POJ 2406：Power Strings

POJ 2406：Power Strings

Power Strings

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.

题意：

给出的一个字符串某个字串链接N次得出的，求最大N；

思想：

KMP匹配算法之next[];模式串结尾指针回溯最小移动次数即最小字串长度，N=s.length%(s.length-next[s.length]);

#include #include char s[1000005];int num[1000005];int main(){ while(scanf("%s",s),s[0]!='.') { int j=-1; num[0]=-1; int n=strlen(s); for(int i=0; i

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。