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.

3 1 3 2
4 1 2 3 4
4 3 3 2 1
0

4
10
3
题意描述：
输入N和N个数字（均是整型数）
计算并输出最大上升子序列之和
解题思路：
该题属于DP水题，第一层循环遍历数组，第二层循环遍历以i结尾的最大上升子序列，判断是否满足上升，是的话选择dp[i]和dp[j]+num[i]的较大值，第二层循环结束后看以i结尾的最大上升子序列是否变大，变大则更新dp[i],并且每
次记录最大值maxr。双层循环结束后输出maxr即可。
代码实现：

 1 #include<stdio.h>
 2 #include<algorithm>
 3 using namespace std;
 4 #include<string.h>
 5 int main()
 6 {
 7     int n,i,j,dp[1010],num[1010],maxr;
 8     while(scanf("%d",&n),n != 0)
 9     {
10         num[0]=0;
11         for(i=1;i<=n;i++)
12             scanf("%d",&num[i]);
13         memset(dp,0,sizeof(dp));
14         maxr=-9999999;
15         for(i=1;i<=n;i++)
16         {
17             for(j=0;j<i;j++)
18             {
19                 if(num[j]<num[i])
20                 dp[i]=max(dp[i],dp[j]+num[i]);
21             }
22             dp[i]=max(dp[i],num[i]);
23             if(dp[i] > maxr)
24             maxr=dp[i];
25         }
26         printf("%d\n",maxr);
27     }
28     return 0;
29  }