poj 2406：Power Strings（KMP）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18825		Accepted: 7857

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

源代码：

#include<iostream>
#include<string.h>
using namespace std;

const int MAX=1000005;
int p[MAX];
char s[MAX];

void PreKMP()
{
	int i,j,len;
	len=strlen(s);
	p[1]=0;
	j=0;
	for(i=2;i<=len;i++)
	{
		while(j>0 && s[j]!=s[i-1]) 
			j=p[j];
		if(s[j]==s[i-1]) j++;
		p[i]=j;
		//cout<<i<<"="<<p[i]<<endl;
	}
	int ans=len-p[len]; //从后面提出字符串a的长度
	if(len%ans==0) printf("%d\n",len/ans);
	else printf("1\n");
}

int main()
{
	while(scanf("%s",s))
	{
		if(strlen(s)==1 && s[0]=='.') break;
		PreKMP();
	}
	system("pause");
	return 0;
}