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: [email protected] > 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 long
using 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;
}