8.24--练习赛D题--Treats for the cows(区间DP)

原题：

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.

题目链接：

https://cn.vjudge.net/contest/181259#problem/D

题意：

给出一列数，从头和尾都可以往外出，第i次出的结果是num[i]*n，问总和最大是多少。

这道题是一道可能比较简单的区间dp的题。。奈何我dp学的如此渣以至于推了半天还推不出来。。感觉学的都忘了。。唉心痛

直接看代码吧。

源代码：

#include <iostream>#include <cstring>#include <stdio.h>#include <algorithm>#include <iomanip>using namespace std;int dp[10000][10000];int main(){    int n,i,j;    int num[10000];    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 = n - 1; i >= 1; i--)        for (j = i + 1; j <= n; j++)            dp[i][j] = max(dp[i + 1][j] + num[i] * (n - j + i),dp[i][j - 1]+num[j] * (n-j+i));    printf("%d\n",dp[1][n]);    return 0;}