POJ-2406 Power Strings（KMP求重复子串出现的最大次数）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50744		Accepted: 21173

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.

/*
题意： 给出一个字符串s，求重复子串出现的最大次数
分析：kmp的next[]数组的应用
要求重复子串出现的最大次数，其实就是要求字符串的最小循环节

这里我们假设这个字符串的长度是len，那么如果len可以被len-next[len]整除的话，我们就可以说
len-next[len]就是那个最短子串的长度


为什么呢？ 假设我们有一个字符串 ababab 那么next[6]=4，  由于next的性质，匹配失配后，下一个能继续进行匹配的位置，
也就是说，把字符串的前四个字母，abab，平移2个单位，这个abab一定与原串的abab重合(否则就不满足失败函数的性质),zhejiu shuomi
题目分析转载于：http://bbezxcy.iteye.com/blog/1377787
*/
#include
#include
#include
#include
using namespace std;

const int maxn =1e6+10;
int next[maxn];     /// 串的next数组 （看这个串的相似度）
char s1[maxn];

void kmp_pre(char x[],int m,int next[])
{
    int i,j;
    j=next[0]=-1;
    i=0;
    while(i