1 solutions

  • 0
    @ 2025-3-21 16:54:34
    #include<bits/stdc++.h>
    using namespace std;
    const int N=1010;
    int f[N][N],v[N],m[N];
    int main()
    {
        int V,M,n;
        cin>>V>>M>>n;
        for(int i=1;i<=n;i++) cin>>v[i]>>m[i]; //每个精灵的体积和价值 
        for(int i=1;i<=n;i++) 
        {
            for(int j=V;j>=v[i];j--) //二维01背包,倒着枚举 
            {
            	for(int k=M;k>m[i];k--)
            	{
            		 f[j][k]=max(f[j][k],f[j-v[i]][k-m[i]]+1);
    			}   
    		}
        }
        cout<<f[V][M]<<" ";
        int k=M;
        while(k>0&&f[V][M]==f[V][k]) k--; //找到收服个数一定,然后精力最小的一个 
        cout<<M-k<<endl;
        return 0;
    }
    
    • 1

    Information

    ID
    993
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    (None)
    # Submissions
    3
    Accepted
    3
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