poj 2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 39097		Accepted: 16230

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

#include<stdio.h>
#include<string.h>
int p[10000010];
char s[10000010];
int len;
void kmp()
{
int i=0,j=-1;
p[i]=j;
while(i<len)
{
if(j==-1||s[i]==s[j])
{
i++;
j++;
p[i]=j;
}
else
j=p[j];
}
}
int main()
{
while(scanf("%s",s)!=EOF)
{
if(strcmp(s,".")==0)
break;
memset(p,0,sizeof(p));
len=strlen(s);
kmp();
if(len%(len-p[len])==0)
printf("%d\n",len/(len-p[len]));//循环次数为
else
printf("1\n");
//printf("%d\n",len-p[len]);//循环节长度为

}
return 0;
}