poj 2406

题目链接：http://poj.org/problem?id=2406

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

题意：求子串的循环次数

解决方法：KMP中的覆盖函数

参考文章：http://www.cppblog.com/suiaiguo/archive/2009/07/16/90237.html            

AC code:

#include<stdio.h>
#include<iostream>
#include<cstring>
#define M 1000008
int next[M];
char str[M];
int length;
void get_next()
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<length)
	{
	 if(j==-1||str[i]==str[j])
	 {
			++i;++j;
			next[i]=j;
	 }
	 else
	 j=next[j];
	}
}
int main()
{
	while(scanf("%s",str),str[0]!='.')
	{
		length=strlen(str);
		get_next();
		int n=length-next[length];             //此为循环节的长度
		if(next[length]!=0&&length%n==0)
		printf("%d\n",length/n);
		else
		printf("1\n");
	}
}
	

附上变式poj 1961 AC code:

#include<iostream>
#include<stdio.h>
using namespace std;
#define M 1000008
int next[M];
char str[M];
int num;
void get_next()
{
	int i=-1;
	int j=0;
	next[0]=-1;
	while(j<num)
	{
		if(i==-1||str[i]==str[j])
		{
			++i;
			++j;
			next[j]=i;
		}
		else
		i=next[i];
	}
}
int main()
{
	int t=0;
	while(scanf("%d",&num),num)
	{
	  int i,temp;
	  for(i=0;i<num;++i)
	  cin>>str[i];
	  get_next();
	  printf("Test case #%d\n",++t);
	  for(i=1;i<=num;++i)
	  {
		temp=i-next[i];
		if(i%temp==0&&i!=temp)
		printf("%d %d\n",i,i/temp);
      }
      printf("\n");
	}
}