dp poj3186

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.

这个地方的动态转移直接转移不好办 所以我们要想办法把这个过程逆向进行 也就是从0个元素开始不停地往里面添加元素  可以从左边添加也可以从右边添加 但是权值是由大到小的 这个地方要注意 dp[i][j]表示从i到j的最大的取值

状态转移方程 dp[i][j]=max(dp[i][j-1]+(n+i-j)*arr[j],dp[i+1][j]+(n+i-j)*arr[i]); 然后所有的全部初始化为 0 直接转移就可以啦 其实还是很好办的 ！

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <string>#include <iostream>#include <algorithm>#include <stack>#include <queue>#include <sstream>#include <ostream>#include <list>#include <ctype.h>#include <cmath>#define inf 2000000007using namespace std;int dp[2002][2002];int arr[2002];int main(){    int n;    cin>>n;    for(int i=1;i<=n;i++)        cin>>arr[i];    for(int i=n;i>=1;i--)    {        for(int j=i;j<=n;j++)        {            dp[i][j]=max(dp[i+1][j]+(n+i-j)*arr[i],dp[i][j-1]+(n+i-j)*arr[j]);        }    }    cout<<dp[1][n]<<endl;}