We've established the signs of values before so we can carefully
jiggle the expressions to be guaranteed overflow-free; the tests
for <0 here were meant to check if the value was "still negative",
i.e. if the sum did not underflow below INT_MIN, and we can
rewrite that using algebra to be overflow-free.
We need an extra case in the Euclidean dvision for
INT_MIN / INT_MIN which is a bit annoying but trivial; finally,
for two's complement platforms, note that abs(x) = (x > 0) ? x : -x
does not work (since for x=INT_MIN, -x still gives INT_MIN),
but the equivalent formulation for _negative_ absolute value
(x < 0) ? x : -x is always in range and overflow-free, so
rewrite the relevant expressions using that negative absolute
value instead.
Fixes issue #741.
