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,…
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…
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): 28136    Accepted Submission(s): 9810 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31798    Accepted Submission(s): 11278 Problem Description Now I think…
转载请注明出处,谢谢:http://www.cnblogs.com/KirisameMarisa/p/4302208.html   ---by 墨染之樱花 dp是竞赛中常见的问题,也是我的弱项orz,更要多加练习.看到邝巨巨的dp专题练习第一道是Max Sum Plus Plus,所以我顺便把之前做过的hdu1003 Max Sum拿出来又做了一遍 HDU 1003 Max Sum 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目描述:…
Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 161294    Accepted Submission(s): 37775 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34541    Accepted Submission(s): 12341 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1081 自己真够垃圾的,明明做过一维的这种题,但遇到二维的这种题目,竟然不会了,我也是服了(ps:猪啊). 最终还是看了题解. 代码如下: #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #define inf 0x3f3f3f3f using namesp…
题目链接:https://cn.vjudge.net/problem/HDU-1024 题意 给n, m和一个序列,找m个不重叠子串,使这几个子串内元素和的和最大. n<=1e6 例:1 3 1 2 3 答:6 (唯一的子串1 2 3) 思路 先顺便记录一下动态规划的一般解题思路: 原问题->子问题->状态->转移->边界 再顺便记录一下最大值最小化这类问题套路解法: 二分 贪心 不能二分的问题,贪心八九不离十. 一般是AB和BA这两个元素的顺序,不影响前后变化时,直接算目标…