1 solutions

  • 0
    @ 2024-3-30 15:21:43
    #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