九度 oj 題目1072：有多少不同的面值組合？

http://ac.jobdu.com/problem.php?pid=1072

#include <cstdio>
const int n = 15;
const int v = 8 * 5 + 10 * 4 + 18 * 6;
const int a[n+1] = { 0,8,8,8,8,8,10,10,10,10,18,18,18,18,18,18 };
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
int count[v + 1];
 
int main()
{
    int ans = 0;
    count[0] = 1;
    rep(i, 1, n)
    {
        per(j, v, a[i])
        {
            if (count[j - a[i]]) count[j] = 1;
        }
    }
    rep(i, 1, v)
    {
        if (count[i]) {
            //printf("%.1lf\n", i / 10.0);
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}