33. Treats for the Cows

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

513152

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 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.

测试输入期待的输出时间限制内存限制额外进程测试用例 1以文本方式显示

1. 5↵
2. 1↵
3. 3↵
4. 1↵
5. 5↵
6. 2↵

以文本方式显示

1. 43↵

1秒64M0

//设Num_dp[i][j]为i-j这一段的最大价值//Num_dp[i][j]=max(dp[i+1][j]+Num_Str[i]*pp, Num_dp[i][j-1]+Num_Str[j]*pp)//其中pp为第pp次拿，pp=n-(j-i)#include<stdio.h>int Num_N, Num_dp[2005][2005] = { 0 }, Num_Str[2005] = { 0 };int fMax(int a, int b){return (a > b ? a : b);}int main(){scanf("%d", &Num_N);for (int i = 1; i <= Num_N; i++)scanf("%d", &Num_Str[i]);for (int i = Num_N; i >= 1; i--)for (int j = i; j <= Num_N; j++){int pp = Num_N - (j - i);Num_dp[i][j] = fMax(Num_dp[i+1][j] + pp*Num_Str[i], Num_dp[i][j-1] + pp*Num_Str[j]);}printf("%d\n", Num_dp[1][Num_N]);return 0;}