Power Strings

Power Strings
Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 64619		Accepted: 26649
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.
Source

Waterloo local 2002.07.01

KMP求循环节

#include 
#include 
#include 
using namespace std;

const int M=1000005;
char a[M];
int p[M],n,j;

void work()
{
	for (int i=1;i<n;i++){
		while (j>0 && a[j+1]!=a[i+1]) j=p[j];
		if (a[j+1]==a[i+1]) j++;
		p[i+1]=j;
	}
	if (n%(n-p[n])==0) printf("%d\n",n/(n-p[n]));
	else printf("1\n");
}

int main(){
    while (1){
        gets(a+1); if (a[1]=='.') break;
        n=strlen(a+1);p[1]=0;j=0;
        work();memset(a,0,sizeof(a));
    }
    return 0;
}