poj 2406 Power Strings

poj 2406 Power Strings

Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 33636 Accepted: 13973

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
題目大意：給出一個字符串，求這個串是有幾個子串構成的
思路：用KMP算法中的next數組，這個字符串的next數組裡邊，字符串的總長度減去最後一個字符所對應的next值就是子串的長度
	具體為什麼，應該是next數組的值是根據這個字符串本身，用一定規則得出的，這個真的好巧妙的說，kmp算法，好好看看。
2014,12,5
#include
#include
char x[1100000];
int next[1100000];
void getnext(char s[]){
	int len1,i,j;
	len1=strlen(s);
	i=0;j=-1;
	next[i]=j;
	while(i