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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;

struct node{
	int id;
	ll val;
	bool operator<(const node&n)const
	{
		return val < n.val;
	}
}a[2005],b[2005];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		a[i].id = i;
		cin >> a[i].val;
		b[i].id = i;
		b[i].val = a[i].val;
	}
	ll ans = 0;
	for (int i = 0; i <= n; i++)
	{
		if( i != 0 ) sort(a+1,a+1+i);
		if( i != n ) sort(a+1+i,a+1+n);
		ll temp = 0;
		for (int j = i; j >= 1; j--)
		{
			ll locate = n - (i - j);
			temp += abs(locate-a[j].id) * a[j].val;
		}
		for (int j = n; j > i; j--)
		{
			ll locate = 1 + n - j;
			temp += abs(locate-a[j].id) * a[j].val;
		}
		ll temp2 = 0;
		for (int j = i; j >= 1; j--)
		{
			ll locate = 1 + (i - j);
			temp2 += abs(locate-a[j].id) * a[j].val;
		}
		for (int j = n; j > i; j--)
		{
			ll locate = n - (n - j);
			temp2 += abs(locate-a[j].id) * a[j].val;
		}
		ans = max(ans,temp);
		ans = max(ans,temp2);
		for (int j = 1; j <= n; j++)
		{
			a[j].id = b[j].id;
			a[j].val = b[j].val;
		}
	}
	cout << ans << '\n';
	return 0;
}