poj2406 连续重复子串

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 41110		Accepted: 17099

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.

/*
poj2406 连续重复子串

给你一个字符串，请问最多能由任意字符串重复多少次得到

最开始用的是DA算法
想的是枚举长度然后进行一下判断即可,只需要判断Rank[0]和Rank[k]是否为n-k
(因为如果相等，0~k = k+1~2*k+1 .... 递推下去 整个串是0~k的子串不断
 重复得到)
但是MLE.估计是处理RMQ时有问题,而且看了别人报告才发现这并不是最优方法。
因为我们求的是所有数到Rank[0],所以直接从rank[0]的位置往两边扫一遍就行了
但是DA算法nlog(n)好像会TLE- -
于是乎重新去搞了DC3的算法，才搞定 2600ms

而且感觉kmp更适合这个(看比人代码很短的样子)

hhh-2016-03-13 18:11:02
*/
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