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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>

using namespace std;
typedef long long unsigned int ll;

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

int main() {


	ll dp[5][3010];

	vector<ll> coin_amount(3);
	vector<ll> coin_value(3);

	coin_value[0] = 500;
	coin_value[1] = 100;
	coin_value[2] = 50;

	//空白があってもいい感じに読み込んでくれる
	for (size_t i = 0; i < 3; ++i) {
		scanf("%d", &coin_amount[i]);
	}
	int n;
	scanf("%d", &n);
	//printf("%d\n", n);

	for (size_t i = 0; i < 5; i++) {
		for (size_t j = 0; j <3010; j++) {
			dp[i][j] = 0;

		}

	}
	dp[0][0] = 1;
	for (size_t i = 0; i < 3; i++) {
		for (size_t j = 0; j < 3010; j++) {
			dp[i + 1][j] += dp[i][j];
			for (size_t k = 1; k <= coin_amount[i]; k++) {

				if (j >= k) {
					dp[i + 1][j] += dp[i][j - k];
				}


			}

		}
	}
	printf("%d\n", dp[3][n]);
	return 0;
}