1 solutions
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0
#include<bits/stdc++.h> using namespace std; int m , n; int a[1001]; long long f[10001]; int main() { scanf("%d%d",&n,&m); for (int i = 1;i <= n;i++) scanf("%d",&a[i]); f[0] = 1; for(int i = 1;i <= n; i++) for (int j = m; j >= a[i]; j--) for (int k = 1;k <= j/a[i]; k++) f[j] += f[j - k*a[i]]; printf("%lld",f[m]); return 0; }
- 1
Information
- ID
- 758
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 21
- Accepted
- 13
- Uploaded By