
a = int(A)
b = int(B)

print(max(a+b, a-b, a*b))
Problem:
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces.  Find the number of chocolate pieces prepared at the beginning of the camp.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
sum=0
for i in range(len(A)):
    n=1
    while 1+(n-1)*A[i]<=D:
        sum+=1
        n+=1
    print(sum)
print(sum+X)
    

Fixed, accepted submission:
N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
sum=0
for i in range(len(A)):
    n=1
    while 1+(n-1)*A[i]<=D:
        sum+=1
        n+=1
print(sum+X)