Removal of a Sequence (Hard Version)

We start with the sequence

$1,2,3,\dots,10^{12}.$

Each operation removes positions $y,2y,3y,\dots$ from the current sequence. The values themselves do not matter during removal; only their current positions matter.

The trap is trying to simulate the sequence forward. That is completely dead on arrival. Even the length is up to $10^{12}$, and $x$ is also up to $10^{12}$. So yeah, no lists, no sets, no "maybe it passes with vectors" nonsense.

Reverse The Operation

Suppose $y>1$.

After one operation, the remaining old positions are exactly the positive integers that are not divisible by $y$:

$1,2,\dots,y-1,y+1,y+2,\dots,2y-1,2y+1,\dots$

So if we want the $k$-th element after one operation, we need the $k$-th positive integer that is not a multiple of $y$.

Every block of length $y$ contributes $y-1$ survivors. Therefore, before one operation, position $k$ after the operation came from position

$f(k)=k+\left\lfloor\frac{k-1}{y-1}\right\rfloor.$

Example with $y=3$:

Surviving old positions are

$1,2,4,5,7,8,\dots$

For $k=5$:

$f(5)=5+\left\lfloor\frac{4}{2}\right\rfloor=7.$

Correct: the 5th survivor is old position 7.

Now repeat this logic. If we want the $k$-th remaining number after $x$ operations, we apply this inverse map $x$ times:

$\text{answer}=f^x(k).$

If this position is greater than $10^{12}$, then the final sequence has fewer than $k$ elements, so the answer is $-1$.

For $y=1$, every position is a multiple of $1$, so the first operation removes everything. Since $x\ge 1$, the answer is always $-1$.

Why We Need Jumps

The formula is simple, but applying it $x$ times is not:

$x\le 10^{12}.$

So we need to accelerate the iteration of

$f(k)=k+\left\lfloor\frac{k-1}{y-1}\right\rfloor.$

Let

$d=y-1.$

Then

$f(k)=k+\left\lfloor\frac{k-1}{d}\right\rfloor.$

Define

$q=\left\lfloor\frac{k-1}{d}\right\rfloor.$

As long as $q$ stays fixed, each application of $f$ simply adds $q$ to $k$:

$k \leftarrow k+q.$

So instead of doing one step at a time, we jump several steps until $q$ changes.

The current $q$ remains valid while

$k \le (q+1)d.$

The next value where $q$ increases is

$k=(q+1)d+1.$

So the number of same-$q$ steps we can safely apply is

$\left\lceil\frac{(q+1)d+1-k}{q}\right\rceil.$

We take the minimum of that and the number of operations still remaining.

There is one more practical cap: we never need to jump farther than the first moment where $k>10^{12}$, because then the answer is already $-1$. This keeps all arithmetic inside long long.

Special Case: $q=0$

If

$q=0,$

then

$f(k)=k.$

That means $k<y$. This position is never moved by the reverse process, so the answer is just $k$.

Example: if $y=5$ and $k=1$, the first survivor is always the original number $1$.

Why This Is Fast

Each jump moves $k$ to a new range of

$\left\lfloor\frac{k-1}{d}\right\rfloor.$

If $d$ is small, $q$ gets big quickly and $k$ grows fast. If $d$ is large, there are not many possible quotient ranges before passing $10^{12}$. The slowest balance is around $d\approx \sqrt{10^{12}}$, which is still only about $10^6$ jumps per test case. That is fine for $t\le 10$.

That is the whole trick: iterate the inverse map, but skip the boring parts.