Suppose that you can use $$$x$$$ operations of type $$$1$$$ and $$$y$$$ operations of type $$$2$$$. Try to reorder operations in such a way that $$$a$$$ becomes the minimum possible.

How many operations do you need in the worst case? ($$$a = 10^9$$$, $$$b = 1$$$)

Iterate over the number of operations of type $$$2$$$.

Notice how it is never better to increase $$$b$$$ after dividing ($$$\lfloor \frac{a}{b+1} \rfloor \le \lfloor \frac{a}{b} \rfloor$$$).

So we can try to increase $$$b$$$ to a certain value and then divide $$$a$$$ by $$$b$$$ until it is $$$0$$$. Being careful as not to do this with $$$b \lt 2$$$, the number of times we divide is going to be $$$O(\log a)$$$. In particular, if you reach $$$b \geq 2$$$ (this requires at most $$$1$$$ move), you need at most $$$\lfloor \log_2(10^9) \rfloor = 29$$$ moves to finish.

Let $$$y$$$ be the number of moves of type $$$2$$$; we can try all values of $$$y$$$ ($$$0 \leq y \leq 30$$$) and, for each $$$y$$$, check how many moves of type $$$1$$$ are necessary. This is a $$$O(\log^2 a)$$$ solution.

If we notice that it is never convenient to increase $$$b$$$ over 6, we can also achieve a solution with better complexity.