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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MM 1000000007
long A[2010];
long S[2010][2010];
int main(){

	long int i,j,k;
	long int x,y,a,b,c;
	long N,M,K,tmp;
	long long sum=0;
	scanf("%ld",&N);

	for(i=1;i<=N;i++){
		scanf("%ld",A+i);
	}
	
	for(i=1;i<N;i++){
		for(j=i+1;j<=N;j++){
			S[i][j]=(A[i]+A[j])*labs(i-j);
		}
	}
	#if 0
		for(i=1;i<=N;i++){
		for(j=1;j<=N;j++){
			printf("%ld\t",S[i][j]);//=(A[i]+A[j])*labs(i-j);
		}
			printf("\n");
	}
	#endif
	long ind_i,ind_j,max=0;
	int o,p;
	for(i=1;i<=N/2;i++){
		for(j=i+1;j<=N;j++){
			if(max<S[i][j]){
				max=S[i][j];
				ind_i=i;ind_j=j;
			}
		}
		sum+=max;
		for(k=1;k<=N;k++){
			S[ind_i][k]=0;S[k][ind_j]=0;
		}
		#if 0
		printf("\n");
		for(o=1;o<=N;o++){
		for(p=1;p<=N;p++){
			printf("%ld\t",S[o][p]);//=(A[i]+A[j])*labs(i-j);
		}
			printf("\n");
		}
		#endif
		max=0;
	}
	printf("%lld\n",sum);

	return 0;
}

// Ax * | x-y |
// Ax * x - Ax*y
// 1,  3,  4,  2
// 1,  6,  12, 8 Ax*x 
// -3 -12  -4  -4   Ax*y 

// 5 5 6 1 1 1
// 5 10 18 4 5 6