2228BUnreviewedunratedGPT-5.5 (high)

Remilia Plays Soku

Progressive hints first, then the full explanation and implementation when you're ready to cash out.

games implementation

Original problem

Review status

AI-generated and still unreviewed. Double-check the details before internalizing them.

Hints

Progressive nudges

Open only as much as you need to keep the solve alive.

Forget the exact labels for a second. On a circle, the only thing that really matters is the current shortest distance $d$ between Reimu and Remilia.

If Remilia stays, Reimu can reduce the shortest distance by $1$ each second. So with $k=0$ , the answer is just the circular distance

\min(|x_1-x_2|,\ n-|x_1-x_2|).

When Remilia spends one move and runs “away”, what happens to $d$ ? Usually it increases by $1$ before Reimu moves, then Reimu decreases it by $1$ , so the distance stays the same. A move is basically a one-second delay.

There is one annoying tiny-circle exception: for $n=2$ or $n=3$ , every possible position of Remilia after her action is reachable by Reimu in one move. So the answer is always $1$ there.

For $n\ge 4$ , every Remilia move can add exactly one extra second, and Reimu can guarantee it adds at most one. Therefore the answer is simply

\min(|x_1-x_2|,\ n-|x_1-x_2|) + k.

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)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int t;
    cin >> t;

    while (t--) {
        ll n, x1, x2, k;
        cin >> n >> x1 >> x2 >> k;

        if (n <= 3) {
            cout << 1 << '\n';
            continue;
        }

        ll diff = llabs(x1 - x2);
        ll d = min(diff, n - diff);

        cout << d + k << '\n';
    }

    return 0;
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int t;
    cin >> t;

    while (t--) {
        ll n, x1, x2, k;
        cin >> n >> x1 >> x2 >> k;

        if (n <= 3) {
            cout << 1 << '\n';
            continue;
        }

        ll diff = llabs(x1 - x2);
        ll d = min(diff, n - diff);

        cout << d + k << '\n';
    }

    return 0;
}

View on Codeforces Back to the list

Remilia Plays Soku

Progressive nudges

Full explanation

Observation 1: distance is king

Observation 2: one move buys at most one second

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

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

Complexity

Generated C++

Remilia Plays Soku

Progressive nudges

Full explanation

Observation 1: distance is king

Observation 2: one move buys at most one second

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

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

Complexity

Generated C++

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

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