Power Strings(kmp next求循环节)

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 45249 Accepted: 18883

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

#include<stdio.h>#include<string.h>#define maxn 1000100int next[maxn];char s[maxn];void get_next(char *s){    int len1=strlen(s);    int i=0,j=-1;    next[0]=-1;    while(i<len1)    {        if(j==-1||s[i]==s[j])        {            next[++i]=++j;        }        else j=next[j];    }}int main(){    int len,i,j;    while(~scanf("%s",s)&&s[0]!='.')    {        len=strlen(s);        get_next(s);        if(len%(len-next[len])==0)        printf("%d\n",len/(len-next[len]));        else            printf("1\n");    }    return 0;}