dp --A - Super Jumping! Jumping! Jumping!

A - Super Jumping! Jumping! Jumping!

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.

InputInput 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.
OutputFor 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：求升序子序列的最大和

 #include<bits/stdc++.h>
 using namespace std;
 
 int dp[];//dp用于存放从数组头到此为止的最大升序子序列和
 int n, a[];
 int maxn;
 
 int main(){
     while(~scanf("%d", &n) && n){
         maxn = ;
         for(int i=; i<n; i++){
             scanf("%d", a+i);
             dp[i] = a[i];//将每个ai的值赋予dpi
         }
         for(int i=; i<n; i++){
             for(int j=; j<i; j++){
                 if(a[i] > a[j] && dp[i] < dp[j]+a[i])
                     dp[i] = dp[j] + a[i];
                 if(dp[i] > maxn)    maxn = dp[i];
             }
         }
 
         printf("%d\n",maxn);
     }
 }

是今天刚学的dp，告诉我们这类题目没有固定的模板，更重要的是思维和思考，借助dp这样的一个数组，去存放你需要求或相关的变量。既然没有固定模板，则需要更多的练习，去接触更多样式的巧妙的题目