Problem:
Score : 500 points

Problem Statement

There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.

Find the maximum total happiness points the children can earn.

Constraints

2 \leq N \leq 2000

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum total happiness points the children can earn.

Sample Input 1

4
1 3 4 2

Sample Output 1

20

If we move the 1-st child from the left to the 3-rd position from the left, the 2-nd child to the 4-th position, the 3-rd child to the 1-st position, and the 4-th child to the 2-nd position, the children earns 1 \times |1-3|+3 \times |2-4|+4 \times |3-1|+2 \times |4-2|=20 happiness points in total.

Sample Input 2

6
5 5 6 1 1 1

Sample Output 2

58

Sample Input 3

6
8 6 9 1 2 1

Sample Output 3

85

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
ll dp[2001][2001],n,a,ans;
vector<P> v;
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a;
        v.push_back({a,i});
    }
    sort(v.begin(),v.end());
    reverse(v.begin(),v.end());
    for(int i=0;i<n;i++){
        dp[i+1][0]=dp[i][0]+v[i].first*abs(v[i].second-i);
        dp[0][i+1]=dp[0][i]+v[i].first*abs(n-v[i].second-1-i);
    }
    for(int i=1;i<n;i++){
        for(int j=1;j<n;j++){
            if(i+j>n)continue;
            dp[i][j]=max(dp[i-1][j]+v[i+j-1].first*abs(v[i+j-1].second-i+1),dp[i][j-1]+v[i+j-1].first*abs(n-v[i+j-1].second-1-j+1));
        }
    }
    for(int i=0;i<=n;i++){
        ans=max(ans,dp[i][n-i]);
    }
    cout<<ans<<endl;
}