描述 Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is de…

Power Strings Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 29663Accepted: 12387 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef…

http://poj.org/problem?id=2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27003   Accepted: 11311 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &q…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28102   Accepted: 11755 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

点击打开链接 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27368   Accepted: 11454 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b =…

题目链接:http://poj.org/problem?id=2406 Time Limit: 3000MS Memory Limit: 65536K Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we thi…

本题是计算一个字符串能完整分成多少一模一样的子字符串. 原来是使用KMP的next数组计算出来的,一直都认为是能够利用next数组的.可是自己想了非常久没能这么简洁地总结出来,也仅仅能查查他人代码才恍然大悟,原来能够这么简单地区求一个周期字符串的最小周期的. 有某些大牛建议说不应该參考代码或者解题报告,可是这些大牛却没有给出更加有效的学习方法,比方不懂KMP.难倒不应该去看?要自己想出KMP来吗?我看不太可能有哪位大牛能够直接自己"又一次创造出KMP"来吧. 好吧.不说"创造…

题意:给一个字符串,求该串最多由多少个相同的子串相接而成. 思路:只要做过poj 1961之后,这道题就很简单了.poj 1961 详细题解传送门. 假设字符串的长度为len,如果 len % (len - next[len])不为0,说明该字符串不能由其他更短的字符串反复相接而成,结果输出1,否则答案为len / (len - next[len]). #include<stdio.h> #include<string.h> #define maxn 1000010 char s[…

传送门 http://poj.org/problem?id=2406 题目就是求循环了几次. 记得如果每循环输出为1.... #include<cstdio> #include<cstring> const int MAXN=1000000+10; char P[MAXN]; int f[MAXN]; int n,m; void getFail() { int i,j; f[0]=f[1]=0; for(i=1;i<n;i++) { j=f[i]; while(j &…

对于数组s[0~n-1],计算next[0~n](多计算一位). 考虑next[n],如果t=n-next[n],如果n%t==0,则t就是问题的解,否则解为1. 这样考虑: 比方字符串"abababab", a  b a b a b a b * next     -1 0 1 2 3 4 5 6  7 考虑这种模式匹配,将"abababab#"当做主串."abababab*"当做模式串.于是进行匹配到n(n=8)时,出现了不匹配: 主串    …

题目传送门 /* 题意:一个串有字串重复n次产生,求最大的n KMP:nex[]的性质应用,感觉对nex加深了理解 */ /************************************************ * Author :Running_Time * Created Time :2015-8-10 10:51:54 * File Name :POJ_2406.cpp ************************************************/ #incl…

连续重复子串问题 poj 2406 Power Strings http://poj.org/problem?id=2406 问一个串能否写成a^n次方这种形式. 虽然这题用kmp做比较合适,但是我们还是用后缀数组做一做,巩固后缀数组的能力. 对于一个串,如果能写出a^n这种形式,我们可以暴力枚举循环节长度L,那么后缀suffix(1)和suffix(1 + L)的LCP应该就是 lenstr - L.如果能满足,那就是,不能,就不是. 这题的话da算法还是超时,等我学了DC3再写上来. 其实这…

[BZOJ2320]最多重复子串 Description 一个字符串P的重复数定义为最大的整数R,使得P可以分为R段连续且相同的子串.比方说,“ababab”的重复数为3,“ababa”的重复数为1. Your Task 对于给定的串S,找出S的一个子串K使得K的重复数最大. Input 第一行T表示数据组数 对于每组数据,一行中一个仅包含小写字母的字符串S Output 对于每组数据,在一行中输出K,如果有多个解,输出字典序最小的那一个 Sample Input 2 ccabababc daa…

题意:重复子串次数 思路:kmp #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 1000005 int next[MaxSize]; void GetNext(char t[]){//求next数组 int j,k,len; j=; k=-; next[]=-; len=strlen(t); while(j<len){ ||t…

Power Strings Problem's Link: http://poj.org/problem?id=2406 Mean: 给你一个字符串,让你求这个字符串最多能够被表示成最小循环节重复多少次得到. analyse: KMP之next数组的运用.裸的求最小循环节. Time complexity: O(N) Source code:  ;;      ;);      ) ;}/* */…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 30069   Accepted: 12553 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28859   Accepted: 12045 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

题目链接:http://poj.org/problem?id=2406 题意:给出一个字符串s,求重复子串出现的最大次数. 分析:kmp的next[]数组的应用. 要求重复子串出现的最大次数,其实就是求字符串的最小循环节. 以下内容转载于:http://bbezxcy.iteye.com/blog/1377787 --------------------------------------------------------------------------------------------…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 37685   Accepted: 15590 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

Power Strings Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 29   Accepted Submission(s) : 14 Problem Description Given two strings a and b we define a*b to be their concatenation. For example,…

F - Power Strings Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2406 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &…

Power Strings   Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 47748   Accepted: 19902 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = &quo…

题目链接:http://poj.org/problem?id=2406 题意:确定字符串最多是多少个相同的字串重复连接而成的 思路:关键是找到字符串的最小循环节 code: #include <cstdio> #include <cstring> ; char s[MAXN]; int next[MAXN]; void GetNext() { int len = strlen(s); ; ; next[] = -; while (i < len) { == j || s[i]…

Description Problem D: Power Strings Given two strings a and b we define a*b to be their concatenation. For example, ifa = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiatio…

题目传送门 Power Strings 格式难调,题面就不放了. 一句话题意,求给定的若干字符串的最短循环节循环次数. 输入样例#1: abcd aaaa ababab . 输出样例#1: 1 4 3 就这样. 分析: 一道思路神奇的题目,需要深入理解$KMP$的$next$数组. 如果自己写几个字符串推一下就可以发现,一个由循环节构成的字符串,从第二个循环节开始$next$值是依次递增的,因为$next$数组的本质是表示$0\~i-1$的最长公共前缀后缀长度.也就不难想到,只要判断一下$nex…

题目链接:https://vjudge.net/problem/POJ-2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 52631   Accepted: 21921 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc"…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 36926   Accepted: 15254 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

                                                                                                  Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 38038   Accepted: 15740 Description Given two strings a and b we define a*b t…

Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 48139   Accepted: 20040 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "…

题目链接 题意:连续重复子串.给定一个字符串 L,已知这个字符串是由某个字符串 S 重复 R 次而得到的(L = S^R ), 求 R 的最大值. 分析:枚举长度,判断条件是能被总长度整除且LCP (suffix (0), suffix (i)) = n - i,预处理出lcp,方法是,lcp[i] = min (height[rank[i]] to height[rank[0]]); DC3算法,C++提交才能AC. #include<cstdio> #include<cstring&…