Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51926 Accepted Submission(s): 24066

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（棋盤）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

先來一段蜜汁翻譯：
超級跳跳跳
一行有n個位置，可以從任意一個開始跳。只能跳向比當前數值大的位置，並且把落點的值加起來，最後輸出最大的和。
清奇的思路
每個位置記錄，這個位置的數值num，以及跳到這個位置時的最大和max。
每個位置i都向前尋找一個最大的位置j作為上一個落點，j要滿足兩個條件：1、位置j的數值num比當前位置i的數值小num。2、在滿足上一個條件的情況下，位置j的最大和max應該也是最大的。然後i.max=j.max+i.num.最後只需要把這些位置中的最大的max輸出就行了

#include<stdio.h>
struct p{
    int num;
    int max;
};
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        struct p a[1001];
        int i;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i].num);
            int j;
            int ans=0;
            for(j=0;j<i;j++)
            {
                if(a[j].num<a[i].num&&ans<a[j].max)
                    ans=a[j].max;
            }
            a[i].max=a[i].num+ans;
        }
        int ans=0;
        for(i=0;i<n;i++)
        {
            if(a[i].max>ans)
                ans=a[i].max;
        }
        printf("%d\n",ans);
    }
}