Max Sum Plus Plus-HDU1024(dp)】的更多相关文章

[题解]最大 M 子段和 Max Sum Plus Plus [Hdu1024] [51nod1052] 传送门:最大 \(M\) 子段和 \(Max\) \(Sum\) \(Plus\) \(Plus\) \([Hdu1024]\) \([51nod1052]\) [题目描述] 给出一个长度为 \(N\) 的序列 ,将这 \(N\) 个数划分为互不相交的 \(M\) 个子段,并使得 \(M\) 个子段的和最大. [样例] 样例输入: 7 2 -2 11 -4 13 -5 6 -2 样例输出:…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17164    Accepted Submission(s): 5651 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41885    Accepted Submission(s): 15095 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are fac…
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.  Given a consecutive number sequence S 1, S 2, S 3,…
Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequ…
Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 207989 Accepted Submission(s): 48681 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36673    Accepted Submission(s): 13069 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" proble…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 #include<iostream> #include<vector> #include<algorithm> #include<cmath> #include<cstring> #include<queue> #include<cstdio> #include<unordered_map> #define in…
这题的意思就是取m个连续的区间,使它们的和最大,下面就是建立状态转移方程 dp[i][j]表示已经有 i 个区间,最后一个区间的末尾是a[j] 那么dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][1..j-1])+a[j]) 看数据范围,1e6 肯定开不下数组,观察发现,dp[i][j]仅和dp[i][j-1]和dp[i-1][1..j-1]中最大值有关,即只和dp[i-1]有关 所以开滚动数组求解   复杂度可以通过开数组mmax[j]表示dp[i-1][1.…