BIT2014级软件学院程序设计-22 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

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.

也是一道DP的题。每次从头取或者从后取，DP方程也很容易。

#pragma comment(linker, "/STACK:102400000,102400000")#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxn 2500int dp[maxn][maxn];int num[maxn];int max(int a, int b){if (a > b)return a;return b;}int main(){int n,i,j,flag;scanf("%d",&n);for (i = 1; i <= n; i++)scanf("%d", &num[i]);for (i = 1;i <= n;i++)dp[i][i] = num[i] * n;for (i = 1; i < n;i++)for (j = 1; j + i <= n; j++){flag = i + j;dp[j][flag] = max(dp[j + 1][flag] + (n - i)*num[j], dp[j][flag - 1] + (n - i)*num[flag]);}printf("%d\n", dp[1][n]);}