i do not know.
The straight forward approach would be:
1) a = x/a;
2) b = (x%a)/b
3) c = ((x%a)%c)/d
But that doesnt work, as you have to be able to "undo" the algorithm, when you...
|CodeGuru Home||VC++ / MFC / C++||.NET / C#||Visual Basic||VB Forums||Developer.com|
Type: Posts; User: skidrive
Search: Search took 0.01 seconds.
Results 1 to 3 of 3
Click Here to Expand Forum to Full Width
This is a CodeGuru survey question.