Problem:
Score : 200 points

Problem Statement

Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes:

Red boxes, each containing R red balls

Green boxes, each containing G green balls

Blue boxes, each containing B blue balls

Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes.
How many triples of non-negative integers (r,g,b) achieve this?

Constraints

All values in input are integers.

1 \leq R,G,B,N \leq 3000

Input

Input is given from Standard Input in the following format:

R G B N

Output

Print the answer.

Sample Input 1

1 2 3 4

Sample Output 1

4

Four triples achieve the objective, as follows:

(4,0,0)

(2,1,0)

(1,0,1)

(0,2,0)

Sample Input 2

13 1 4 3000

Sample Output 2

87058

Solution in PyPy3 (2.4.0) (judge verdict: Runtime Error)

# -*- coding: utf-8 -*-
"""
Created on Sat Apr 20 20:47:47 2019

@author: Owner
"""

# -*- coding: utf-8 -*-
"""
Created on Sat Apr 13 09:53:12 2019

@author: Owner
"""
import bisect

"""
import numpy as np
import sys
import collections
import scipy.misc
import math
from operator import itemgetter
import itertools
import copy
"""
#素因数を並べる
def prime_decomposition(n):
  i = 2
  table = []
  while i * i <= n:
    while n % i == 0:
      n /= i
      table.append(int(i))
    i += 1
  if n > 1:
    table.append(int(n))
  return table   
# 桁数を吐く
def digit(i):
    if i > 0:
        return digit(i//10) + [i%10]
    else:
        return []
    
def getNearestValueIndex(list, num):
    """
    概要: リストからある値に最も近い値のインデックスを取得する関数
    @param list: データ配列
    @param num: 対象値
    @return 対象値に最も近い値
    """

    # リスト要素と対象値の差分を計算し最小値のインデックスを取得
    idx = np.abs(np.asarray(list) - num).argmin()
    return idx

def find_index(l, x, default=False):
    if x in l:
        return l.index(x)
    else:
        return default


"""
N, X = map(int, input().split())

x = list(map(int, input().split()))

P = [0]*N
Y = [0]*N
for n in range(N):
    P[n], Y[n] = map(int, input().split())

# 多次元配列の宣言（あとでintにすること。）（タプルにすること。）
dp = np.zeros((N+1, 4,4,4))
    
all(nstr.count(c) for c in '753')

# 複数配列を並び替え
ABT = zip(A, B, totAB)
result = 0
# itemgetterには何番目の配列をキーにしたいか渡します
sorted(ABT,key=itemgetter(2))
A, B, totAB = zip(*ABT)
A.sort(reverse=True)

# 2進数のbit判定
(x >> i) & 1

# dp最小化問題
dp = [np.inf]*N
for n in range(N):
    if n == 0:
        dp[n] = 0
    else:
        for k in range(1,K+1):
            if n-k >= 0:
                dp[n] = min(dp[n], dp[n-k] + abs(h[n]-h[n-k]))
            else:
                break
# 累積和
add = 1 # 問題によって決まる
res = 0
sums = [0]*(len(nums)+1)
for i in range(len(nums)):
    sums[i+1] = sums[i] + nums[i]
for i in range(0, len(nums), 2):
    left = i
    right = min(i+add, len(nums))
    tmp = sums[right] - sums[left]
    res = max(tmp, res)

#２分探索
li, ri = bisect.bisect_left(p_ac, l[i]-1), bisect.bisect_right(p_ac, r[i]-1)    

#ソート関数
org_list = [3, 1, 4, 5, 2]
new_list = sorted(org_list)
print(org_list)
print(new_list)
# [3, 1, 4, 5, 2]
# [1, 2, 3, 4, 5]

"""
C = [0]*3
C[0], C[1], C[2], N = list(map(int, input().split()))
C.sort(reverse=True)

def main():
    cnt = 0
    CI = [N//C[0], N//C[1], N//C[2]] 
    K  = C[2]*list(range(CI[2]+1))
    for i in range(CI[0]+1):
        for j in range(CI[1]+1):            
            if C[0]*i +C[1]*j > N:
                break
            if (N - (C[0]*i+C[1]*j)) == K[bisect.bisect_left(K, N - (C[0]*i+C[1]*j))]:
                cnt+=1
    print(cnt)
                
                
                
            
            
                

    
if __name__ == '__main__':
    main()