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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define int long long int
int dp[2001][2001]={};
unordered_map<string,int>m;
int arr[2001]={};
int helper(int i,int n,string s){
 if(i>n)
    return 0;
 if(m.count(s)>0)return m[s];
    int ans=0;
for(int j=0;j<n;j++){
 if(s[j]=='0'){
    string s1=s;
    s1[j]='1';
    ans=max(ans,dp[i][j]+helper(i+1,n,s1));
 }
}
m[s]=ans;
return ans;
}

signed main(){

ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t=1;
//cin>>t;
while(t--){
m.clear();
int n;
cin>>n;
for(int i=0;i<n;i++)
    cin>>arr[i];

for(int i=0;i<n;i++){

    for(int j=0;j<n;j++){
        dp[i][j]=arr[i]*(abs(j-i));
    }
}
string s;
for(int i=0;i<n;i++)
    s+='0';
int ans=helper(0,n,s);

cout<<ans<<endl;

}

}