POJ2406(next原理理解)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 40448		Accepted: 16828

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=;
char s[MAXN];
int len;
int next[MAXN];
void getnext()
{
    int i=,k=-;
    next[]=-;
    while(i<len)
    {
        if(k==-||s[i]==s[k])
        {
            i++;
            k++;
            next[i]=k;
        }
        else    k=next[k];
    }
}
int main()
{
    while(gets(s)&&*s!='.')
    {
        len=strlen(s);
        getnext();
        int ans=;
        if(len%(len-next[len])==)    ans=len/(len-next[len]);
        printf("%d\n",ans);
    }

    return ;
}