Eddy's research I

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最新推荐文章于 2022-04-12 17:17:45 发布

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Eddy's research I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7605 Accepted Submission(s): 4620

Problem Description

Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

Output

You have to print a line in the output for each entry with the answer to the previous question.

Sample Input

11
9412

Sample Output

11
2*2*13*181

上定理：算术基本定理：
      又称为正整数的唯一分解定理,即:每个大于1的自然数均可写为质数的积,而且这些素因子按大小排列之后,写法仅有   一种方式。
.
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
using namespace std;
bool is_prime(int n)
{
	int k=sqrt(n);
	for(int i=2; i <= k; i ++)
	    if(n%i==0) return false;
    return true;
}
int a[65536];
int b[60000];
int main()
{
    memset(a,0,sizeof(a));
    for(int i = 2;i<=65535;i ++)
       if(is_prime(i)) a[i]=i;
   	int n;
	while(cin>>n)
	{
		int j=0;
		if(a[n]!=0) {
			cout<<n<<endl;
		}
	    else{
	    	int sum=n;
	    	for(int i = n; sum!=1; i--)
			{
	    		if(a[i]!=0 && sum%a[i]==0)
				{
				   b[j++]=a[i];
				   sum=sum/a[i];
				   while(sum%a[i]==0)
				   {
				   	 b[j++]=a[i];
				   	 sum=sum/a[i];
				   }
	    		} 
	    	}
	    	for(int i=j-1;i >0;i --)
	    	     cout<<b[i]<<"*";
  	        cout<<b[0]<<endl;
	    }
	}
	
}