Observation 1: We first notice that no matter how the judge shifts $$$V$$$ (the value we send), there are only 2 kinds of situations.
the odd bits of $$$V$$$ xor the odd bits of $$$X$$$, and the even bits of $$$V$$$ xor the even bits of $$$X$$$
the odd bits of $$$V$$$ xor the even bits of $$$X$$$, and the even bits of $$$V$$$ xor the odd bits of $$$X$$$
Here "odd bits" represents the bits in position 1, 3, 5, 7, and "even bits" represents the bits in position 0, 2, 4, 6
Observation 2: If odd bits of $$$X$$$ are all 1s, and even bits of $$$X$$$ are all 0s (or the opposite), then we can send value $$$01010101$$$ to change $$$X$$$ to $$$00000000$$$ or $$$11111111$$$. It means that we only need to make odd bits of $$$X$$$ be all the same, and also for even bits of $$$X$$$.
Observation 3: Let $$$last$$$ be the number of 1s after last query, and if we send value $$$01010101$$$ and receive $$$k$$$, we can calculate the number of 1s for odd bits and for even bits. That is $$$x = \frac{last + k}{2}$$$, $$$y = k - x$$$. Although we can't determine which value represents odd bits or even bits, it doesn't matter. It means that we can use 1 query to get the number of 1s for odd bits and for even bits. Let's call this operation $$$getNum$$$.
The solution makes no more than 20 queries.
