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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.PrintWriter;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int[] result = new int[n];
        boolean[] took = new boolean[n];
        int[] a = new int[n];
        for (int i = 0; i < a.length; i++) {
            a[i]= in.nextInt();
        }
        int l = 0;
        int r = n - 1;
        long ans = 0;
        for (int i = 0; i < n; i++) {
            int id = i;
            long mx = 0;
            boolean left = true;
            int value = -1;
            int who = -1;
            for (int j = 0; j < n; j++) {
                if (took[j])
                    continue;
                int v = a[j];
                if (v * 1l * (r - j) > mx) {
                    mx = v * 1l * (r - j);
                    id = r;
                    left = false;
                    value = v;
                    who = j;
                }
                if ( v * 1l * (j - l) > mx) {
                    mx = v * 1l * (j - l);
                    id = l;
                    left = true;
                    value = v;
                    who = j;
                }
            }
            if (mx != 0) {
                took[who] = true;
                result[id] = value;
                ans += mx;
                if (left) {
                    l++;
                } else {
                    r--;
                }
            }
        }
        out.println(ans);
        out.close();
    }
}