귀가 장치

The bitwise OR operation cannot turn a bit that has become 1 back to 0. Therefore, any bit that is already 1 in $A$ must also be 1 in $B$. That is, $(A \mid B) = B$ is required. Also, if pressing a certain button $i$ includes a bit in $a_i$ that is not in $B$, that bit will be turned on, and it can never become $B$ ultimately. Therefore, the only usable buttons are those satisfying $(a_i \mid B) = B$. Now, we need to turn on each bit that is 1 in $B$ but 0 in $A$. For each bit, if there is at least one button that includes that bit and satisfies $(a_i \mid B) = B$, we just need to press that button. Since the range of integers is up to $10^9$, there are only 30 bits to check. By looking at each button and saving one button that can turn on each bit, it can be solved in $O(N \log 10^9)$. The number of selected buttons is less than or equal to the number of required bits, which is at most 30, so it always satisfies the 100 times limit.