题目链接:https://vjudge.net/problem/HDU-3746 Cyclic Nacklace Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10985    Accepted Submission(s): 4692 Problem Description CC always becomes very depresse…

Problem Description CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Bein…

题目链接:HDU 3746 Sample Input 3 aaa abca abcde Sample Output 0 2 5 Author possessor WC Source HDU 3rd "Vegetable-Birds Cup" Programming Open Contest Solution 题意 给定一个字符串,问至少需要在末尾添加多少个字符使得字符串循环. 思路 KMP 设前缀函数为 \(\pi()\),字符串长度为 \(n\),下标从 \(1\) 开始. 最小循环…

题意:给出一段字符串  求最少在最右边补上多少个字符使得形成循环串(单个字符不是循环串) 自己乱搞居然搞出来了... 想法是:  如果nex[len]为0  那么答案显然是补len 否则  答案为循环节的长度-nex[len] 小于0变0 写了一个假算法居然能过    ... ababca就错了 kmp重要性质: 重要的性质len-next[i]为此字符串的最小循环节(i为字符串的结尾),另外如果len%(len-next[i])==0,此字符串的最小周期就为len/(len-next[i]);…

https://www.bnuoj.com/v3/contest_show.php?cid=9147#problem/F [题意] 给定一个字符串,问在字符串后最少添加多少个字母,得到的新字符串能是前缀循环的字符串. [思路] 这道题的关键是要理解KMP中的nxt数组什么含义. nxt[i]就是以i结尾的字符串中前缀和后缀匹配的最长长度. 所以len-nxt[len]就是最小循环节. 如abcdab,nxt[len]就是2,最小循环节就是4(abcd). 理解了这个这道题就迎刃而解了. [网上解…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3746 题意:问在一个字符串末尾加上多少个字符能使得这的字符串首尾相连后能够循环 题解:就是利用next的性质求最小循环节 kmp求最下循环节http://www.cnblogs.com/wuyiqi/archive/2012/01/06/2314078.html ans=len-next[len];ans就是最小循环节长度证明看那个博客. 然后就简单了. #include <iostream> #…

Cyclic Nacklace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15422 Accepted Submission(s): 6425 Problem Description CC always becomes very depressed at the end of this month, he has checked his…

题目链接:https://vjudge.net/problem/HDU-1358 Period Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9652    Accepted Submission(s): 4633 Problem Description For each prefix of a given string S with…

Common Divisors CodeForces - 182D 思路:用kmp求next数组的方法求出两个字符串的最小循环节长度(http://blog.csdn.net/acraz/article/details/47663477,http://www.cnblogs.com/chenxiwenruo/p/3546457.html),然后取出最小循环节,如果最小循环节不相同答案就是0,否则求出各个字符串含有的最小循环节的数量,求这两个数量的公因数个数(也就是最大公因数的因子个数)就是答案.…

KMP算法: 一:next数组:next[i]就是前面长度为i的字符串前缀和后缀相等的最大长度,也即索引为i的字符失配时的前缀函数. 二:KMP模板 /* pku3461(Oulipo), hdu1711(Number Sequence) 这个模板 字符串是从0开始的 Next数组是从1开始的 */ #include <iostream> #include <cstring> using namespace std; ; int _next[maxn]; char S[maxn],…