Treats for the Cows

Description

FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.

The treats are interesting for many reasons:

• The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
• Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
• The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
• Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.

Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?

The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.

Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 contains the value of treat v(i)

Output

Line 1: The maximum revenue FJ can achieve by selling the treats

Sample Input

5
1
3
1
5
2

Sample Output

43

Hint

Explanation of the sample:

Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).

FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1×1 + 2×2 + 3×3 + 4×1 + 5×5 = 43.

Source

USACO 2006 February Gold & Silver    

/*
POJ  3186
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=2020;
int a[MAXN];
int dp[MAXN][MAXN];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
          scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
          dp[i][i]=a[i]*n;
        for(int k=1;k<n;k++)
          for(int i=1;i+k<=n;i++)
          {
              int j=i+k;
              dp[i][j]=max(dp[i+1][j]+(n-k)*a[i],dp[i][j-1]+(n-k)*a[j]);
          }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}