Power Strings （循环节问题）

POJ - 2406 

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.

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1e6+7;
char str[MAXN];
int nxt[MAXN];
void getnxt(int m){
//	int i=0,j=-1;
//	nxt[0]=-1;
//	while(i<m){
//		if(j==-1||str[i]==str[j]){
//			i++;
//			j++;
//			nxt[i]=j;
//		}
//		else j=nxt[j];
//	}
	int i=1,j=0;
	nxt[0]=0;
	while(i<m){
		if(str[i]==str[j]) nxt[i++]=++j;
		else if(!j) {
			nxt[i]=0;
			i++;
		}
		else j=nxt[j-1];
	}
}
int main(){
	while(~scanf("%s",str) && str[0]!='.'){
		int len=strlen(str);
		getnxt(len);
		int l=len-nxt[len-1];   //l为最小循环节 
		//printf("%d\n",nxt[len]); 
		if(len%l==0) printf("%d\n",len/l);
		else printf("1\n");
	}
	return 0;
}