poj 2406 Power Strings

Posted 2020-09-02 日拱一卒 功不唐捐

Power Strings

http://poj.org/problem?id=2406

Time Limit: 3000MS		Memory Limit: 65536K

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.

求最短循环节的出现次数

求最短循环节长度的方法:

a=len-next[len]

若len%a==0 ，最短循环节长度为a

否则，没有循环节

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n;
char s[1000001];
int f[1000010];
void getnext()
{
    for(int i=1;i<n;i++)
    {
        int j=f[i];
        while(j&&s[i]!=s[j]) j=f[j];
        f[i+1]= s[i]==s[j] ? j+1:0;
    }
}
int main()
{
    while(cin>>s)
    {
        if(s[0]==‘.‘) break;
        n=strlen(s);
        getnext();
        int ans=n-f[n];
        if(n%ans) printf("1\n");
        else printf("%d\n",n/ans);
    }
    
}