poj2406

maksyuki 发表于 oj 分类，标签:
15 9月 2015
0

Power Strings

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

 

题目类型：KMP中Next数组应用

算法分析：直接计算出输入字符串的Next数组，然后判断len % (len – Next[len])是否为0，如果为0，则len / (len – Next[len])即为最后的答案，否则结果为1