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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <math.h>
#include <algorithm>
#include <set>
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <string>
#include <cstring>
#include <functional>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <chrono>
#include <climits>
#include <bitset>
using namespace std ;
#define ld long double
#define int long long
#define ii pair<int,int>
#define FAST_IO ios_base::sync_with_stdio(false) ;cin.tie(NULL);cout.tie(NULL); clock_t beg = clock()
#define dbg(x) cerr << #x << " is " << x << endl;
#define ms(x,a) memset(x,(int)(a),sizeof(x))
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define vi vector<int >
#define pb push_back
#define endl '\n'
#define ff first
#define ss second
const int N = 2e3+7 ;
//const int INF = 1e18 ;
const int MOD = 1e9 + 7 ;

int dp[N][N];

void maxi(int&a , int b){
    a = max(a,b) ;
}
int add(int a, int b){
    a += b ;
    while(a>=MOD) a-= MOD ;
    return a ;
}
int mult(int a, int b){
    a *= b ;
    a %= MOD ;
    return a ;
}

void solve(){
    int n ;
    cin >> n ;
    vector<ii> a(n) ;
    for(auto& x : a) cin >> x.ff ;
    for(int i = 0 ; i < n ; i++){
        a[i].ss = i ;
    }
    sort(all(a));reverse(all(a)) ;
    dp[0][n] = 0 ; // base 
    // ans = max(dp[i][i])
    for(int i = 0 ; i < n ; i++){
        ii cur = a[i] ; // this is best a[i].ff 
        // can be used to put either front or at back
        for(int l = 0 ; l <= i ; l++){
            int r = l+n-i ;
            // l = 0 , r = n
            // l = 0 , r = n-1 ;; l = 1 , r = n 
            // i = 0 ; r = n-2 ;; l = 1 , r = n-1 ;; l = 2 , r = n
            maxi(dp[l+1][r], add(dp[l][r],mult(cur.ff,llabs(cur.ss-l))));
            maxi(dp[l][r-1], add(dp[l][r],mult(cur.ff,llabs((r-1LL)-cur.ss))));
        }
    }
    int ans = 0 ;
    for(int i = 0 ; i <= n ; i++){
        ans = max(ans, dp[i][i]) ;
    }
    cout << ans ;
}

int32_t main(){
    FAST_IO ;
    int t = 1 ;
//    cin >> t ;
    while(t--){
        solve() ;
    }
    cerr << "\nExecution time: " << (clock() - beg) / 1000 << '\n';
}