ＰＯＪ ２４０６ ＫＭＰ求字符串循环节

　　还是一道求字符串循环节的问题，用ＫＭＰ算法可以轻松解决，只要注意是以什么结束的就可以了。。。。。。题目：

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 19573		Accepted: 8173

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

ａｃ代码；

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
const int N=1000010;
int len,nextt[N];
char str[N];
int get_next()
{
  nextt[0]=0;
  for(int i=1;i<len;++i){
    int temp=nextt[i-1];
	while(temp&&str[temp]!=str[i])
		temp=nextt[temp-1];
	if(str[i]==str[temp])
		nextt[i]=temp+1;
	else
		nextt[i]=0;
  }
  int x=len-nextt[len-1];
  if(len%x==0)
  {
    int y=len/x;
    return y;
  }
  else
	  return 1;
}
int main(){
	while(scanf("%s",str)){
		if(str[0]=='.')
			break;
	  len=strlen(str);
	  int x=get_next();
	  printf("%d\n",x);
	}
  return 0;
}