analysis

本人表示非常兴奋，毕竟这是我第一个没有看题解自己推出来的方程：
$f[i][j]=min(f[i-1][j-k]+D[i]\times C[j-k+1])$f[i][j]=min(f[i−1][j−k]+D[i]×C[j−k+1])
其中
$min(0,j-m)&lt;=k&lt;=j+1-i$min(0,j−m)<=k<=j+1−i

code

#include<bits/stdc++.h>
using namespace std;
#define loop(i,start,end) for(register int i=start;i<=end;++i)
#define clean(arry,num); memset(arry,num,sizeof(arry));
#define min(a,b) ((a>b)?b:a)
int n,m;
const int maxn=1000,maxm=1000;
int D[maxn];
int C[maxm];
int f[maxn][maxm];
inline int read()
{
    int _ans=0;bool _neg=false;char _r=getchar();
    while(_r>'9'||_r<'0'){if(_r=='-')_neg=true;_r=getchar();}
    while(_r>='0'&&_r<='9'){_ans=_ans*10+_r-'0';_r=getchar();}
    return (_neg)?-_ans:_ans;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("datain.txt","r",stdin);
    #endif
    n=read();
    m=read();
    clean(f,0x7f);
    clean(f[0],0);
    loop(i,1,n)D[i]=read();
    loop(i,1,m)C[i]=read();
    loop(i,1,n)
    {
    	loop(j,i,m)
    	{
    		loop(k,j-m,j+1-i)
    		{
                if(k<=0)continue;
                f[i][j]=min(f[i][j],f[i-1][j-k]+D[i]*C[j-k+1]);
            }
        }
    }
    int res=INT_MAX;
    loop(i,n,m)res=min(res,f[n][i]);
    printf("%d",res);
    return 0;
}