POJ 2406 kmp求循环节个数

题目链接：

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

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 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

abcd
aaaa
ababab
.

Sample Output

1
4
3

/*************************************************************************    > File Name: poj_2406.cpp    > Author: dulun    > Mail: [e-mail]    > Created Time: 2016年03月16日 星期三 13时19分57秒 ************************************************************************/#include<iostream>#include<stdio.h>#include<cstring>#include<cstdlib>#include<algorithm>#define LL long longusing namespace std;const int N = 1000086;char a[N];int nxt[N];void getnxt(){    int m = strlen(a);    int k = 0;     memset(nxt, 0, sizeof(nxt));    for(int i = 1; i < m ; i++)    {        while(k&& a[k] != a[i]) k = nxt[k-1];        if(a[k] == a[i]) k++;        nxt[i] = k;    }}int main(){    while(~scanf("%s", a) && a[0] != '.')    {        getnxt();        int m = strlen(a);        int t = m - nxt[m-1];        if( m % t == 0)            printf("%d\n", m / t);        else            printf("1\n");        memset(a, 0, sizeof(0));    }    return 0;}