Super Jumping! Jumping! Jumping! HDU

 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 24 1 2 3 44 3 3 2 10

Sample Output

4103

题意：用之前的每一个数和当前的数比较，比当前数小的就加上dp【之前数】，不然就等于当前数

#include<stdio.h>#include<string.h>#define INF 0x3f3f3f3fint num[1010];int dp[1010];int max(int a,int b){    return a > b ? a : b;}int main(){    int n,i,j;    while(~scanf("%d",&n)&&n)    {         memset(dp,0,sizeof(dp));         num[0] = 0;         for(i=1;i<=n;i++)         {            scanf("%d",&num[i]);         }         int Max = -INF;         for(i=1;i<=n;i++)         {             for(j=0;j<i;j++)             {                if(num[j]<num[i])                {                   dp[i] = max(dp[i],dp[j]+num[i]);//之前数比当前数小，就加上当前数到之前数的和上面                }             }             dp[i] = max(dp[i],num[i]);//之前数都不比当前数小的情况下直接就是当前数             if(dp[i]>Max)//记录最大的那个和             Max = dp[i];         }         printf("%d\n",Max);    }    return 0;}