HDU 1087 Super Jumping! Jumping! Jumping! (DP)

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 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. 

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

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define    MAX    1005
 5 
 6 int    main(void)
 7 {
 8     int    n;
 9     int    s[MAX];
10     int    dp[MAX],max,ans;                        //dp[i]存的是以i为起点的最大升序子串的和
11 
12     while(scanf("%d",&n) && n)
13     {
14         for(int i = 0;i < n;i ++)
15             scanf("%d",&s[i]);
16 
17         ans = dp[n - 1] = s[n - 1];
18         for(int i = n - 2;i >= 0;i --)            
19         {
20             max = 0;
21             for(int j = i + 1;j < n;j ++)         //看s[i]后面哪一个子串是s[i]可以加进去的,找出所有这样的子串,取DP值最大那个
22                 if(s[i] < s[j] && max < dp[j])
23                     max = dp[j];
24             dp[i] = s[i] + max;
25 
26             ans = ans < dp[i] ? dp[i] : ans;
27         }
28         ans = ans < 0 ? 0 : ans;                 //注意可以从起点直接跳终点,此时值为0,若是算出最终值是负数的话要选此方案
29         
30         printf("%d\n",ans);
31     }
32 
33     return    0;
34 }