AtCoder Beginner Contest 087 B

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

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.

Constraints

• 0≤A,B,C≤50
• A+B+C≥1
• 50≤X≤20 000
• A, B and C are integers.
• X is a multiple of 50.

---

Input

Input is given from Standard Input in the following format:

A
B
C
X

Output

Print the number of ways to select coins.

---

Sample Input 1

Copy

2
2
2
100

Sample Output 1

Copy

2

There are two ways to satisfy the condition:

• Select zero 500-yen coins, one 100-yen coin and zero 50-yen coins.
• Select zero 500-yen coins, zero 100-yen coins and two 50-yen coins.

---

Sample Input 2

Copy

5
1
0
150

Sample Output 2

Copy

0

Note that the total must be exactly X yen.

---

Sample Input 3

Copy

30
40
50
6000

Sample Output 3

Copy

213

直接暴力枚举

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>
using namespace std;
int a,b,c,x,d;
 
int main()
{
    cin>>a>>b>>c>>x;
    for(int i = 0;i <= a;i ++)
    {
        for(int j = 0;j <= b;j ++)
        {
            for(int k = 0;k <= c;k ++)
                if(i * 500 + j * 100 + k * 50 == x)d ++;
        }
    }
    cout<<d<<endl;
}