题目链接&＃xff1a;hdu 1087

Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45600    Accepted Submission(s): 21127

Problem Description

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&＃xff08;棋盘&＃xff09;and some chessmen&＃xff08;棋子&＃xff09;, 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

题意&＃xff1a;找最大的上升的子序列的和是多少&＃xff0c;不一定最长。

题解&＃xff1a;定义 dp [ i ] 为以 i 位置为结尾的最大的上升的序列的和的值&＃xff0c;O(n^2) 遍历序列即可。

AC代码&＃xff1a;

#include
#include
#include
#include using namespace std;long long dp[10010];
long long a[10010];
int n;int main(){ios::sync_with_stdio(false);while(cin >> n && n){for(int i&＃61;1;i<&＃61;n;i&＃43;&＃43;){cin >> a[i];dp[i]&＃61;a[i];}for(int i&＃61;1;i<&＃61;n;i&＃43;&＃43;){for(int j&＃61;1;j<&＃61;i;j&＃43;&＃43;){if(a[i]>a[j]){ //如果i位置的大于j位置的&＃xff0c;那么更新dp[i]的最大值dp[i]&＃61;max(dp[i],dp[j]&＃43;a[i]); //状态转移更新最大值}}}long long ans&＃61;0;for(int i&＃61;1;i<&＃61;n;i&＃43;&＃43;){ans&＃61;max(ans,dp[i]); //找以i位置结尾的和的最大值}cout <}

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