Power Strings
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions: 32810 |
|
Accepted: 13670 |
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
解题报告：正常做法是用KMP，求失败指针什么的。我这里直接用字符串哈希和矩阵快速幂搞定了。
首先从小到大枚举可能的重复串的长度，使用矩阵快速幂递推最后一位的哈希值，与实际计算的哈希值比较，相同则说明当前子串为最小的重复串，此时n最大。
代码如下：
#include
#include
#include
#include
#include
#include
#include
#include
