Problem  Description：

 

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.

Your task is to output the maximum value according to the given chessmen list.

Input：

Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N 
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.

A test case starting with 0 terminates the input and this test case is not to be processed.

Output：

For each case, print the maximum according to rules, and one line one case.

Sample  Input：

3 1 3 2
4 1 2 3 4
4 3 3 2 1

0

Sample  Output：

4
10

3

题意：棋盘上有n个棋子，所有棋子都用正整数标记，玩家只能从一个棋子跳到另一个更大的棋子上，并且玩家可以穿过多个棋子，但不能后退，求玩家所能得到的最高分。

思路：这道题是dp的思想。

My  DaiMa：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
    int n,a[1003],b[1003],flag,sum;//b[i]数组是用来存前i个棋子能得的最高分
    while(~scanf("%d",&n))
    {
        sum=-9999;
        if(n==0) break;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(b,0,sizeof(b));
        for(int i=0;i<n;i++)//此for循环是dp的核心代码
        {
            flag=0;
            for(int j = 0; j < i; j ++)
            {
                if(a[j]<a[i]&&flag<b[j])//找出前i个棋子中比a[i]棋子小并且分数最高的棋子，记录它的分值
                    flag=b[j];
            }
            b[i]=flag+a[i];//前i-1个棋子的最高分加上第i个棋子的分，求得即是前i个棋子的最高分
        }
        for(int i=0;i<n;i++)
        {
            if(sum < b[i])
                sum = b[i];
        }
        printf("%d\n",sum);
    }
    return 0;
}