poj-2406-Power Strings（KMP）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38544		Accepted: 16001

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.

/*一个裸的KMP，最后判断一下是否循环*/
#include<stdio.h>
#include<string.h>
int p[1000010];
char str[1000010];
int len;
void getp()
{
	int i=0,j=-1;
	p[i]=-1;
	while(i<len)
	{
		if(j==-1||str[i]==str[j])
		{
			i++,j++;
			p[i]=j;
		}
		else
		j=p[j];
	}
}
int main()
{
	while(scanf("%s",str)!=EOF)
	{
		memset(p,0,sizeof(p));
		len=strlen(str);
		getp();
		if(strcmp(str,".")==0)
		break;
		if(len%(len-p[len])==0)
		printf("%d
",len/(len-p[len]));
		else printf("1
");
	}
	return 0;
}