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: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <iterator>
//#include <cmath>
#include <set>
#include <deque> 
#include <string>


using namespace std;

int n;
long long f[2010][2010];	//i--left，j--right左区，右区

struct Elem
{
	int oriPos;
	int val;
};

Elem a[2010];

bool compare(const Elem &a, const Elem &b) {
	if (a.val == b.val) {
		return a.oriPos < b.oriPos;
	} else {
		return a.val > b.val;
	}
}


int main()
{
	scanf("%d", &n);
	for (int i = 0 ; i < n ; i++) {
		scanf("%d", &(a[i].val));
		a[i].oriPos = i;
	}
		
	sort(a, a + n, compare);
	
	f[0][0] = 0;
	
	int count = 0;
	for (int i = 0 ; i < n ; i++) {
		count++;
		for (int l = 0 ; l <= count ; l++) {
			int r = count - l;
			
			long long curMax = 0;
			if (l == 0) {
				int rPos = n - r;
				curMax = f[l][r - 1] + abs(rPos - a[i].oriPos) * a[i].val;
			} else if (r == 0) {
				int lPos = l - 1;
				curMax = f[l - 1][r] + abs(lPos - a[i].oriPos) * a[i].val;
			} else {	//都不为0 
				int rPos = n - r;
				int lPos = l - 1;
				
				long long lVal = f[l][r - 1] + abs(rPos - a[i].oriPos) * a[i].val;
				long long rVal = f[l - 1][r] + abs(lPos - a[i].oriPos) * a[i].val;
				
				if (lVal > rVal)
					curMax = lVal;
				else
					curMax = rVal;
			}
			
			if (f[l][r] < curMax)
				f[l][r] = curMax;
		}
	}
	
	long long realMax = 0;
	for (int l = 0 ; l <= n ; l++) {
		int r = n - l;
		if (realMax < f[l][r])
			realMax = f[l][r];
	}
	
	printf("%lld\n", realMax);
		
	return 0;
}