[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher :G - Power Strings POJ

[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher :G - Power Strings POJ - 2406（kmp简单循环节）...

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发布时间：2019-06-08

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[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher

G - Power Strings

POJ - 2406

题目：

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 defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcdaaaaababab.

Sample Output

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题意：求字符串的循环节个数

思路：要注意如果该字符串如果不满足都是循环组成的则输出-1，若是都是循环节组成的，那么n-nextt[n]就是循环节长度，用总长度除以这个长度就是循环节长度

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                    using namespace std;const int maxn=1e7+10;char str[maxn];int nextt[maxn];void getnextt(){ int i=0,j=-1; nextt[0]=-1; int n=strlen(str); while(i

转载于:https://www.cnblogs.com/Vampire6/p/11360900.html

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