Solution:

For every $c>0$ (not necessarily an integer either), the Lucchese Dedalus, of Genoese origin, can imprison the Minotaur while spending less than $c$ cents of a drachma.

Lemma. For every pair of positive integers $k$ and $m$, the expense of buying all strings of length $mk$ in which $k$ or more consecutive zeros never occur is less than or equal to
$$\left(1-\frac{1}{2^{k}}\right)^{m}$$
drachmas.

To prove this, observe first of all that the cost of each of these strings is $2^{-mk}$ drachmas. This expense must be multiplied by the total number of strings of length $mk$ that do not contain $k$ (or more) consecutive zeros. The exact number of these strings is hard to compute, but in order to obtain an upper bound on the price Dedalus pays, it suffices to find an upper estimate.

To do this, we divide the string of length $mk$ into $m$ blocks of length $k$, and observe that each block of length $k$ can be any binary string except the one made up of only 0 digits. Since the binary strings of length $k$ number $2^{k}$, it follows that for each block there are $2^{k}-1$ possible choices. Having $m$ blocks, the total number of strings is at most $\left(2^{k}-1\right)^{m}$, hence the total cost does not exceed
$$\frac{1}{2^{mk}} \cdot\left(2^{k}-1\right)^{m}=\left(1-\frac{1}{2^{k}}\right)^{m}$$
as required.

Given now any $c>0$, we choose $k$ such that $1 / 2^{k}<c / 200$, and once $k$ is fixed we choose $m$ such that
$$\left(1-\frac{1}{2^{k}}\right)^{m}<\frac{c}{200}$$
At this point Dedalus will buy the string consisting of $k$ digits 0, which is able to block all sequences that contain, somewhere, at least $k$ consecutive digits 0, plus all strings of length $mk$ in which $k$ or more consecutive zeros never occur, able to block all the remaining possible sequences simply by considering the first block of $mk$ characters.

All this while spending less than $c / 200+c / 200=c / 100$ drachmas!

Solution:

Alternative solution

We present an alternative version of the preceding Lemma, which requires the purchase of a smaller number of strings, and still allows us to conclude in an analogous manner. For simplicity, we call a $k$-sequence any infinite sequence in which $k+1$ (or more) consecutive zeros never occur.

Lemma. For every pair of integers $k \geq 0$ and $m \geq 0$, Dedalus can imprison all $k$-sequences by buying $(k+1)^{m}$ strings, all ending with the digit 1, spending in total
$$\left(1-\frac{1}{2^{k+1}}\right)^{m}$$
drachmas.

To prove this, suppose $k$ is a fixed parameter and proceed by induction on $m$. The case $m=0$ is easy, since it suffices to buy the string 1.

Suppose now the statement is true for some $m$, and let us prove it for $m+1$. Let $S_{1}$, $\ldots$, $S_{(k+1)^{m}}$ be strings capable of imprisoning any $k$-sequence. We now replace each of these strings, say the string $S_{i}$, with the $k+1$ strings obtained by appending to the right of $S_{i}$ the strings
$$1, \quad 01, \quad 001, \quad \ldots, \quad \underbrace{0 \ldots 0}_{k} 1$$
The key observation is that, if $S_{i}$ is able to block a certain $k$-sequence, then one of the new strings is also able to block the same $k$-sequence, simply by positioning itself at the same point. Indeed, a $k$-sequence that contains $S_{i}$ can continue to the right of $S_{i}$ in only one of the $k+1$ indicated ways.

Letting $L_{i}$ denote the length of $S_{i}$, the cost of the $k+1$ strings that replace $S_{i}$ is
$$\frac{1}{2^{L_{i}+1}}+\cdots+\frac{1}{2^{L_{i}+k+1}}=\frac{1}{2^{L_{i}}}\left(\frac{1}{2}+\cdots+\frac{1}{2^{k+1}}\right)=\frac{1}{2^{L_{i}}} \cdot\left(1-\frac{1}{2^{k+1}}\right)$$
and hence it equals that of $S_{i}$ multiplied by a fixed factor less than one.

If we now apply the same procedure to all $(k+1)^{m}$ strings that were suitable at step $m$, we obtain $(k+1)^{m+1}$ strings that are still suitable and have a total cost equal to the previous one multiplied by the common fixed factor $\left(1-1 / 2^{k+1}\right)$. Moreover, they still all end with the digit 1.