Power Strings（KMP）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 45008		Accepted: 18794

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

Hint

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

Source

Waterloo local 2002.07.01

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#define N 1000010
using namespace std;
int next[N];
char str[N];
void getnext(char str[],int next [])
{
    next[0]=next[1]=0;
    int len=strlen(str);
    for(int i=1;i<len;i++)
    {
        // cout<<len<<endl;
        int k=next[i];
        while(k&&str[i]!=str[k]) 
        {
            k=next[k];
            //cout<<k<<endl;
        }    
        next[i+1]=(str[i]==str[k])?k+1:0;
        
    }
}
int main()
{
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    while(scanf("%s",&str)!=EOF)
    {
        //cout<<str<<endl;
        int m=strlen(str);
        if(m==1)
            break;
        getnext(str,next);
        // for(int i=0;i<m;i++)
            // cout<<next[i]<<" ";
        // cout<<endl;
        if(m%(m-next[m])==0)
            printf("%d
",m/(m-next[m]));
        else
            puts("1");
    }
    return 0;
}