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

输入

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.

输出

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

样例输入

abcd

aaaa

ababab

.

样例输出

1

4

3

提示

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

思路：

就是问一个字符串写成(a)^n的形式，求最大的n.

根据KMP的next函数的性质，已知字符串t第k个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串。

#include
     
      using namespace std;const int M=1e6+5;char t[M];int next[M],tlen;void getNext(){    int i=0,j=-1;    next[0]=-1;    while(i