3 ms·
Actually, e is the inverse of the xor, so (e & c1) is guaranteed to be c1, and you still need popcount(c1 & ((e & c2) + c3)).
by hairtuq 5y ago
Actually, e is the inverse of the xor, so (e & c1) is guaranteed to be c1, and you still need popcount(c1 & ((e & c2) + c3)).