UVA10298 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: a0 = ‘’ (the empty string) and a(n+1) = a∗(an).
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 = an for some string a.
Sample Input
abcd

aaaa

ababab

.

Sample Output
1

4

3

翻译：

POJ 2406:Power Strings
    题意：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。

传统朴素算法时间复杂度：O(m*n)

KMP算法时间复杂度：O(m+n)

code:

#include
#include
#include
#include

using namespace std;
const int N=1000005;
char p[N];
int nextl[N];
void getnext()
{
    int j=0;
    int k=-1;
    nextl[0]=-1;
    while(j