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 <bits/stdc++.h>
#define ll long long
using namespace std;
const int M=2010;
int n;
ll dp[M][M],ans;
struct elem{
    ll val,pos;
    bool operator<(elem tmp)
    {return val>tmp.val;}
}a[M];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lld",&a[i].val),a[i].pos=i;
    sort(a+1,a+n+1);
    for(int i=1;i<=n;i++)
        for(int L=0,R=n-i+2;R<=n+1;L++,R++)
        {
            dp[L+1][R]=max(dp[L+1][R],dp[L][R]+abs(L+1-a[i].pos)*a[i].val);
            dp[L][R-1]=max(dp[L][R-1],dp[L][R]+abs(R-1-a[i].pos)*a[i].val);
            //printf("%d %d %d %d\n",L,R,dp[L+1][R],dp[L][R-1]);
        }
    for(int i=1;i<n;i++)
        ans=max(dp[i][i+1],ans);
   // for(int i=0;i<=n+1;i++)
   // {
   //     for(int j=0;j<=n+1;j++)
   //         printf("%lld ",dp[i][j]);
   //     puts("");
   // }
    printf("%lld\n",ans);
    return 0;
}