maksyuki 发表于 oj 分类,标签:

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.


Line 1: A single integer, N

Lines 2..N+1: Line i+1 contains the value of treat v(i)


Line 1: The maximum revenue FJ can achieve by selling the treats

Sample Input







Sample Output



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.


USACO 2006 February Gold & Silver



算法分析:dp[i][j]表示[i,j]范围内的子序列所能够到达的最大值,则状态转移方程为dp[i][j] = max (dp[i+1][j] + aa[i] * (n - len), dp[i][j-1] + aa[j] * (n - len)),初始化dp[i][i] = aa[i] * n,最后直接输出dp[1][n]即可