POJ - 2406 Power Strings

Power Strings

POJ - 2406

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.

思路：

从子串中求出能完全与此串重合的最大循环次数

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

int nex[1000005];
int n,m,ans;
char b[1000005];

void get_next(){
	int j=-1,i=0;
	int m=strlen(b);
	nex[0]=-1;
	while(i<m){
		if(j==-1 || b[i]==b[j])nex[++i]=++j;
		else j=nex[j];
	}
}

int main()
{
	while(~scanf("%s",b)){
		if(strcmp(b,".")==0) break;
		ans=0;
		m=strlen(b);
		get_next();
		int num=m-nex[m];//循环节长度 
		if(m%num==0)
			printf("%d\n",m/num);//求有几个循环节 
		else // 防止出现错误答案，例 12121
			printf("1\n");
	}
	return 0;
}