Main topic
Interactive questions.
The coordinates of a point on the output plane, which the interaction program gives to this point (white or black).
So repeat $n $ ($ 1\le n \le 30$).
The $n $ points required for the output are different, and no matter how they are dyed by the interactive program, they can find a straight line separating the white and black dots (the dividing line cannot pass through the staining point).
The coordinates of two points on the output divider line.
Requirement: All coordinates of the output must be an integer between $0$ and $10^9$.
Solution
It is not difficult to think of $n $ points can be selected in a line parallel to the $x $ axis.
Each time you take the midpoint of the line of the middle two adjacent black and white dots.
Since the ordinate is a constant, we only consider the horizontal axis below.
The first point of the (horizontal) coordinates selected as $0$, it is advisable to set the first point is given white, at this time we assume that the $10^9+1$ is a (virtual) black spot.
Each time you select the two dots of the adjacent black-and-white point $ L + (R-L)/2$ or $ (l+r)/2$ (note: This notation has the risk of overflow).
My train of thought was roughly right, but at the end of the output divider, I chose a straight line perpendicular to the $x $ axis, making it with $x $ (l + 1, 0) $, $l $ is the horizontal axis of the rightmost white point.
But I didn't think of a situation where the distance between the two black and white dots adjacent to the middle is probably $1$ (this is due to $2^{30} > 10^9+1$. In fact, if the $x = 10^9 + 1$, will be $x \gets \lceil x/2 \rceil$ repeatedly execute $29$ times, there must be $x = 1$; will $x \gets \lfloor x/2 \rfloor$ repeated $29$ times also necessarily There is $x = 1$), at which point the dividing line passes through the leftmost black dot. It is easy to see that if the divider is perpendicular to the $x $ axis, it occupies a position on the horizontal line where the black and white dots are located ($10^9 + 1$ positions in total). We can choose a diagonal line as the divider line, because eventually there must be $l < R $, we can make the black and white point of the ordinate is $1$, so the separation line is advisable to pass $ (l, 0) $ and $ (r, 2) $ two points straight line.
Codeforces #1063C dwarves, Hats and extrasensory abilities