http://acm.hdu.edu.cn/showproblem.php?pid=4763 Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2129    Accepted Submission(s): 997 Problem Description It's time for music! A lot of…

题目大意:给你一个串,从这个串里面找出一个前缀后缀中间相等的串的最大长度也就是 EAEBE,每个字母都代表一个串,E出现了三次,找出最长的那个E.   分析:我们知道KMP里面保存的就是前缀和后缀的最大匹配度,如果在这匹配度中间再找一个串也让他等于这个匹配度,那么一定就是最大值,如果找不到就回朔next,其实还是挺暴力的做法,不过很意外时间用的并不多.   代码如下: ================================================================…

题意: 求出最长公共前后缀 不能重叠  而且 这个前后缀 在串的中间也要出现一次 解析: 再明确一次next数组的意思:完全匹配的最长前后缀长度 求一遍next 然后暴力枚举就好了 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set>…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4876    Accepted Submission(s): 2439 Problem Description It's time for music! A lot of popular musicians are invited to join us in…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 574    Accepted Submission(s): 308 Problem Description It's time for music! A lot of popular musicians are invited to join us in th…

http://acm.hdu.edu.cn/showproblem.php?pid=4763 Theme Section Problem Description   It's time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the pro…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 271    Accepted Submission(s): 121 Problem Description It's time for music! A lot of popular musicians are invited to join us in the…

题目链接:https://vjudge.net/problem/HDU-4763 Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4264    Accepted Submission(s): 2056 Problem Description It's time for music! A lot of popu…

题意:在一个串中找 EAEBE 的形式的最长的E,其中E为一个字符串,也就是说找到前缀与后缀相同,并且串中还存在相同的一段,它们不能重复. 思路:利用next数组,next[len]代表的即是最大的相同的前缀与后缀,然后让 i 从len-1往前遍历找到 i>=2(前面部分最少要有2个字符),在过程中更新最长的长度ans即可. #include<iostream> #include<stdio.h> #include<string.h> using namespac…

Theme Section Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1995 Accepted Submission(s): 943 Problem Description It's time for music! A lot of popular musicians are invited to join us in the mus…