Overview

This is a problem where you simulate a single-elimination tournament and determine in which round each player is eliminated.

Analysis

Understanding the Structure of the Problem

The tournament bracket has the structure of a complete binary tree. Let’s consider the case where \(N = 8\).

Round 1: (Player 1 vs Player 2), (Player 3 vs Player 4), (Player 5 vs Player 6), (Player 7 vs Player 8)
Round 2: (Winner of Match 1 vs Winner of Match 2), (Winner of Match 3 vs Winner of Match 4)
Round 3: (Winner of Match 1 vs Winner of Match 2) → Champion is determined

In each round, two adjacent players (two consecutive winners in the list) compete against each other, and the stronger one advances to the next round. The losing player’s elimination round is the current round number.

Is a Straightforward Simulation Sufficient?

Since the number of players is halved each round, the total number of matches across all rounds is:

\[\frac{N}{2} + \frac{N}{4} + \cdots + 1 = N - 1\]

In other words, there are only \(N - 1\) matches in total. Since \(N \leq 2^{19} = 524288\), a straightforward simulation runs fast enough.

Algorithm

1. Initialization: Initialize the list of remaining players current_players with \([0, 1, 2, \ldots, N-1]\) (0-indexed).
2. For each round \(r = 1, 2, \ldots, \log_2 N\):
   • Pair up players from the beginning of the list, two at a time (the \(i\)-th pair: current_players[2i] and current_players[2i+1]).
   • For each pair, compare their strengths \(S\), and record the elimination round of the weaker player as \(r\).
   • Add only the winner to the next round’s list next_players.
3. The last remaining player is the champion, and their elimination round is set to \(0\) (left as the initial value).

Concrete Example (\(N = 4\), \(S = [3, 1, 4, 2]\))

• Round 1: Player 1 (\(S=3\)) vs Player 2 (\(S=1\)) → Player 1 wins, Player 2 is eliminated in Round 1. Player 3 (\(S=4\)) vs Player 4 (\(S=2\)) → Player 3 wins, Player 4 is eliminated in Round 1.
• Round 2: Player 1 (\(S=3\)) vs Player 3 (\(S=4\)) → Player 3 wins, Player 1 is eliminated in Round 2.
• Result: 2 1 0 1 (Player 3 is the champion)

Complexity

• Time Complexity: \(O(N)\) — Since the total number of matches across all rounds is \(N - 1\).
• Space Complexity: \(O(N)\) — Required for storing the player list and the result array.

Implementation Notes

• You can concisely compute \(\log_2 N\) as an integer using N.bit_length() - 1. For example, if \(N = 8\), then bit_length() is \(4\), so \(R = 3\).

• Player numbers are 1-indexed, but arrays are managed as 0-indexed, with the appropriate conversion applied during output.

• By reading all input at once using sys.stdin.read(), I/O does not become a bottleneck even for large inputs.

  Source Code

import sys

def main():
    input_data = sys.stdin.read().split()
    N = int(input_data[0])
    S = [int(x) for x in input_data[1:N+1]]
    
    # result[i] = round in which player i+1 (0-indexed) loses, 0 for champion
    result = [0] * N
    
    # Simulate the tournament
    # current_players[i] = index (0-based) of the player currently in position i
    current_players = list(range(N))
    
    R = N.bit_length() - 1  # log2(N)
    
    for r in range(1, R + 1):
        num_matches = len(current_players) // 2
        next_players = []
        for i in range(num_matches):
            p1 = current_players[2 * i]
            p2 = current_players[2 * i + 1]
            if S[p1] > S[p2]:
                next_players.append(p1)
                result[p2] = r
            else:
                next_players.append(p2)
                result[p1] = r
        current_players = next_players
    
    print(' '.join(map(str, result)))

main()

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