POJ-3186-Treats for the Cows

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Treats for the Cows
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5496 Accepted: 2863

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.

给出长度为n的序列,每次从最左端或最右端取一个数,获得的价值为a[i]*step,a[i]为该数的值,step为该数被取得次序,要求使最后获得的值最大

逆序取数

定义dp[i][j]为从i到j获得的最大值

dp[i][j] = max(dp[i+1][j]+a[i]*(n-k), dp[i][j-1]+a[j]*(n-k));

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <vector>#include <map>#include <cmath>#include <stdlib.h>using namespace std;const double PI = acos(-1.0);const double eps = 0.1;const int MAX = 2010;const int mod = 1e9+7;int dp[MAX][MAX];int a[MAX], n;int main(){    scanf("%d", &n);    for(int i = 1; i<=n; ++i)        scanf("%d", &a[i]);    memset(dp, 0, sizeof(dp));    for(int i = 1; i<=n; ++i)        dp[i][i] = n*a[i];    for(int k = 1;k<n; ++k)        for(int i = 1; i<=n; ++i)            if(i+k<=n)            {                int j = i+k;                dp[i][j] = max(dp[i+1][j]+a[i]*(n-k), dp[i][j-1]+a[j]*(n-k));            }    printf("%d\n", dp[1][n]);    return 0;}



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