Hi

My assignment is to find the unique prime factors of a user inputed number. So for 54, the uniqe prime factors would be 2 and 3. If we display the factors as 2 3 3 3 we get extra credit. I figured out how to do this without manipulating bit by bit (not the extra credit). We were never told if it had to be bit by bit so my question is: is it easy to display these factors using bit manipulation? The reason i ask is if its 10,000 lines or something crazy i doubt my teacher would make us do it. If it is simple and I just can't get my head around it then please tell me so and please provide a starting point.

If you have a way for me to get the extra credit with my current code I would appreciate it cuz i cant figure it out.

My source code is below


Thanks


Code:
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <cmath> 
#include <bitset> 

using namespace std;

void getFactors (int n)
{	
	for ( int i = 1 ; i <= n ; i++ )
	{
		int j = i - 1;	
		while ( j > 1 )
		{
			if ( i % j == 0 )
				break;
			else
				j--;
		}
		if ( j == 1 )
		{
			if ( n % i == 0 )
			{										
				cout<<i<<"   "; 
			}
		}	
	}
	cout<<endl;
}

int _tmain(int argc, _TCHAR* argv[])
{
   const int SIZE = 1024;
   int value;
   std::bitset< SIZE > sieve; // create bitset of 1024 bits
   sieve.flip(); // flip all bits in bitset sieve
   sieve.reset( 0 ); // reset first bit (number 0)         
   sieve.reset( 1 ); // reset second bit (number 1)        

   // perform Sieve of Eratosthenes
   int finalBit = sqrt( static_cast< double > ( sieve.size() ) ) + 1;

   // determine all prime numbers from 2 to 1024
   for ( int i = 2; i < finalBit; i++ )
   {
      if ( sieve.test( i ) ) // bit i is on
      {
         for ( int j = 2 * i; j < SIZE; j += i ) 
            sieve.reset( j ); // set bit j off
      } // end if
   } // end for

   cout << "The prime numbers in the range 2 to 1023 are:\n";

   // display prime numbers in range 2-1023
   for ( int k = 2, counter = 1; k < SIZE; k++ )
   {
      if ( sieve.test( k ) ) // bit k is on
      {
         cout << setw( 5 ) << k;

         if ( counter++ % 12 == 0 ) // counter is a multiple of 12
            cout << '\n';
      } // end if          
   } // end for    
   
   cout << endl;

   // get value from user 
   cout << "\nEnter a value from 2 to 1023 (-1 to end): ";
   cin >> value;

   // determine whether user input is prime
   while ( value != -1 ) 
   {
      if ( sieve[ value ] ) // prime number
	  {
         cout << value << " is a prime number\n";
	  }
      else // not a prime number
	  {
		  cout << value << " is not a prime number\n The unique prime factors are: ";
		  getFactors(value);
	  }
      
      cout << "\nEnter a value from 2 to 1023 (-1 to end): ";
      cin >> value;
   } // end while

	return 0;
}