## 1652: [Usaco2006 Feb]Treats for the Cows

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 271  Solved: 209
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## 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.

•零食按照1．．N编号，它们被排成一列放在一个很长的盒子里．盒子的两端都有开口，约翰每
天可以从盒子的任一端取出最外面的一个．
•与美酒与好吃的奶酪相似，这些零食储存得越久就越好吃．当然，这样约翰就可以把它们卖出更高的价钱．
•每份零食的初始价值不一定相同．约翰进货时，第i份零食的初始价值为Vi(1≤Vi≤1000)．
•第i份零食如果在被买进后的第a天出售，则它的售价是vi×a．
Vi的是从盒子顶端往下的第i份零食的初始价值．约翰告诉了你所有零食的初始价值，并希望你能帮他计算一下，在这些零食全被卖出后，他最多能得到多少钱．

## 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

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

## Sample Output

43

OUTPUT DETAILS:

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.

## Source

Silver

dp的过程都写得出来，然后就是一直不明白到底边界要怎么弄，其实按一般情况弄就好了qwq。

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int nmax=2001;
int a[nmax],f[nmax][nmax];
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
for(int j=1;j+i-1<=n;j++){      //其实这里f[i][i]=a[i]*n;
int tmp=n-i+1;
int temp=i+j-1;
f[j][temp]=max(f[j+1][temp]+tmp*a[j],f[j][temp-1]+tmp*a[temp]);
}
printf("%d\n",f[1][n]);
return 0;
}

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