All editorial codes contain solutions wrapped in classes. I'll only paste the important part (namespace solution in C++, class Solution in Python 3) as plaintext in the solution code section (along with the corresponding AC submission link). You might want to see the code template here.

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

namespace solution {
    bool hasMultipleTests = true;

    void preprocess() {
        // something
    }

    void input(int testcase) {
        // something
    }

    void solve(int testcase) {
        // something
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    solution::preprocess();

    int t = 1;
    if (solution::hasMultipleTests) cin >> t;

    for (int testcase=1; testcase<=t; testcase++) {
        solution::input(testcase);
        solution::solve(testcase);
    }

    return 0;
}

import sys
input = sys.stdin.readline


class Solution:
    hasMultipleTests = True

    @classmethod
    def preprocess(cls):
        pass

    @classmethod
    def input(cls, testcase):
        pass

    @classmethod
    def solve(cls, testcase):
        pass

# end Solution

if __name__ == '__main__':
    Solution.preprocess()

    t = int(input()) if Solution.hasMultipleTests else 1
    for testcase in range(1, t+1):
        Solution.input(testcase)
        Solution.solve(testcase)

Also, per my own custom ever since 2019, solution codes will only be published after system testing, to ensure all codes used here guarantee AC. Thank you for your patience.

In which cases would a light be on?

Imagine that you have $$$x$$$ white socks and $$$y$$$ black socks. What would you do to make the maximum pairs of matched socks?

Imagine that you have $$$x$$$ white socks and $$$y$$$ black socks. What would you do to make the maximum pairs of unmatched socks?

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem

There is always exactly one solution for $$$n = 1$$$. For $$$n \ge 3$$$, for which $$$k$$$ should there be no solution at all?

Try to find a solution with $$$k = 2$$$ and $$$k = 4$$$, assuming $$$n$$$ allows at least one solution to exist. Can you find a pattern and generalize it for all even $$$k$$$?

Try to find a solution with $$$k = 3$$$ and $$$k = 5$$$, assuming $$$n$$$ allows at least one solution to exist. Can you find a pattern and generalize it for all odd $$$k$$$?

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem

Without loss of generality, you can rewrite the triangle inequality into a much more compacted form if the side lengths are properly ordered.

Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(a, b, c)$$$.

Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(a, c, d)$$$.

Given four integers $$$a \le b \le c \le d$$$, if $$$(a, b, d)$$$ are side lengths of a non-degenerate triangle, so does $$$(b, c, d)$$$.

To make a sorted array $$$[b_1, b_2, \ldots, b_m] \space (m \ge 3)$$$ satisfy the problem, we must ensure that $$$b_1 + b_2 > b_m$$$.

Fix the result array to have two minimas and one maxima at respectively $$$a_x$$$, $$$a_y$$$ and $$$a_z$$$. How many operations are required to do so?

Is there a fast and optimal way to try all possible $$$(a_x, a_y, a_z)$$$ triplets?

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem

There are many specific constraints on the system tree.

First, notice that $$$1 \le x \le y \le n-1$$$ if and only if $$$p_x \le p_y$$$. Draw a few trees with this constraint and start pointing your finger on the nodes from $$$0$$$ to $$$n-1$$$. Notice the pattern your finger is making.

Incorporate hint #1 with the fact that the "tentacles" of the Genokraken are just paths. Try to draw more detailed conclusion about that pattern.

Is there a way to know how many tentacles there are?

What use could we make of the fact that node $$$1$$$ has exactly two adjacent nodes?

The query limit is very small. Is there a way to query that the information obtained in each query should (at least partially) determine a node?

For each "tentacle", can we know when its path has ended? How many queries should it take?

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem

TBD

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem

TBD

Submission link: TBD

// TBD

Submission link: TBD

# TBD

• Didn't attempt
• Great problem
• Nice problem
• OK problem
• Bad problem
• Terrible problem