B - Power Strings

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 <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int next[1100000] ;
char str[1100000] ;
void getnext(int len)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||str[i]==str[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }
}
int main()
{
    int l , m ;
    while(scanf("%s", str)!=EOF)
    {
        if( str[0] == '.' ) break;
        l = strlen(str);
        getnext(l) ;
        m = next[l];
        //if(整个串是用循环节组成的),(l-m)为最小循环节 
        if( l % (l-m) != 0 )//判断最小循环节会被n整除 
            printf("1\n");
        else
        {
            m = l / ( l-m );
            printf("%d\n", m);
        }
        memset(str,0,sizeof(str));
    }
    return 0;
}