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Competitive Programming Session Editorial

This editorial compiles the problems and solutions discussed during this session, reordered according to the official contest problem set.

Problem A: Lecture Capacity

Problem Statement: A contest has \(n\) teams, and each team must have exactly 3 participants. Calculate the total number of participants.

Solution: Since every team has exactly 3 members, the total number of participants is simply \(3 \times n\). Read the integer \(n\) and output \(3n\).

Problem B: Pseudo Palindrome

Problem Statement: Can array \(a\) be rearranged such that \(|a_i - a_{n+1-i}| \le d\) for all \(i\)?

Solution: This is a matching problem. We need to pair elements such that their difference is \(\le d\). Sort the array \(a\). Optimal pairings will always occur between elements that are relatively close in the sorted version. Use Dynamic Programming:

  • dp[i][0]: Can prefix \(i\) be perfectly matched? True if dp[i-2][0] is true AND \(|a_i - a_{i-1}| \le d\).

  • dp[i][1]: Can prefix \(i\) have exactly one unpaired element? True if dp[i-1][0] is true OR (dp[i-2][1] is true AND \(|a_i - a_{i-1}| \le d\)). If \(n\) is even, check dp[n][0]. If \(n\) is odd, check dp[n][1].

Problem C: XOR LCM

Problem Statement: Find positive integers \(a, b \le 10^{17}\) such that \((a \oplus c) + (b \oplus c) = \text{lcm}(a, c) + \text{lcm}(b, c)\) for a given \(c \le 10^7\).

Solution: We need \((a \oplus c) - \text{lcm}(a, c) + (b \oplus c) - \text{lcm}(b, c) = 0\). Set \(a = c\). Then \((a \oplus c) - \text{lcm}(a, c) = 0 - c = -c\). We need the second part to equal \(+c\). Set \(b = c \cdot 2^k\) for sufficiently large \(k\) (e.g., \(k=30\) since \(c \le 10^7 < 2^{30}\)). Then \(b \oplus c = b + c\) and \(\text{lcm}(b, c) = b\). The second part becomes \((b+c) - b = c\). Result: \(-c + c = 0\). Valid pair is \((c, c \cdot 2^{30})\).

Problem D: Make Empty

Problem Statement: Given a permutation \(p\) of even length \(n\). In one operation, remove a subsequence \(t\) of length \(2k\) if \(\max(t_1 \dots t_k) < \min(t_{k+1} \dots t_{2k})\) (Type 1) or \(\min(t_1 \dots t_k) > \max(t_{k+1} \dots t_{2k})\) (Type 2). Find the minimum operations to empty \(p\).

Solution: The minimum number of operations is always 1 or 2. Let \(S = \{1, \dots, n/2\}\) (Small values) and \(L = \{n/2+1, \dots, n\}\) (Large values).

Detailed Explanation

The core insight is partitioning numbers into Small (\(S\)) and Large (\(L\)).

  • Type 1 operation (\(\max < \min\)) works if we pick Small numbers that appear before Large numbers.

  • Type 2 operation (\(\min > \max\)) works if we pick Large numbers that appear before Small numbers.

Case 1: 1 Operation If the input naturally has all \(S\) before all \(L\), or all \(L\) before all \(S\), we can clear it in one go.

Case 2: 2 Operations (The Guarantee) If they are interleaved, we need 2 operations. We can always achieve this by splitting the array indices exactly in half: \([1, n/2]\) and \([n/2+1, n]\). We define four sets based on where elements are (Index) and what they are (Value):

  • \(A\): First half, Small value

  • \(B\): First half, Large value

  • \(C\): Second half, Large value

  • \(D\): Second half, Small value

Why does this work perfectly? Every permutation has exactly \(n/2\) Small values. Total Small values = \(|A| + |D| = n/2\). Total positions in first half = \(|A| + |B| = n/2\). From these two equations, we must have \(|B| = |D|\). Similarly, \(|A| = |C|\). This exact size match allows us to perfectly pair them up without leftovers:

  1. Op 1 (Type 1): Pair all of \(A\) with all of \(C\). (\(A\) is first half, \(C\) is second half \(\implies\) valid order. \(A\) is Small, \(C\) is Large \(\implies\) valid values).

  2. Op 2 (Type 2): Pair all of \(B\) with all of \(D\). (\(B\) is first half, \(D\) is second half \(\implies\) valid order. \(B\) is Large, \(D\) is Small \(\implies\) valid values).

Problem E: Counting is Fun

Problem Statement: Given a tree, label edges with \(1 \dots n-1\). Traversal starts at node 1 with state \(x=1\). If current edge label matches \(x\), increment \(x\) (Cost +1). Find number of label permutations yielding total cost \(c\).

Solution: The answer is \(N(c) - N(c+1)\), where \(N(k)\) is the number of permutations with cost \(\ge k\).

Understanding the Cost

Think of \(x\) as a key and edge labels as locks. You start with Key 1. You can open any door except Door 1. To get cost \(\ge 1\), Door 1 must "trap" you at the root (it must be incident to node 1). You are forced to upgrade to Key 2. To get cost \(\ge 2\), you must then travel deeper and get trapped by Door 2. This means Door 2 must be in the subtree 'behind' Door 1. This implies that to have cost \(\ge k\), the edges labeled \(1, 2, \dots, k\) must form a continuous chain starting from the root.

Combinatorial Breakdown of N(k)

Formula: \(N(k) = W_k \times 2^{k-1} \times (n-1-k)! \pmod{998244353}\).

  1. \(W_k\) (Choosing the path): We need to choose \(k\) edges that form a root-starting chain. For any node \(v\), the path to the root has \(dep[v]\) edges. Any \(k\) edges from this path (must include the one touching \(v\)) form a valid chain ending at \(v\). \(W_k = \sum_{v=2}^n \binom{dep[v]-1}{k-1}\).

  2. \(2^{k-1}\) (Labeling the chain): We must assign labels \(\{1, \dots, k\}\) to force sequential increments. This means these labels must form a single contiguous block on the chain, built one by one:

  3. Label 1 can be anywhere (1 choice).

  4. Label 2 must be immediately adjacent to 1 (above or below it \(\to\) 2 choices).

  5. Label 3 must be immediately adjacent to the \(\{1,2\}\) block (above or below it \(\to\) 2 choices).

  6. ...and so on for all \(k\) labels. Total orderings \(= 1 \times \underbrace{2 \times 2 \times \dots \times 2}_{k-1 \text{ times}} = 2^{k-1}\).

  7. \((n-1-k)!\) (The rest): The remaining \(n-1-k\) edges can be labeled arbitrarily with the remaining values.

Problem F: Non Unique

Problem Statement: Let \(F(a)\) be the set of indices with the maximum number of inversions to their left. Find the number of ways to achieve \(|F(a)| \ge 2\) in the minimum number of adjacent swaps.

Solution: Let \(v_i = f(a, i)\) be the count of \(j < i\) such that \(a_j > a_i\).

  1. Initial State: Calculate all \(v_i\) using a Segment Tree or Fenwick Tree. Identify the maximum value \(v_{max}\), its frequency, and the position pos if unique.

  2. Case 0 Operations: If the frequency of \(v_{max}\) is initially \(\ge 2\), the minimum operations needed is 0. The number of ways is 1 (do nothing).

  3. Case \(x\) Operations: If there is a unique maximum at pos, we must perform swaps.

  4. Find the second largest value, \(f_2 = \max_{i \neq pos}(v_i)\).

  5. The minimum number of operations is \(x = v_{pos} - f_2\).

  6. We need to find the number of valid swap sequences of length \(x\) that result in another index \(i\) having the same \(f\)-value as pos.

  7. Counting Ways:

  8. Iterate through all other indices \(i\) that currently have \(v_i = f_2\). These are candidates to tie with pos.

  9. For each candidate \(i\), determine how many valid swap sequences exist. This depends on:

    • The "gap" \(x\) we need to close.

    • The distance between pos and \(i\).

    • Constraints on movement (handled by finding the Next Greater Element to limit valid swap ranges).

  10. Use precomputed combinatorial prefix sums (e.g., \(\sum \binom{x}{k}\)) to efficiently count the valid paths for each candidate and sum them up.