题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1087

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24452    Accepted Submission(s):
10786

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

题目大意：求解最大递增子序列和，这里要特别注意一点：这个题目不要求是连续的最大递增子序列。但是一定要注意是递增的！！

题目思路：dp数组表示是包含当前这个数的最大递增子序列和。dp[i]表示的是前i个并且包含第i个的最大递增子序列和！给个数据：3 1 4 显然dp[1]=3,dp[2]=1表示两个数的最大值。因为分两种情况讨论，如果第二个数大于第一个数，就加上，即dp[2]=dp[1]+num[2]；如果不大，dp[2]=num[2];dp[3]=7表示三个数的最大值。首先比较num[3]和num[1],如果num[3]>num[1],dp[3]=7先存下来,如果num[3]>num[2],dp[3]=5依旧存下来;还有一种如果num[3]比前两个值都小，dp[3]=num[3];最后在存下来的dp[3]中找到一个最大的！呼呼~终于解释差不多了，小伙伴们也屡一下思路吧0.0

详见代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4
 5 using namespace std;
 6
 7 int num[1010],dp[1010];
 8
 9 int main ()
10 {
11     int n,Max;
12     while (~scanf("%d",&n))
13     {
14         if (n==0)
15             break;
16         Max=0;
17         memset(dp,0,sizeof(dp));
18         for (int i=0;i<n;i++)
19         {
20             scanf("%d",&num[i]);
21         }
22         dp[0]=num[0];
23         for (int i=1;i<n;i++)
24         {
25             for (int j=0;j<i;j++)
26             {
27                 if (num[i]>num[j])
28                 dp[i]=max(dp[i],dp[j]+num[i]);
29             }
30             dp[i]=max(dp[i],num[i]);
31         }
32         for (int i=0;i<n;i++)
33         {
34             Max=max(Max,dp[i]);
35         }
36         printf ("%d\n",Max);
37     }
38     return 0;
39 }