, 2018

Solution:

The first keystroke must be $c$ and the last keystroke must be $p$. If there are $k$ $c$'s pressed in total, let $n_{i}$ denote one more than the number of $p$'s pressed immediately following the $i$'th $c$, for $1 \leq i \leq k$.

Then, we have that the total number of keystrokes is
$$s := \sum_{i=1}^{k} n_{i}$$
and the total number of $a$'s is
$$r := \prod_{i=1}^{k} n_{i}$$
We desire to minimize $s$ with the constraint that $r \geq 2018$. We claim that the minimum possible $s$ is $s=21$.

This value of $s$ is achieved by $k=7$ and $n_{1}=n_{2}=n_{3}=n_{4}=n_{5}=n_{6}=n_{7}=3$, so it remains to show that $s=20$ is not possible.

Suppose it were for some $k$ and $n_{i}$. By the AM-GM inequality,
$$\left(\frac{n_{1}+n_{2}+\cdots+n_{k}}{k}\right) \geq \sqrt[k]{n_{1} n_{2} \cdots n_{k}}$$
implying that
$$\begin{aligned} 2018 & \leq n_{1} n_{2} \cdots n_{k} \\ & \leq \left(\frac{n_{1}+n_{2}+\cdots+n_{k}}{k}\right)^{k} \\ & = \left(\frac{20}{k}\right)^{k} \end{aligned}$$
which is satisfied by no positive integers $k$. More rigorously, the function $f(x)=x^{\frac{1}{x}}$ is well known to have a maximum at $x=e$. Making the substitution $u=\frac{20}{k}$, we obtain
$$\begin{aligned} \left(\frac{20}{k}\right)^{k} & = u^{\frac{20}{u}} \\ & = \left(u^{\frac{1}{u}}\right)^{20} \end{aligned}$$
which is maximized by setting $u=e$. However, $e^{\frac{20}{e}} \approx 1568.05$, meaning that $s=20$ is not possible.