poj2406 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

这题就是求最小的循环节，直接用len-next[len] &&　next[len]!=0就行，对于len%(len-next[len])!=0要输出0.

#include<stdio.h>
#include<string.h>
char s[1000006];
int len,next[1000006];
void nextt()
{
	int i,j;
	i=0;j=-1;
	memset(next,-1,sizeof(next));
	while(i<len){
		if(j==-1 || s[i]==s[j]){
			i++;j++;next[i]=j;
		}
		else j=next[j];
	}
}

int main()
{
	int n,m,i,j;
	while(scanf("%s",s)!=EOF)
	{
		if(strcmp(s,".")==0)break;
		len=strlen(s);
		//printf("%d
",len);
		nextt();
		if(len%(len-next[len])==0 && next[len]!=0){
			printf("%d
",len/(len-next[len]));
		}
		else printf("1
");
	}
}