歪んだ道の果て、輝きの終わり。(The end of the twisted road kills the shine.) —— Final Verdict's cover

Testify — void (Mournfinale) feat. 星熊南巫 marks a climax in the plotline of Arcaea. The song was released in Final Verdict, a 2022 DLC for the 2017 rhythm game.

We can easily find out that the overall average of the array will not change throughout the whole operation phase. This means only when the average of the given array equals to $x$ will the array meet the demands.

public static void solve() throws Exception {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    int x = scanner.nextInt();
    int[] a = new int[n];
    int sum = 0;
    for (int i = 0; i < n; i++) {
        a[i] = scanner.nextInt();
        sum += a[i];
    }
    if (sum == n * x) {
        System.out.println("YES");
        return;
    }
    else
        System.out.println("NO");
    return;
}

歪められた彼女の未来の先には…　(Ahead of her distorted future is...) —— Vicious Labyrinth's cover

Axium Crisis — ak+q was released in Vicious Labyrinth, a Main Story song pack released in version 1.5.0 of Arcaea. The song is also the theme song of Axium (or Tairitsu), a character in the game.

The teleporters should make as many labyrinth residents travel to the exit as they can. So: When $k$ is odd, just equip all cells other than the exit cell itself with teleporters that fly the cell occupier to the exit, and equip the exit cell with a teleporter to its nearest cell. This will make all people other than the exit occupier able to exit, and the initial owner of the exit will be only one cell away from the exit, achieving the minimum total distance.

When $k$ is even, equip all cells other than cell $n−1$ with teleporters to the excluded cell, and the $n−1$-th cell flies people to the exit as the interchange. The configuration can make all people other than the interchange's initial owner to the exit, minimizing the total distance.

public static void solve() throws Exception {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    int k = scanner.nextInt();
    if (k % 2 != 0) {
        for (int i = 1; i < n; i++) {
            System.out.print(n + " ");
        }
        System.out.println(n - 1);
    }
    else {
        for (int i = 1; i < n - 1; i++) {
            System.out.print(n - 1 + " ");
        }
        System.out.print(n + " ");
        System.out.println(n - 1);
    }
}

神の揺り籠、還らぬ日の導きへ…　(The cradle of God, guiding to the day that will not return...) —— Lucent Historia's cover

Breach of Faith — Supire feat.eili's name got straightly highlighted from the problem's title. The song is from Lucent Historia, another Main Story pack of Arcaea which just got released in November 2024 that details a world apart from Arcaea.

We know that $a_1=a_2−a_3+a_4−a_5+⋯+a_{2n}−a_{2n+1}$. The equation can be transformed to $a_1+a_3+⋯+a_{2n+1}=a_2+a_4+⋯+a_{2n}$, meaning that the sum of all odd-indexed elements in the array a equals to that of all even-indexed ones.

There are $n+1$ elements in the left formula and n in the right of the equation above. Therefore, if we choose the largest $n+1$ elements of the given array $b$ as all the odd-indexed elements and the rest for the even-indexed, indicating the lost element is with an even index in $a$, the sum of odd-indexed elements is, beyond no doubt, larger than the sum of the even-indexed ones (the lost element excluded of course), ensuring that the lost element is positive. Any of odd-indexed elements, also, will be larger than all the even-indexed ones other than the lost element (in other words, $a_{2i}<a_{2j−1}$ where $i$ and $j$ are integers). Suppose the element lost is $a_2$, and think about whether $a_2$ is bound to be larger than the largest element of $b$ (let's index the odd-indexed elements in the order of $a_1>a_3>⋯>a_{2n+1}$). The given equation can be transformed to be $a_2=a_1+(a_3−a_4)+(a_5−a_6)+⋯+(a_{2n−1}−a_{2n})+a_{2n+1}$, thus it can be reached that $a_2>a_1$ because of $a_{2i−1}−a_{2i}>0$ mentioned in our considerations in the paragraph above, which means the lost element IS the largest element of $a$, securing the pairwise distinct highlighted by the problem.

So what we need to do is just sort the array $b$, index the $n+1$ largest elements of $b$ with odd numbers and the rest with even numbers in $a$, and calculate out the lost element $a_2$.

But scarcely when the contest was only one single minute away from finishing had I reached this aha moment :(

public static void solve() throws Exception {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    long[] b = new long[2 * n];
    for (int i = 0; i < 2 * n; i++) {
        b[i] = scanner.nextLong();
    }
    Arrays.sort(b);
    long lostElement = 0;
    for (int i = 0; i < n - 1; i++) {
        lostElement -= b[i];
    }
    for (int i = n - 1; i < 2 * n; i++) {
        lostElement += b[i];
    }
    System.out.print(b[2 * n - 1] + " " + lostElement + " ");
    for (int i = 0; i < n - 1; i++) {
        System.out.print(b[2 * n - i - 2] + " " + b[i] + " ");
    }
    System.out.println(b[n - 1]);
}