F - Power Strings（kmp）

Description

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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.

法一：可以利用KMP的最小循环节的性质求解

对于str2，cyc = len2 - next[len2]是它的最小循环节长度.
如果len2 % cyc == 0，表示str2能够由完整的几个循环节组成，输出len2 / cyc。
如果len2 % cyc != 0，说明最后不是一个完整的循环节。在这道题里面就需要直接输出1，表示它自己由它本身这样一个循环节组成，而不是len2 / cyc。

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int n=1e6;

int next[n+5];
char str1[n+5],str2[n+5];
int len1,len2;

void get_next()//求next数组 
{
	int i1=0,i2=-1;
	next[0]=-1;
	while(i1<len2)//遍历完str2时结束 
	{
		if(i2==-1||str2[i1]==str2[i2])//如果是str2中下标为1的元素，或者能够匹配上 
			next[++i1]=++i2;
		else//不能匹配上，回跳 
			i2=next[i2];
	}
}

int main(void)
{
	int cyc;
	while(scanf("%s",str2)!=EOF&&strcmp(str2,".")!=0) 
	{
		len2=strlen(str2);
		get_next();
		cyc=len2-next[len2];//循环节 
		if(len2%cyc==0)
			printf("%d\n",len2/cyc);
		else
			printf("1\n");
	}
	return 0;
}

法二：利用hash算法来求解

yfy同学的代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

typedef long long LL;

const int N=1000100,P=131;

int n,m,len;
char str[N];
LL p[N],h[N];
int mod=10009;

LL cal(int x,LL y)
{
	LL re=1;
	while(x)
	{
		if(x&1)
		{
			re=(re*y)%mod;
		}
		x>>=1;
		y=(y*y)%mod;
	}
	return re;
}

bool check(int x)
{
	LL cc=cal(x,LL(P));
	for(int i=x;i<=len;i+=x)
	{
		if(h[x]!=(h[i]-(h[i-x]*cc)%mod+mod)%mod)
		{
			return false;
		}
	}
	return true;
}

int main(void)
{
	while(scanf("%s",str+1) != EOF && str[1] != '.')
	{
	    len=strlen(str+1);

	    for(int i=1;i<=len;i++)
	    {
		    h[i]=(h[i-1]*P+str[i])%mod;
	    }
	    int flag=0;
	    // 枚举循环节长度 
	    for(int i=1;i<=len;i++)
	    {
	    	if(len%i==0&&check(i))
	    	{
	    		printf("%d\n",len/i);
	    		flag = 1;
	    		break;
			}
		}
		if(flag==0)
			printf("1\n");
	} 
	
	return 0;
}