素数筛选法

Eddy's research I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2659    Accepted Submission(s): 1644

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
   
   
   
   

 

Author

eddy

 

Recommend

JGShining

 


#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int prime[50000],num,is[70000];
void prm(int n)
{
  int i,j,k=0;
  int s;
  int e=(int)(sqrt(0.0+n)+1);
  memset(is,1,sizeof(is));
  prime[k++]=2;
  is[0]=is[1]=0;
  for(i=4;i<n;i+=2) is[i]=0;
  for(i=3;i<e;i+=2)
    {
      if(is[i]){
    prime[k++]=i;
      for(s=i*2,j=i*i;j<n;j+=s)
    {
      is[j]=0;
    }
      }
    }
  for(;i<n;i=i+2)
    if(is[i])
      prime[k++]=i;
}
int main()
{
  int n;
  prm(70000-1);
  /*  for(int i=0;i<=100;i++)
    {
      printf("%d ",prime[i]);
      }*/
  while(scanf("%d",&n)!=EOF)
    {
      int temp=0;
      int k=0;
      while(n!=1)
    {
      if(n%prime[k]==0)
        {
          n=n/prime[k];
          if(temp==0)
        {
          temp++;
          printf("%d",prime[k]);
        }
          else
        {
          printf("*%d",prime[k]);
        }
        }
      else k++;
    }
      printf("\n");
    }
  return 0;
}