Strange Machine

The problem looks like it wants something clever with cycles, binary search, or precomputation. It really does not. The constraint $n \le 20$ is basically the problem whispering: “simulate me, but don’t be dumb about the one bad case.”

We have a circular string of operations:

• A: $x := x - 1$
• B: $x := \left\lfloor x/2 \right\rfloor$

For each query, we start at position $0$ and keep applying operations until $x=0$.

The trap

The value $a$ can be as large as $10^9$, so simulating one second at a time is obviously bad if the string is all A. For example:

n = 1
s = A
a = 1000000000

That would take $10^9$ seconds, and your program would also take approximately forever. Not ideal.

But that is the only bad case.

Case 1: no B exists

If every machine is A, then every second decreases $a$ by exactly $1$.

So the answer is simply:

No simulation needed.

Case 2: at least one B exists

Now the important observation:

Since the machines are arranged in a circle and $n \le 20$, if the string contains at least one B, then while moving around the circle, we will encounter a B at least once every $n$ steps.

Each B changes:

So after about $\log_2 a$ visits to B, the value must become $0$ unless earlier A operations already finished it off.

Because:

there can only be around $30$ meaningful halvings. Between two halvings, we do at most $n$ operations, and $n \le 20$.

So the total number of simulated steps per query is bounded by roughly:

Tiny. Absolutely harmless.

Simulation details

For each query:

1. Set pos = 0 because we always start at machine $1$.
2. Set ans = 0.
3. While a > 0:
   • If s[pos] == 'A', do a--.
   • Otherwise, do a /= 2.
   • Increment ans.
   • Move to the next machine: pos = (pos + 1) % n.
4. Print ans.

This exactly matches the statement: update first, then move.

Why this is correct

For the all-A case, each operation subtracts exactly $1$, so starting from $a$, it takes exactly $a$ operations to reach $0$.

For the case with at least one B, the simulation follows the process literally: it applies the current machine, counts one second, and advances clockwise. Since a B appears at least once per full cycle, the value is repeatedly halved. With $a \le 10^9$, after at most about $30$ such halvings the value reaches $0$, and A operations only make it smaller faster. Therefore the simulation finishes quickly and produces the exact number of seconds.

Complexity

For an all-A string, each query is answered in $O(1)$.

Otherwise, each query takes $O(n \log a)$ time, which is tiny here.

Memory usage is $O(1)$ besides the input string.