//	Check weather a number is  a perfect number
//	Variant 2:  i  is  a factor of  <=>  a/i  is a factor of a
//			code is much much faster
//		g++ -Wno-deprecated PerfectNum2.cc 

#include <iostream.h>
#include <math.h>

main()
{
 long long int a,		// number to investigate
 	       i,n,		// loop index and last index
	       sum_a;		// sum of all factors

 cout << endl << "  Input  a = ";
 cin  >> a;
 cout << endl;
 
 n = floor(sqrt(a));		// NEW: last potential factor in Var.2
 
 sum_a = 1;			// NEW: Don't forget 1
 for ( i=2; i<=n; i++ )		// NEW: we will start with 2
  {
    if ( ( a % i ) == 0 )	// is var.  i  a factor?
     {
       sum_a += i + a/i;	// NEW: add also  a/i
     }
  }

 cout << " a =  " << a << "   is ";
 if ( sum_a != a)		// sum of factors not equal number  a ?
  {
    cout << "not ";
  }
 cout << "a perfect number" << endl << endl;
 
}








/*
Perfect numbers are:
6

28

496

8128

33550336		// Laptop: user    0m0.002 / 0m0.004s  fuer 4/8 Byte-int

8589869056		// Here, we need 8 Byte integers (long long int)
			// Laptop: user    0m0.013s  [old: 7m25.344s]


*/