Cake Assignment

Let the total number of cakes be

$T = 2^{k+1}.$

Chocola starts with $T/2 = 2^k$ cakes. Since Vanilla always has the rest, we only need to track one value:

$a = \text{number of cakes Chocola has}.$

The two operations become:

• Operation $1$: Chocola gives half away, so $a \to \frac a2.$
• Operation $2$: Vanilla gives half of her cakes to Chocola. Vanilla has $T-a$, so $a \to a + \frac{T-a}{2} = \frac{T+a}{2}.$

So every move replaces $a$ by either the midpoint between $0$ and $a$, or the midpoint between $a$ and $T$.

Walk backward

Forward construction looks a bit annoying, because at each state there may be two choices. But backward from the target, the last move is forced.

Suppose the current target value is $x$.

If the last operation was operation $1$, then before that operation Chocola had:

$2x.$

If the last operation was operation $2$, then:

$x = \frac{T+a}{2} \implies a = 2x - T.$

Now look at where $x$ lies:

• If $x < T/2$, the last operation cannot be operation $2$, because operation $2$ always produces something bigger than $T/2$. So the last operation was forced to be operation $1$, and we replace $x$ by $2x$.
• If $x > T/2$, the last operation cannot be operation $1$, because operation $1$ always produces something smaller than $T/2$. So the last operation was forced to be operation $2$, and we replace $x$ by $2x-T$.
• If $x = T/2$, we are already at the initial state, so we stop.

That is the whole trick. The process is deterministic. No BFS, no DP, no bitmask drama despite the tag trying to look fancy.

Why this gives the minimum

Every valid sequence ending at $x \ne T/2$ has some last operation. From the position of $x$ relative to $T/2$, that last operation is uniquely determined.

So there is no choice to optimize at the end. After undoing that forced last operation, the same argument applies again. This means the entire sequence of operations is forced when read backward.

Therefore, the number of backward steps until reaching $T/2$ is exactly the minimum number of forward steps. Any shorter sequence would also need to have the same forced suffix, then the same forced suffix before that, and so on, which is impossible once we count them all.

Building the answer

Maintain a vector ops of operations discovered backward.

While $x \ne 2^k$:

• If $x < 2^k$, append 1, then set $x = 2x$.
• Otherwise, append 2, then set $x = 2x - 2^{k+1}$.

These appended operations are in reverse order, because each one is the last operation of the remaining path. Reverse ops, print its size, then print the operations.

All values fit in long long because $k \le 60$, so $2^{k+1} \le 2^{61}$, well below the signed 64-bit limit. The statement promises the answer length is at most $120$, so this loop is tiny.