Pizza Time

Let Hao's total be $H$ and Alex's total be $A$.

The whole problem is secretly about comparing those two numbers. The daily splitting rule says:

Hao eats $m_1$, Alex eats $m_2$, and $m_3$ survives.

So on every day where Hao gets anything, Alex gets at least as much as Hao. Then, when the remaining pile has at most $2$ slices, Alex eats all of it. Since the pile is positive, Alex gets at least one final slice that Hao never gets.

Therefore Alex's total is always at least one more than Hao's total:

Also, every slice is eventually eaten by exactly one of them:

Combine them:

So:

That gives an upper bound. Now we need to prove it is actually reachable, because an upper bound alone is just math cosplay.

Why the bound can be reached

Define $f(n)$ as the maximum number of slices Hao can eat from $n$ slices. We claim:

The cases $n=1,2$ give $0$, and $n=3$ works with the split $(1,1,1)$.

For larger $n$, choose the carried-over pile size $c$ so that:

This makes it possible for $c$ to be the largest group while still leaving at least one slice for both Hao and Alex.

Now split the remaining $n-c$ slices as evenly as possible:

Because $c\ge \lceil n/3\rceil$, we have $m_2\le c$, so the split is valid.

After this day, Hao gets

and then the game continues from $c$ slices. By induction, the future can give him

So the total becomes:

The two numerators add to $n-1$. This equals

as long as we avoid the annoying case where both $n-c$ and $c-1$ are odd. That bad case happens exactly when $n$ is odd and $c$ is even.

So choose $c=\lceil n/3\rceil$, but if $n$ is odd and that $c$ is even, use $c+1$ instead. This is still a valid carried pile, and now the floors add perfectly.

Thus the upper bound is reachable, so the answer is exactly:

In C++, integer division already floors, so print:

(n - 1) / 2

That's it. The pizza story is doing a lot of dramatic acting for a one-line formula.