Yes, it is.

So you can wait 16 years for the Spice and Wolf remake but cannot wait a few years for C++23?

Please refer to the official editorial.

The case $$$A = B = t$$$ is easy. For the case $$$A = t + 1$$$ and $$$B = t$$$, note that $$$p \cdot A + q \cdot B = (p + q) \cdot t + p$$$, so the goal is to generate all numbers $$$X = a \cdot t + b$$$, such that $$$a \geq b$$$. Let us fix $$$b$$$, then $$$a \cdot t + b \leq N$$$; $$$a \leq \left\lfloor \frac{N - b}{t} \right\rfloor$$$. So basically for each remainder $$$b$$$ in $$$[0, t - 1]$$$ you will find this value $$$a$$$. But observe that $$$a$$$ has at most only two possible values. So find the range of remainders $$$b$$$ in $$$[0, \text{lo} - 1]$$$, for which this value is $$$q$$$, and another range $$$[\text{lo}, t - 1]$$$ for which this value is $$$q - 1$$$. From here it shall be straightforward.

Here is a more visual solution.

For the case $$$A = m, B = m + 1$$$, consider the non-negative integers arranged in a grid of width $$$m$$$, where the $$$(i, j)$$$th cell represents the integer $$$mi + j$$$. For the cell labeled by $$$i$$$, we write $$$1$$$ if we can make a sum of $$$i$$$, and $$$0$$$ otherwise. Observe that if we can make a sum of $$$k$$$, then we can make a sum of $$$k + lm$$$ for any $$$l\geq 1$$$. Therefore, if we have a $$$1$$$ in some cell, then every cell below it in the same column has $$$1$$$. Observe that we can make $$$0$$$, implying that the first column is all $$$1$$$s. We can make $$$m + 1$$$ but not $$$1$$$ implying that the second column is $$$1$$$ from the second row onwards. Proceeding similarly shows that we get the following pattern:

Example for $$$m = 5$$$:

The problem thus reduces to some casework to count the number of integers $$$\leq n$$$ that you miss.