HDU 1087 题目大意:给定一个序列,只能走比当前位置大的位置,不可回头,求能得到的和的最大值.(其实就是求最大上升(可不连续)子序列和) 解题思路:可以定义状态dp[i]表示以a[i]为结尾的上升子序列的和的最大值,那么便可以得到状态转移方程 dp[i] = max(dp[i], dp[j]+a[i]), 其中a[j]<a[i]且j<i; 另外每个dp[i]可以先初始化为a[i] 理解:以a[i]为结尾的上升子序列可以由前面比a[i]小的某个序列加上a[i]来取得,故此有dp[j]+a[…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24452    Accepted Submission(s): 10786 Problem Description No…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47055    Accepted Submission(s): 21755 Problem Description N…

Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2017-04-13) Description Nowadays, a kind of chess game called “Super…

Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47017    Accepted Submission(s): 21736 Problem DescriptionNowadays, a kind of chess game called “Super Jumping! Ju…

Super Jumping! Jumping! Jumping! 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 ca…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 59516    Accepted Submission(s): 27708   Problem Description Nowadays, a kind of chess game called “Super Jumping…

Super Jumping! Jumping! Jumping! 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 ca…

题意: 求解最大递增子序列. 例如:3 1 3 2 输入 3  个数 1 3 2 则递增子序列有 {1} {3} {2} {1 3} {1 2} ,故输出子序列的最大和 4 解题思路: x[n](n个数) 数组存储 输入的数据 dp[i] 用来记录 前 i+1 {数据从下标为0开始存储}个数的 最大结果 遍历dp[] 找到最大值即可. 因 G20 网站暂停提交 明天测试代码. 2016.08.27 代码提交成功! 代码如下: #include<bits/stdc++.h> using name…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32564    Accepted Submission(s): 14692 Problem Description Nowadays, a kind of chess game called “Super Jumping!…