Power Strings（POJ2406）（KMP） - c++编程基础

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33623		Accepted: 13966

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

Waterloo local 2002.07.01 KMP，next表示模式串如果第i位(设str[0]为第0位)与文本串
第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。
则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。
如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

#include
  
   
#include
   
     #include
    
      using namespace std; int next[1000002]; char s[1000002]; int len; int get_next(char *s) { int i=0,j=-1; next[0]=-1; while(i