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（棋盤）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

思路：
求最長上升子序列的和，可以先看前三項：

第一項：dp[1] = num[1]

第二項：若num[2]>num[1]，則dp[2] = dp[1] + num[2]，否則dp[2] = num[2]

第三項：若num[3]>num[2]，則dp[3] = dp[2] + num[3]；
? ? ? ? ? ??? 若num[3]>num[1]，則dp[3] = dp[1] + num[3]；
? ? ? ? ? ??? 若num[3]≯num[2]且num[3]≯num[1]，則dp[3] = num[3]

綜上：用之前的每一個數和當前的數比較，比當前數小的就加上dp（之前數），否則就等于當前數。

Source Program

#include<iostream> #include<cstring> using namespace std; int main() {int n;int num[1001],dp[1001];int max_num;int i,j;while(cin>>n && n){/*清零工作*/max_num=-1;memset(dp,0,sizeof(dp));for(i=1;i<=n;i++)cin>>num[i];for(i=1;i<=n;i++){for(j=0;j<=i;j++)//之前數比當前數小，就將當前數與之前數相加if(num[i]>num[j])dp[i]=max(dp[i],dp[j]+num[i]);dp[i]=max(dp[i],num[i]);//若之前數不比當前數小，直接就是當前數max_num=max(max_num,dp[i]);//尋找最大和}cout<<max_num<<endl;}return 0; }

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總結

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