Power Strings

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 51222 Accepted: 21388

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

abcdaaaaababab.

Sample Output

143

思路：循环节只要不被长度整出就输出1.

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;char a[1000010];int f[1000010];int len;void getnext(){    int i,j=0;    f[1]=0;    for(i=2; i<=len; i++)    {        while(j>0&&a[j+1]!=a[i])            j=f[j];        if(a[j+1]==a[i])            j++;        f[i]=j;    }}int main(){    while(gets(a+1))    {       if(!strcmp((a+1) , "."))        break;        len=strlen(a+1);        getnext();        int l=len-f[len];        if(len%l==0)        {            printf("%d\n",len/l);        }        else        {            printf("1\n");        }    }    return 0;}