POJ Power Strings 2406

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 44948 Accepted: 18770

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.

Source

Waterloo local 2002.07.01

题意：kmp求循环节

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>using namespace std;#define maxx 1123456int next[maxx];char t[maxx];void get_next(char t[]){    int l=strlen(t);    int i=0,j=-1;    next[0]=-1;    while(i<l)    {        if(j==-1 || t[i]==t[j])        {            i++;j++;next[i]=j;        }        else        {            j=next[j];        }    }}int main(){    int n,m,i,j;    while(~scanf("%s",t))    {        if(strcmp(t,".")==0)        {            break;        }        get_next(t);        int l=strlen(t);        if(l%(l-next[l])==0)        printf("%d\n",l/(l-next[l]));        else            printf("1\n");    }    return 0;}