We can always choose $$$x = 1$$$ so that when performing $$$a_i = a_i + \lfloor{\frac{a_i}x\rfloor}$$$ is the same as $$$ a_i = 2a_i$$$.

Let $$$b_i$$$ is the minimum number of the above operation to get from $$$1$$$ to $$$i$$$. Sense $$$i <= 10^3$$$ so we need at most $$$\lceil \log_2{10^3} \rceil$$$ operations, $$$b_i \le 10$$$.

Update:

Ops! forget about it, in the last operation we can't increase by all values in the range $$$[1, a_i)$$$. Only some of them are available.

The editorial provided a DP approach for it, check it out.