poj 2406 Power Strings【KMP】

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Power Strings

Time Limit: 3000MS

Memory Limit: 65536K

Total Submissions: 33548

Accepted: 13935

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

Sample Output

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题目翻译意：给一个字符串S长度不超过10^6，求最大的n使得S由n个相同的字符串a连接而成，如："ababab"则由n=3个"ab"连接而成，"aaaa"由n=4个"a"连接而成，"abcd"则由n=1个"abcd"连接而成。

解题思路：假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]

利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，则最大循环次数n为len/(len - next[len])，否则为1