Tab Closing

Let $m$ be the number of tabs currently open. After one click, $m$ decreases by $1$, and the tabs are redrawn from scratch.

The length of each tab is

$$\text{len}=\min\left(b,\frac{a}{m}\right),$$

so the x-positions are

$$\text{len},2\text{len},3\text{len},\ldots,m\text{len}.$$

We count mouse moves, not clicks. If your cursor is already on an x after the layout changes, you can click again without paying anything. That is the whole game.

The two useful cursor positions

There are basically two phases.

Set

$$q=\left\lfloor \frac{a}{b}\right\rfloor.$$

Phase 1: too many tabs

If $m>q$, then $m>a/b$, so

$$\frac{a}{m}<b.$$

Therefore

$$\text{len}=\frac{a}{m}.$$

The rightmost x-position is

$$m\cdot \frac{a}{m}=a.$$

So while there are too many tabs, position $a$ always works. Move the cursor to the right edge once and keep clicking the rightmost tab.

Phase 2: tabs have max width

If $m\le q$, then $mb\le a$, so

$$b\le \frac{a}{m}.$$

Therefore

$$\text{len}=b.$$

Now the x-positions are

$$b,2b,3b,\ldots,mb.$$

So position $b$ always works. Move there once and keep clicking the first tab until everything is gone.

That immediately proves the answer is at most $2$:

• click at $a$ while $m>q$;
• click at $b$ while $m\le q$.

When is one move enough?

If only one cursor position is used, then it must also work for the very last tab. When $m=1$, the only x-position is $b$. So a one-move strategy must use position $b$.

Now check when $b$ works for every needed value of $m$.

If

$$n\le q,$$

we start directly in Phase 2. Every layout has tab length $b$, so clicking at $b$ closes all tabs. Answer: $1$.

There is one extra special case: if

$$a=b,$$

then positions $a$ and $b$ are the same point. In the too-many-tabs phase, $a$ works; in the fixed-width phase, $b$ works; but since $a=b$, you never actually move again. Answer: $1$.

Otherwise, suppose $b<a$ and $n>q$. Then we start in Phase 1 and eventually need to finish Phase 2. Could one position still do everything? No.

A one-move strategy would have to use $b$, because the last remaining tab ends at $b$. But at the first too-many-tabs state, $m=q+1$, the tab length is $a/(q+1)$, and $b$ is not one of those endpoints when $b<a$. So $b$ cannot even handle that state.

So in this case one move is impossible, and the two-move strategy above is optimal.

Therefore the answer is simply

$$\boxed{ \begin{cases} 1, & n\le \left\lfloor a/b\right\rfloor \text{ or } a=b,\\ 2, & \text{otherwise.} \end{cases}}$$

No simulation. No floating-point nonsense. Just integer division and a tiny edge case that exists specifically to make people trip over their shoelaces.