rounding of 4 unsigned char numbers

Hi. I've encountered a problem with rounding which I cannot resolve
elegantly and I has hoping that somebody could help me out. Here's the
problem description:

We are given 4 numbers a, b, c, and d in the range [0255]. These are
8-bit numbers (unsigned char), but I assume that the solution should
exists for any non-negative integer. We are supposed to determine x =
x(a, b, c, d) so that the following 2 formulas give identical results:

(a + b + c + d + 2) >2 = (((a + c + 1) >1) + ((b + d + 1) >1) +
x) >1

This means that the ceiling of quarter sum of 4 numbers should be
accomplished in 2 steps.

Similar formula should be applied for flooring of quarter sum of 4
numbers:

(a + b + c + d) >2 = (((a + c) >1) + ((b + d) >1) + x) >1

Any ideas how can we do this elegantly? It looks like the solution
inspects if a, b, c, d are odd or even, plus if they are divisible by
4.

Thanks in advance for your help! I'd really appreciate it.

Regards,
Aleksandar