Replace "(a * (1 << shift) * b + c) >> shift" by "a * b + (c >> shift)".
It is equivalent to the old code because a is in the range of uint16_t,
shift is 12 and b is effectively a signed 4-bit number, so that no
overflow/truncation of high bits happens during the multiplication
(overflow would be undefined anyway).
Signed-off-by: Andreas Rheinhardt <andreas.rheinhardt@gmail.com>
Signed-off-by: Michael Niedermayer <michael@niedermayer.cc>