POJ

 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.

题意：

给定n个数，按输入顺序排成一排，每次从外面取一个数，再乘以k（第几个取出的数）。每取一个就总和加上这个值。求最后取完后，总和最大为多少。

思路：

反向思考，相当于往里面加数，先加一个数，其乘以n，接着循环相加左右情况.

代码有具体解释

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint s[2005];long long dp[2005][2005];int num[2005];int main(){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&s[i]);for(int i=1;i<=n;i++)//区间长度为0时 dp[i][i]=s[i]*n;//表示第 i 个数最后一个取出 //从i=1假设每一个数都是最后一个被取出 ，所以乘以n.//接着下面的二层循环区间长度为2开始依次枚举所有数的情况 for(int l=1;l<n;l++){//区间长度 ，从2开始，l=1但是长度为2for(int i=1;i+l<=n;i++){//区间开始值 int j=i+l;//区间末尾值 dp[i][j]=max(dp[i+1][j]+(n-l)*s[i], dp[i][j-1]+(n-l)*s[j]);//               右边的进                左边的进 //区间由小往大推 }}printf("%lld\n",dp[1][n]);return 0;}