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<math.h>
#include<stdlib.h>

#define TEST

typedef struct{
  int num;
  long long int Happy;
}Human;
  
int koujun(const void*a,const void*b){
  return ((Human*)b)->Happy-((Human*)a)->Happy;
};

int main(){
  long long int chi_n,sum=0;
  scanf("%lld",&chi_n);
  Human child[2000];
  for(int i=0;i<chi_n;i++){
    child[i].num=i;
    scanf("%lld",&child[i].Happy);
  };
#ifndef  TEST
  for(int y=0;y<chi_n;y++){
    printf("%lld\n",child[y].Happy);
  };
#else
#ifdef TEST
  qsort(child,chi_n,sizeof(Human),koujun);  
#else   
  for(int y=0;y<chi_n;y++){
    printf("%lld\n",child[y].Happy);
  };
#endif  
#endif
  int posi[2000]={};
  int j=0,k=0;
  for(int i=0;i<chi_n;i++){
    if(child[i].num>chi_n-1-child[i].num){
      j=0;
      while(1){
        if(posi[j]==0){
          sum+=(child[i].num-j)*child[i].Happy;
          posi[j]=1;
          break;
        };
        j++;
      };
    }
    else if(child[i].num<chi_n-1-child[i].num){
       k=0;
       while(1){
        if(posi[chi_n-1-k]==0){
          sum+=(chi_n-1-k-child[i].num)*child[i].Happy;
          posi[chi_n-1-k]=1;
          break;
        };
         k++;
       };
    }
    else{
      j=k=0;
       while(1){
        if(posi[j]==0){
          sum+=(child[i].num-j)*child[i].Happy;
          posi[j]=1;
          break;
        }
        else if(posi[chi_n-1-k]==0){
          sum+=(chi_n-1-k-child[i].num)*child[i].Happy;
          posi[chi_n-1-k]=1;
          break;
        };
         j++;
         k++;
      };
    };
  };
  printf("%lld",sum);
};