POJ 2406 Power Strings

Power Strings

 

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.

 

 

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

const int maxn=1000005;

int next[maxn];

void getnext(char str[])
{
    int i,j;
    int len=strlen(str);
    i=0;j=-1;
    next[i]=j;
    while(i<=len)
    {
        if(j==-1||str[i]==str[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
        j=next[j];
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int cnt,i;
    char s[maxn];
    while(scanf("%s",s)&&strcmp(s,".")!=0)
    {
        getnext(s);
        int len=strlen(s);
        if(len%(len-next[len])==0)
        printf("%d
",len/(len-next[len]));
        else
        printf("%d
",1);
    }
    return 0;
}