题意：

有一串字符串,问求出有多少个循环节连续重复组成，即可以用KMP直接求出循环节有多少个字符组成。答案就是l/next[l]（刚开始理解错题意，认为是找出最多的重复子串）

题目

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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.

思路

1.即可以用KMP直接求出循环节有多少个字符组成。答案就是l/next[l]。
2.若不连续则循环节个数为一。

ac代码

#include <cstdio>
#include <cstring>
#include <algorithm>
const  int M=1e6+10;
int next[M] ;
char str[M] ;
void getnext(int l)
{int j = 0 , k = -1 ;next[0] = -1 ;while(j < l){if( k == -1 || str[j] == str[k] )next[++j] =++k ;elsek = next[k] ;}
}
int main()
{int l , m ;while(~scanf("%s", str)&&str[0]!= '.'){l = strlen(str);getnext(l) ;m = next[l];if( l % (l-next[l]) != 0 )printf("1\n");else{int k= l/( l-next[l]);printf("%d\n",k);}memset(str,0,sizeof(str));}return 0;
}/**题目要求有循环节重复连续，即整个串是用循环节组成的*/