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>
using namespace std;

using ll = long long;
const int MAXN = 2005;
pair<ll, int> a[MAXN];

int n;
ll memo[MAXN][MAXN];

ll dp(int L, int R) {
    if(L > R) return 0;
    if(memo[L][R] != -1) return memo[L][R];
    int i = L + (n-1-R);
    ll& ans = memo[L][R];
    ll on_left = a[i].first * abs(a[i].second - L);
    ll on_right = a[i].first * abs(a[i].second - R);
    ans = max(on_left + dp(L+1, R), on_right + dp(L, R-1));
    return ans;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    memset(memo, -1, sizeof memo);

    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> a[i].first;
        a[i].second = i;
    }
    sort(a, a+n);
    reverse(a, a+n);
    cout << dp(0, n-1) << endl;
}