Treats for the Cows POJ

//dp[i][j]表示第i次从左边取，第j次从右边取的价值，所以我们可以得到状态方程
//dp[i][j]=max(dp[i-1][j]+(i+j)*a[i],dp[i][j-1]+(i+j)*a[n-j+1]) (i > 0 && j > 0 )
//dp[i][0]=dp[i-1][0]+i*a[i],dp[0][i] dp[0][i-1]+i*a[n-i+1];
#include<cstdio>
#include<cmath>
#include<string>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=2e3+10;
int a[maxn];
ll dp[maxn][maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>a[i];
    ll ans=0;
    dp[0][0]=0;
    for(int i=1;i<=n;++i)
    {
        dp[i][0]=dp[i-1][0]+a[i]*i;
        dp[0][i]=dp[0][i-1]+a[n-i+1]*i;
        ans=max(ans,dp[i][0]);
        ans=max(ans,dp[0][i]);
    }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j + i <= n; ++j)
        {
            dp[i][j]=max(dp[i-1][j]+(i+j)*a[i],dp[i][j-1]+(i+j)*a[n-j+1]);
            ans=max(ans, dp[i][j]);
        }
    cout<<ans<<endl;
    return 0;
}