poj2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33273		Accepted: 13825

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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int next[1000010],m;
char s[1000010];

void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<m)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else j=next[j];
    }
}
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        if(s[0]=='.')break;
        m=strlen(s);
        get_next();
        if(m%(m-next[m])==0)printf("%d
",m/(m-next[m]));
        else printf("1
");
    }
    return 0;
}