Finding lowest common multiple / highest common factor

I've got these two functions, homegrown obviously, which find the lowest common multiple of two numbers (or least common denominator for two fractions, for example 3 and 6 would be 3) and the highest common factor of two numbers (for example, 15 and 10 would be 5).

void CProblem::SolveFindHCF (__int64 n1, __int64 n2, __int64 *hcf)
{ __int64 i, range;

  if (n1 <= -1)
  { n1 *= -1; }
  if (n2 <= -1)
  { n2 *= -1; }

  *hcf = 1;
  if (n1 > n2)
  { range = n2; }
  { range = n1; }

  for (i = 1; i <= range; i++)
  { if (n1 % i == 0 && n2 % i == 0)
    { *hcf = i; }

void CProblem::SolveFindLCD (__int64 n1, __int64 n2, __int64 *lcd)
{ __int64 tden, bden;

  for (tden = n1; tden < n1 * 10000; tden += n1)
  { for (bden = n2; bden <= tden; bden += n2)
    { if (tden == bden)
      { *lcd = tden; // That's that

The problem is the speed.  When using large numbers, these functions are very slow.  I would like tips on improving the speed of these functions.  Sometimes millions of iterations are necessary for any given problem, and this is unacceptable as anyone with half a brain could realize.

Please post ways to enhance the performance and eliminate any bugs in these functions.  Whoever contributes the most will get the points.  Thanks!
Who is Participating?
glebspyConnect With a Mentor Commented:
Dear Helpmealot,
 No, sorry - that doesn't look correct.
 Also I made a mistake: it should be B>A, not A>B. Or you have to take greater number mod smaller number, anyway.

So it looks like this:
 Choose numbers A,B

 Sort so B>A

 C=B%A (that guarantees that C<A<B)

 if(C==0)return A

 else {B=A;A=C;}

end loop

Should work.

 I also thought of a good way of using this to determine the LCM of A,B (your second problem) - I think this is pretty fast, too:


Do the multiplication,division in the above order so there is less risk of an overflow, although the expression LCM(A,B)=A*B/HCF(A,B) is neater.
I can give you a good tip for the highest common factor. Use this algorithm:

Numbers A,B


 Switch if necessary so that A>=B
 C=B mod A  (C=B%A in c++)

 if(C==0) Answer is B, break;

 otherwise repeat with numbers B,C

end loop

This is a really fast algorithm. From what I know, it could be the fastest in existence.

I guess you can probably turn that into code yourself but if you need help just say.

helpmealotAuthor Commented:
Does this implementation appear to be correct?

__int64 CProblem::SolveFindHCF (__int64 A, __int64 B)
{ if (A < B)
  { __int64 temp;
    temp = A;
    A = B;
    B = temp;

  if (A == 0 || B == 0)
  { return 1; }

  __int64 C;
  for (;;)
  { C = B % A;
    if (C == 0)
    { return B; }

    A = B;
    B = C;
helpmealotAuthor Commented:
Anybody out there?  This question couldn't be that tough. :)
helpmealotAuthor Commented:
Thank you, it works now, and is much faster than before.
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