There are \(O(R)\) lines that intersects with the circle and that can be represented as \(x = c\) (where \(c\) is an integer).
For each of them, we can apply Pythagorean theorem to find the range of \(y\) coordinates of the intersection of the line and the circle, so we can find the number of lattice points that is within the circle and on the line \(x = c\).

However, the floating-point error is a big issue here.
For example, given \((X, Y) = (-0.0001, 0), R = 100000\), when calculating the upper bound of the intersection of line \(x = 0\) and the circle, we have to calculate \((10^5)^2 - (10^{-4})^2\), which requires far more precision than double type.
To resolve the issue, multiply \(X, Y, R\) by \(10^4\) to make them integers. Then, we can count the number of points where both \(x\) and \(y\) are multiples of \(10^4\).
Just like before, there are \(O(R)\) lines that intersects with the circle and that can be represented as \(x = 10^4 c\) (where \(c\) is an integer), so it can be solved in the similar way. Since entire calculation are done with integer, we don’t have to care about precision errors.

The complexity is \(O(R \log(R))\) in total, with assumption that calculating the square root requires an \(O(\log(R))\) time.
Since \(X, Y,\) and \(R\) cannot be inputted precisely in most cases, just multiplying them by \(10^4\) may not precisely results in integers. Therefore, use proper functions to round into integers, such as round function in C++. (In C++, a double value is rounded to int value in the direction of \(0\), so an attempt to convert to int after adding \(0.5\) fails is it is negative)