poj 2406 Power Strings 【KMP算法】

【题意】给个字符串s，求它的类似循环次数，即它的字串看成x ,s=k*x,s是由几个x 叠加而成的 求k

【思路】kmp求 循环周期；一些KMP算法结论附在代码中

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.
Select Code

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=1e6+10;
int nex[maxn];
char s[maxn];
int  getnex(int m){
	int i=1,j=0;//初始化nex[0]=0,则j=nex[j-1]与其他人写的不一样，若初始化为-1，则j=nex[j] 
	memset(nex,0,sizeof(nex));
	while(i<m){
		if(s[i]==s[j])
		nex[i++]=++j;
		else if(!j)
		++i;
		else
		j=nex[j-1];		 //nex[]数组失配是总是要回溯到最近的循环节
	}					//所以j-nex[j-1]就是最小的循环长度 
	return i;			//总结一下 求循环长度 若i%(i-nex[i-1]==0)&&nex[i-1]!=0 
}						//则说明字符串循环，长度为i/(i-nex[i-1]);(i为while(i<m)结束的值) 
					//详情可见https://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html 
int kmp(int n){
	int i=0,j=0;  
	while(i<n&&j<n){
		if(s[i]==s[j]){
			++i;
			++j;
		}
		else if(!j)
		++i;
		else
		j=nex[j-1];
	}
	
}
int main(){
	
	while(scanf("%s",s)&&s[0]!='.'){
		int n=strlen(s);
		int i=getnex(n);
		if(i%(i-nex[i-1])==0&&nex[i-1]!=0)
		printf("%d\n",i/(i-nex[i-1]));
		else printf("%d\n",1);
		
	}
	
}