Sicily 11600. Pick It

11600. Pick It

Constraints

Time Limit: 5 secs, Memory Limit: 256 MB

Description

You are playing a game of Pick It. You are given a list of positive integers, and you are allowed to select any number other than the first or last number in this list. When you pick a number, that number is removed from the board, and your score increases by the sum of the number that you picked and the sum of the neighbouring numbers.

For example, if the list contained 1 2 3 4 5, and you picked 3, your score would be2+3+4=9. On the next turn, your list would be 1 2 4 5, and if you picked 4 next, your score would be 9+2+4+5 = 20, leaving you with the list 1 2 5. The game concludes when there are only two numbers remaining.

Given a list of numbers, what is the maximum score that you can obtain?

Input

The input will consist of a number of test cases (at most 200 test cases). A test case is of the form n k₁ k₂ ... k_n where n (n ≤ 200) is the number of numbers in the list, and each integer k_i satisfies 1 ≤ k_i ≤ 100). In all test cases n ≥ 3, except in the case where n=0, which indicates the end of input.

Output

For each test case, output the maximum score attainable.

Sample Input

5 1 2 3 4 55 2 1 5 3 46 30 20 40 50 70 600

Sample Output

3031570

Problem Source

2014年每周一赛第八场

http://blog.csdn.net/gwq5210/article/details/40949033

// Problem#: 11600// Submission#: 3550130// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <string.h>unsigned short dp[200][200], w[200], N;inline int max(int a, int b) {if (a > b) return a; return b;}int main() {    while (scanf("%d", &N) && N) {        for (int i = 0; i < N; i++) scanf("%d", w + i);        N--;        memset(dp, 0, sizeof(dp));        for (int i = 1; i < N; i++)             for (int j = 1; j + i <= N; j++)                 for (int k = j; k < j + i; k++)                     dp[j][j + i] = max(dp[j][j + i], dp[j][k] + dp[k + 1][j + i] + w[j - 1] + w[k] + w[j + i]);        printf("%d\n", dp[1][N]);    }    return 0;}