【kmp】poj 2406 Power Strings

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 73134		Accepted: 30196

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.

next数组很重要的应用——求循环节

当n%(n-next[n])==0 为有循环节 个数为n/(n-next[n])

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
char c[1000005];
int p[1000005];
void next(int len)
{
	p[0]=-1;
	int i=0,j=-1;
	while(i<len)
	{
		if(j==-1 || c[i]==c[j]) i++,j++,p[i]=j;
		else j=p[j]; 
	}
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	while(scanf("%s",c) && c[0]!='.')
	{
		memset(p,0,sizeof(p));
		int len=strlen(c);
		next(len);
		int ans=1;
		if(len%(len-p[len])==0)
			ans=len/(len-p[len]);
		printf("%d\n",ans);
	}
	return 0;
}