其实就是用一个int数组记录其中一个循环中的循环个数,


//   KMP算法

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;

char s[2000000];
int go[2000000];


int main()
{
	//freopen("fuck.txt","r",stdin);
	int i,j,k;
	
	
	while (cin>>s)
	{
		if (s[0]=='.')
			break;
		int n=strlen(s);
		go[0]=0;
		j=-1;        //这里注意
		for (i=1;i<n;i++)
		{
			while (j>0&&s[j+1]!=s[i])    //循环中的一个数没有满足要求,则返回原来的满足要求的一个数.
				j=-1;    //或者用 j=go[j];    其实后者不易理解,而且效率感觉 更慢,还是自己用的这个简单粗暴,
			if (s[j+1]==s[i])
				j++;
			go[i]=j;
			//cout<<go[i]<<endl;
		}
		if (n%(n-go[n-1]-1)==0)   //这里也注意
			cout<<n/(n-go[n-1]-1)<<endl; 
		else
			cout<<'1'<<endl;
	}

	return 0;
}




Power Strings


Time Limit: 3000MS

 

Memory Limit: 65536K

Total Submissions: 21745

 

Accepted: 9128


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