Lawn Mower

We need choose as many removed boards as possible while avoiding $w$ consecutive removed boards.

Call a removed board 1 and a board left standing 0. The condition is simple: the binary string of length $n$ must not contain $w$ consecutive 1s. So the whole problem is: maximize the number of 1s without making a run of length $w$.

First, get a hard upper bound. Split the first $\left\lfloor \frac{n}{w} \right\rfloor \cdot w$ boards into disjoint blocks of length $w$:

$$[1,w], [w+1,2w], [2w+1,3w], \dots$$

Inside any full block of $w$ consecutive boards, we cannot remove all $w$ boards. If we did, the mower has a hole exactly $w$ meters wide, and it's gone. So every full block must contain at least one board that remains.

There are

$$\left\lfloor \frac{n}{w} \right\rfloor$$

such full blocks, so at least that many boards must stay. Therefore the number of removed boards is at most

$$n - \left\lfloor \frac{n}{w} \right\rfloor.$$

Now we show this bound is actually reachable, because otherwise it's just math cosplay.

Leave boards numbered

$$w, 2w, 3w, \dots, \left\lfloor \frac{n}{w} \right\rfloor w.$$

Remove every other board. Between two kept boards, there are exactly $w-1$ removed boards. Before the first kept board, there are also $w-1$ removed boards. After the last kept board, there are $n \bmod w$ removed boards, which is less than $w$.

So the longest consecutive removed segment has length at most $w-1$. The mower cannot escape, and we removed exactly

$$n - \left\lfloor \frac{n}{w} \right\rfloor$$

boards.

Edge cases behave nicely:

• If $w=1$, even one removed board is enough for the mower to escape, so the answer is $0$. The formula gives $n-\lfloor n/1\rfloor=0$.
• If $w>n$, no hole can ever be wide enough, so all boards can be removed. The formula gives $n-0=n$.
• Large values are fine because we only do integer division.

For each test case, the answer is:

$$\boxed{n - \left\lfloor \frac{n}{w} \right\rfloor}$$

Time complexity: $O(1)$ per test case.

Memory complexity: $O(1)$.