POJ2406

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23530		Accepted: 9880

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>

using namespace std;

char str[1000010];
int next[1000010];

int main()
{
    int i,j;
    int len;
    int ans;

    while(scanf("%s",str)!=EOF)
    {
        if(!strcmp(str,"."))
            break;
        memset(next,0,sizeof(next));
        //kmp -- getNext()
        i=-1;
        j=0;
        len=strlen(str);
        next[0]=-1;
        while(j<len)
        {
            if(i==-1 || str[i]==str[j])
            {
                i++;
                j++;
                next[j]=i;
            }
            else
            {
                i=next[i];
            }
        }

        ans=1;
        i=len-next[len];
        if(len%i==0)
        {
            ans=len/i;
        }
        printf("%d\n",ans);
    }
    return 0;
}