【KMP】Power Strings

描述

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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).

输入

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.

输出

For each s you should print the largest n such that s = a^n for some string a.

样例输入

abcd
aaaa
ababab
.

样例输出

1
4
3

提示

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

Waterloo Jun.1 2002

分析：
此题作为理解KMP算法的练习。
代码：
#include
using namespace std;
int ll;
int next[1000010];
void kmp(string s)//KMP算法
{
int l=0,r=-1;
next[0]=-1;
while(l {
if (r==-1||s[l]s[r])
{
l++;
r++;
next[l]=r;
}
else
r=next[r];//回到初位置
}
}
int main()
{
string s;
int l;
while(cin>>s)
{
if (s[0]’.’) break;
ll=s.length();
kmp(s);
l=ll-next[ll];
if (ll%l==0)//最短子串的出现次数
cout< else
cout<<1< }
return 0;
}