Poj-2406-Power Strings【KMP】

传送门：http://poj.org/problem?id=2406常发生

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50464		Accepted: 21048

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

Hint

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

Source

题意：拿第三组数据来看 找到子串ab 则s为ababab  a为ab  n为s除以a   

第二组数据 匹配子串为一个a（这里就是next数组的运用）s为aaaa  a为a  n为4

#include
#include
#define M 1000005
char s[M];
int next[M];
int len;
void Next()
{
	int i=0;
	int j=-1;
	next[0]=-1;
	j=next[i];
	while(i