KMP入门（周期）

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.

 

 

解题思路：题目大意：给定一个字符串由某一个子串重复得来，求出重复的次数。

利用fail数组fail[len],子串循环的次数满足if(len%(len-fail[len])==0)
           ans=len/(len-fail[len]);

否则ans=1

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

char s[];
int fail[];
int len;

void get_fail();

int main()
{
    while(scanf("%s",s+),s[]!='.')
    {
       len=strlen(s+);
       get_fail();
       int ans=;
       //ÀûÓÃfail[len]
       if(len%(len-fail[len])==)
           ans=len/(len-fail[len]);
       printf("%d\n",ans);
    }
}
void get_fail()
{
    memset(fail,,sizeof(fail));
    fail[]=-;
    for(int i=;i<=len;i++)
    {
        int p=fail[i-];
        while(p>=&&s[p+]!=s[i])
        p=fail[p];
        fail[i]=p+;
        //cout<<i<<" "<<fail[i]<<endl;
    }
}