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…

题目描述 给定一个字符串S,要求你找到一个最长的子串,它既是S的前缀,也是S的后缀,并且在S的内部也出现过(非端点) 题解 CF原题不解释....http://codeforces.com/problemset/problem/126/B KMP的失配函数fail[i]的值就是s[0..i]的最长前缀且是后缀的长度~~~,因此我们从S的末尾位置开始沿着失配函数跑即可,对于当前fail[i],判断前缀s[0-i]是否在s[i+1..length(s)-i]是否出现即可~~~~如果存在则是最长子串的…

题意:求最长的子串E,使母串满足EAEBE的形式,A.B可以任意,并且不能重叠. 题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=4763 思路:这题对next数组可以说是考察的非常的细,也是通过这道题,也让我对next数组有了更深刻的了解.或者说之前只是云里雾里.(题外话就到这).首先我们求出该字符串的Next数组,我们都知道对于Next[i]是从串首开始长度为i的子串的前缀与后缀相同的最大长度.那么我们标记字符串的中间位置,就是已该位…

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…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1114    Accepted Submission(s): 579 Problem Description It's time for music! A lot of popular musicians are invited to join us in t…

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…

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…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=4763 题目: Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3491    Accepted Submission(s): 1623 Problem Description It's time for music!…

Theme Section Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5091    Accepted Submission(s): 2550 Problem Description It's time for music! A lot of popular musicians are invited to join us in t…

kmp中next数组的含义是:next[i]表示对于s[0]~s[i-1]这个前缀而言,最大相等的前后缀的长度是多少.规定next[0]=-1. 迭代for(int i=next[i];i!=-1;i=next[i]) 就可以得到某个前缀所有长度相等的前后缀的长度. 这题你就暴力枚举整个字符串的所有相等的前后缀,然后暴力判中间的部分是否存在即可.当然使用next数组会很简便. #include<cstdio> #include<cstring> using namespace st…