3求最大字串和

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.<br><br><center><img src=/data/images/1087-1.jpg></center><br><br>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.<br>Your task is to output the maximum value according to the given chessmen list.<br>

 

Input

Input contains multiple test cases. Each test case is described in a line as follow:<br>N value_1 value_2 …value_N <br>It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.<br>A test case starting with 0 terminates the input and this test case is not to be processed.<br>

 

Output

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

 

Sample Input

3 1 3 24 1 2 3 44 3 3 2 10

 

Sample Output

4103

 

题意：给一串数，求其上升子序列和最大是多少

思路：动态规划，遍历。

设dp[i]为以num[i]为结尾时，这个字串的和，然后依次求出数串中每个元素结尾时的字串和，求出最大值便可。例如

1,2,3,5,,4

字串有

1

1,2

1,2,3

1,2,3,5

1,2,3,4

显然1,2,3,5字串和最大为11故输出便可

#include<iostream>

#include<cstring>
using namespace std;
int num[1001];
int dp[1001];//表示以i为结尾的字串的和
int main()
{


    int n,max;
    while(cin>>n)
    {
        memset(dp,0,sizeof(dp));
     if(n==0)
        break;
    for(int i=1;i<=n;++i)
    {
        cin>>num[i];
        dp[i] = num[i];
    }


    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<i;++j)
        {
            if(num[j] < num[i]&&dp[j] + num[i] > dp[i])//比前一个数小，并且这个数之前的字串和加上这个数时要大于这个数
            {
              dp[i] = dp[j] + num[i];
            }


        }
    }
    int max = -1000000;
    for(int i=1;i<=n;++i)//求最大值
    {
        if(dp[i] > max)
            max = dp[i];
    }
     cout<<max<<endl;
    }
}