The problem is here.
The solution I've learned is:
Color all the integer points in black and white, a point (x, y) is colored black when 2|(x+y).
If there are both black and white points in the n points, divide them by color.
Otherwise, assuming all the points are black, we rotate and zoom all n points, transforming (x,y) into ((x+y)/2, (x-y)/2).
Then, we solve the sutask, there could be O(logV) subtasks in total before the points don't change.
My problem is:
I know the subtask is equivalent to the original problem, but why is the solution always found in this way when there is a solution?
Thank you for helping!
Let $$$k$$$ be the maximum integer such that $$$2^k \mid (x_i+y_i)$$$ for all $$$i$$$. Then color $$$(x_i, y_i)$$$ black iff it is divisible by $$$2^{k+1}$$$ (this is equivalent to your solution), and by construction at least one point is white. Then for any two points which are the same color, their Euclidean distance is a multiple of $$$2^{k+1}$$$, and for any two points which are different colors, their Euclidean distance is equal to $$$2^k\pmod{2^{k+1}}$$$.
By the way, I think your solution might die if all of the points are white.
Edit: Actually I don't know what your zoom is doing. Anyways, it still doesn't seem to work on all white points.
I mean, if all the points are white, we let all the x[1...n] plus 1, then all of them are black.
The lowest k bits of x[i]+y[i] should be equal for all i, and No Solution doesn't exist.
I understand this, thank you!!!