HDU2441ACM(Array Complicated Manipulation)题解数论

本文介绍了一道名为ArrayComplicatedManipulation的ACM题目，通过对无限整数数组的操作，探讨如何判断一个特定数值是否存在于经过一系列复杂变换后的数组中。通过分析素因子及其组合的变化来确定数值的状态。

ACM(Array Complicated Manipulation)

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 204 Accepted Submission(s): 48

Problem Description

Given an infinite array of integers 2,3,.... Now do some operations on it.

The operation is to choose a minimum number from the array which is never been chosen, then change the status of its multiples excluding itself, i.e remove the multiples of the chosen number if they are in the array , otherwise add it to the array.keep the order after change.

For instance, the first step, choose number 2, change the status of 4, 6, 8, 10... They are all removed from the array. The second step, choose 3, change the status of 6, 9, 12, 15...

Pay attention: 9 and 15 are removed from the array while 6 and 12 are added to the array.

Input

Every line contains an integer n. The zero value for n indicates the end of input.

Output

Print "yes" or "no" according whether n is in the array.

Sample Input

Sample Output

  
   yes
yes
no


   
    
     Hint
    The number n never has a prime factor greater than 13000000, but n may be extremely large.

Source

2008 Asia Harbin Regional Contest Online

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gaojie

我先打表看小数据找规律

判断是否有重复的素因子
当m的素因子没有重复的时候，

能改变m状态的数的个数有c(n,1)+c(n,2)+...+c(n,n-1)=2^n-c(n,0)-c(n,n)=2^n-2是一个偶数，改变偶数次回到原状态。

c(n,n)表示从m的n种素因子中选n种，当m的素因子没有重复的时候，选n个乘积就等于m了，所以不能选n个；

当m有重复素因子时，可以选n个不同的素因子，此时他们的乘积小于m，所以能选n个。
当m有重复素因子时，能改变n状态的数的个数加上c(n,n)=1变成了奇数，

所以状态改变了，数m不存在了。

比如30=2*3*5 能改变它状态的数有(2,3,5),(6,10,15)偶数个，所以30存在

60=2*2*3*5 能改变它状态的数有(2,3,5),(6,10,15),(30)奇数个，所以60不存在

n表示m的素因子种类数。
注意1和高精度

import java.io.*;
import java.math.*;
import java.util.*;

public class Main
{
	public static boolean factor(int p[],BigInteger n)
	{
		int i=0;
		BigInteger x=BigInteger.valueOf(p[0]*p[0]);	
		for(i=0;x.compareTo(n)<1;)
		{
			BigInteger y=BigInteger.valueOf(p[i]);
			BigInteger z=n.mod(y);
			if(z.compareTo(BigInteger.ZERO)==0)
			{
				int s=0;
				while(z.compareTo(BigInteger.ZERO)==0)
				{
					n=n.divide(y);					
					s++;
					if(s>1)
						return false;
					z=n.mod(y);
				}
			}
			i++;
			x=BigInteger.valueOf(p[i]*p[i]);
		}
		return true;
	}
    public static void main(String[] args)   
	{
	        Scanner cin = new Scanner (new BufferedInputStream(System.in));
	        int []p=new int[130005];
	        int i,j,m=0,N=1300000;
	        boolean []f=new boolean[1300005];
	        f[0]=f[1]=true;
	    	for(i=4;i<=N;i+=2)
	    		f[i]=true;
	    	for(i=3;i*i<=N;i+=2)
	    		if(!f[i])
	    			for(j=i*i;j<=N;j+=i*2)
	    				f[j]=true;			
	    	for(i=2;i<=N;i++)
	    		if(!f[i])
	    			p[m++]=i;
	    	
	        BigInteger n=BigInteger.valueOf(0);
	        while(cin.hasNext())
	        {
	        	n=cin.nextBigInteger();
	        	if(n.compareTo(BigInteger.ZERO)==0)
	        		break;
	        	if(n.compareTo(BigInteger.ONE)==0) 
	        		System.out.println("no");
	        	else if(factor(p,n))
	        		System.out.println("yes");
	        	else
	        		System.out.println("no");
	        }       
	}
}