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 <stdio.h>
#include <stdlib.h>

#define max(a, b) ((a)>(b)?(a):(b))


struct node {
	struct node *left;
	struct node *right;
	int index;
	int activity;
};

struct node *
insert(struct node *p, int act, int index)
{
	int cmp;
	if (p == NULL) {
		p = (struct node *)malloc(sizeof(struct node));
		p->left = (struct node *)NULL;
		p->right = (struct node *)NULL;
		p->index = index;
		p->activity = act;
		return p;
	} else {
		if ( act < p->activity) {
			p->left = insert(p->left, act, index);
			return p;
		} else if (act >= p->activity) {
			p->right = insert(p->right, act, index);
			return p;
		} else {
			printf("This case is not exist");
		}
	}
}

int depthNode(struct node *p, int n, int *idx) {
	if (p->left == NULL) {
		idx[n] = p->index;
		n++;
	} else {
		n = depthNode(p->left, n, idx);
		idx[n] = p->index;
		n++;
	}
	
	if (p->right == NULL) {
		return n;
	} else {
		n = depthNode(p->right, n, idx);
	}
	return n;
}

int main()
{
	int i, j, N;
	int ix, iy;
	long *A;
	long **DP;
	int *Index;
	int ret;
	int act;
	int n;
	
	struct node *root = (struct node *)NULL;
	
	ret = scanf("%d", &N);
	A = (long *) malloc(sizeof(long)*N);
	DP = (long **) malloc(sizeof(long *)*N);
	
	for (ix = 0; ix < N; ix++) {
		DP[ix] = (long *) malloc(sizeof(long)*N);
		for (iy = 0; iy < N; iy++)
			DP[ix][iy] = 0;
	}
	
	Index = (int *) malloc(sizeof(int)*N);
	
	for (i = 0; i < N; i++) {
		ret = scanf(" %ld", &A[i]);
		root = insert(root, A[i], i);
	}

	n = depthNode(root, 0, Index);
	
	for (i = 0; i < N; i++)
		DP[i][i] = A[Index[0]]*abs(Index[0] - i);
		
	for (i = 1; i < N; i++) {
		for (ix = 0; ix + i < N;ix++) {
			iy = ix + i;
			DP[ix][iy] = 0;
			DP[ix][iy] = max(DP[ix][iy],
							 max(DP[ix+1][iy] + A[Index[i]]*abs(Index[i]-ix),
								 DP[ix][iy-1] + A[Index[i]]*abs(Index[i]-iy)));
		}
	}
	printf("%ld\n",DP[0][N-1]);
	fflush(stdout);
	free(A);
	free(Index);
	return EXIT_SUCCESS;
}