Watch out that the low 16-bit part should again be treated as unsigned (or else you'll get weird stuff when bit 15 is 1). Technically, I think that method still loses some precision in the ((a1 / b) << 16) when compared to a higher-precision divide such as ((long long)a1 << 16) / b), due to the fact that you're shifting in 16 zeros whereas the long long division actually calculates the low 16 bits.

If you're that worried about performance you can always use the assembly routines I posted (the division routine takes about the same number of steps as a normal 32-bit division). I understand if you don't want to use them because it's for a contest, though. Then again, I'd let anyone use this code anyway.

I would just use the routines with long long, but then I get compiler errors with newlib. Using your asm routines in a contest won't be a big deal, right? I might not have to use them, though, depending on how fast it runs.

You'd have to ask the people in charge of the contest, I guess. I consider these routines to be like the built-in arithmetic operations except with fixed-point instead of integer. After all, the normal 32-bit division routine used by GCC was written by someone else, heh

Yay, multiplication and division work now! 

Now, I have to get the engine to stop dividing by zero or I will be responsible for many black holes created inside calcs.

Actually, it's represented as FFFF.FFF0, which is FFFF.0000 (signed -1) plus 0.FFF0 (unsigned). That's why the upper 16 bits is treated as signed and the lower 16 bits is unsigned.

Oh, right. But anyways, wouldn't the fact that the bottom half is still signed, but not treated so in the multiplication routine mess stuff up?

That's the thing, the bottom half isn't signed. Really, what determines signedness is whether the top bit means the entire value is offset by some 2^N. Of course, only bit 0x80000000 of the whole 32-bit value determines this, which means that any parts of the value not including that bit are unsigned, and parts that include it are signed. You can think of signed values having their top bit repeated infinitely to the left.