There are many observations to make in order to get to a working solution.

• For every lattice point (i, j), the ray that intersects it is unique and it's identified by the pair , where g is the gcd of i and j.
• The problem is now reduced to counting the number of irreducible fractions such that a ≤ N and b ≤ M. This is the same as counting for every i between 1 and N, the amount of numbers in the range [1, M] that are coprime with i.
• Consider a certain number x with prime factors p₁, p₂. How do we know how many numbers in range [1, M] are coprime with it? That's equal to M minus the amount of multiples of p₁ minus the amount of multiples of p₂ plus the amount of multiples of p₁ * p₂. This is inclusion-exclusion, and in general, if the amount of elements is even, we add, otherwise, we subtract.
• So now we have a working (but slow) solution: Iterate over every i in the range [1, N] and for every i, factorize it, try out all combinations of primes and then, for every combination that results in a number k, add if the amount of primes is even or subtract if the amount of primes is odd.
• The previous approach is very slow for two reasons: You'll be factorizing each number every time and you'll be doing a lot of repeated work. Every combination of primes you try out at each step will result in a certain number k. A crucial observation is that the higher exponent of that number k will be 1, because we're trying combinations of different primes. Another crucial observation is that this number k will be seen times in total. Finally, each time we see it, it will contribute by to the final answer (or if the amount of primes is odd).
• Knowing all this, we can precalculate a lot of stuff and then solve each test case in O(N). We should precalculate the amount of prime factors of every number in the range [1, 10⁶] (this can be done with a simple sieve), and we should cross out numbers that have some prime with an exponent higher than 1 (in other words, multiples of some square). Once we have precalculated all that, we simply iterate from 1 to N and for every number x that we didn't cross out, we add (or subtract) to our answer.
• Final observations: We should add 2 to our answer (the two borders). If N = 0, the answer is 1, except M = 0 too, in which case the answer is 0.

Here's the simple C++ code: Ray Gun