POJ 2406【KMP+next数组+循环节】

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 57771		Accepted: 23994

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

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int maxn=1000005;
char s[maxn];
int Next[maxn];
int minx=INF;
void get_next(){
	int j=0,k=-1;
	int len=strlen(s);
	Next[0]=-1;
	while(j<len){
		if(k==-1||s[j]==s[k]){
			j++;
			k++;
			if(s[j]==s[k]){
				Next[j]=Next[k];
			}else 
				Next[j]=k;
		}else 
			k=Next[k];
	}
}
int main(){
	int ans;
	while(~scanf("%s",s)&&strcmp(s,".")){
		ans=1;
		minx=INF;
		get_next();
		int len=strlen(s);
		if(len%(len-Next[len])==0)
			ans=len/(len-Next[len]);
		printf("%d\n",ans);
	}
    return 0;
}