Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

H - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2015-11-18) Description Nowadays, a kind of chess game called “Su…

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[…

Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2015-09-05) Description Nowadays, a kind of chess game called “Super…

http://acm.hdu.edu.cn/showproblem.php?pid=1087   Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Description Nowadays, a kind of chess game called “Super Ju…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1087 Online Judge Online Exercise Online Teaching Online Contests Exercise Author F.A.QHand In HandOnline Acmers Forum |DiscussStatistical Charts Problem ArchiveRealtime Judge StatusAuthors Ranklist       C/C…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1087 设dp[i]表示去到这个位置时的最大和值.(就是以第i个为结尾的时候的最大值) 那么只要扫描一遍dp数组,就能得到ans,因为最后一步可以无条件到达终点. 那么可以用O(n^2)转移,枚举每一个位置,其中要初始化dp值,dp[i] = a[i],意思就是到达第i个位置的时候,和值最小也是 a[i]把,因为无论如何也可以一步到达a[i]这个值. #include <cstdio> #include &l…