题干：

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

解题报告：

     就是个最大递增子序列。

附一个别人的解题报告：

      求最大的递增子序列的和，不用连续，例如（1,5,2），那么（1,2）也算他的递增子序列。　

　　我刚开始做了一遍，忽略了不用连续这个问题，结果发现好简单，提交就是不对，应当注意这里。

　　这个还是得用动态规划去做，最后的最优结果是由上一步的结果加上上一次的决策。n个数，由n-1个数的结果加上第n个数。n-1个数由n-2个数的结果。。。。。

　　推倒最后，前两个数的的结果就很容易求了.以数列（3,2,4,2,3,6）为例。​​

 


下标i

0

1

2

3

4

5

a[i]

3

2

4

2

3

6

sum[i]

3

2

7

2

5

13

ans

0

3

7

7

5

13


 

 

 

 

AC代码：

//dp需要赋初值！
#include<bits/stdc++.h>

using namespace std;
int n;
int a[1000 + 5];
int dp[1000 + 5];
int main()
{
    while(~scanf("%d",&n)) {
        if(n == 0) break;
        memset(dp,0,sizeof(dp));
        for(int i = 1; i<=n; i++) scanf("%d",&a[i]);
        for(int i = 1; i<=n; i++) dp[i] = a[i];
        for(int i = 1; i<=n; i++) {
            for(int j = 1; j<i; j++) {
                if(a[i] > a[j] )
                    dp[i] = max(dp[j] + a[i],dp[i] );
            }
        }
        printf("%d\n",*max_element(dp+1,dp+n+1));
    }
    return 0 ;
}