Remilia Plays Soku

Observation 1: distance is king

Let

$$d = \min(|x_1-x_2|,\ n-|x_1-x_2|)$$

be the initial shortest circular distance between Reimu and Remilia.

If Remilia never moves, Reimu just walks along a shortest path and catches her after exactly $d$ seconds. Nothing fancy, no anime footsies.

So the whole problem is: how much extra time can Remilia buy with her at most $k$ moves?

Observation 2: one move buys at most one second

Suppose the current shortest distance after Reimu's previous move is $D$.

If Remilia stays:

• if $D=1$, Reimu catches immediately this second;
• otherwise Reimu moves closer, and the distance becomes $D-1$.

If Remilia moves to an adjacent position, her distance from Reimu can increase by at most $1$. Then Reimu moves after seeing this, so he can decrease the distance by $1$ again.

Thus one Remilia move can only prevent the distance from decreasing for that second. It cannot make Reimu “lose progress” overall.

A nice potential-function way to say this is:

$$\Phi = D + r,$$

where $r$ is Remilia's remaining number of moves.

Each non-catching second decreases $\Phi$ by at least $1$:

• if Remilia stays, then $D$ decreases by $1$;
• if Remilia moves, then $r$ decreases by $1$, and $D$ does not increase after Reimu's response.

Initially,

$$\Phi = d+k.$$

Since while not caught we have $D\ge 1$, Reimu catches by time at most

$$d+k.$$

Observation 3: Remilia can actually achieve this for $n\ge 4$

For $n\ge 4$, whenever the distance is not maximal in the circle, Remilia can move to a neighbor that is farther from Reimu.

If the current distance is $D$, she moves to make it $D+1$. Since $D+1\ge 2$, Reimu cannot catch immediately. After Reimu moves, the distance is at least $D$ by triangle inequality.

So each spent move really does buy one full extra second.

If Remilia starts exactly opposite Reimu on an even cycle, or at maximum distance in general, moving may not help. She can simply stay for one second first; Reimu reduces the distance, and then she starts burning her $k$ moves to stall.

Therefore for all $n\ge 4$:

$$\text{answer}=d+k.$$

Tiny circles: $n=2$ and $n=3$

For $n=2$ or $n=3$, every pair of distinct positions is adjacent. After Remilia acts, Reimu can always move or stay to her position immediately.

So:

$$\text{answer}=1.$$

Complexity

Each test case is just a few arithmetic operations:

$$O(1)$$

Total complexity:

$$O(t)$$

Memory usage:

$$O(1)$$