結果

問題 No.582 キャンディー・ボックス３
ユーザー qwewe
提出日時 2025-05-14 13:09:14
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 2,240 bytes
コンパイル時間 593 ms
コンパイル使用メモリ 68,260 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-05-14 13:10:10
合計ジャッジ時間 1,448 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 12 WA * 5
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <iostream> // Include the standard input/output library

// The main function where the program execution begins
int main() {
    // Optimize standard input/output operations for faster execution.
    // std::ios_base::sync_with_stdio(false) disables synchronization with C standard streams.
    // std::cin.tie(NULL) unties cin from cout, meaning flushing cout won't be forced before cin operations.
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    int N; // Declare an integer variable N to store the number of boxes.
    std::cin >> N; // Read the number of boxes from standard input.
    
    long long C_i; // Declare a long long variable C_i to store the number of candies in the current box.
                   // Use long long because the number of candies can be up to 10^9.
    int count_ones = 0; // Initialize an integer variable count_ones to 0.
                        // This variable will count the number of boxes containing exactly one candy.
    
    // Loop N times, once for each box.
    for (int i = 0; i < N; ++i) {
        std::cin >> C_i; // Read the number of candies for the i-th box.
        // Check if the current box contains exactly 1 candy.
        if (C_i == 1) {
            count_ones++; // If it contains exactly 1 candy, increment the counter.
        }
    }
    
    // The winning strategy depends on the parity of the number of boxes with exactly 1 candy.
    // This is based on the analysis that player B has more control over the game state,
    // particularly regarding the parity of the count of 1-candy boxes.
    // B can often force the parity to be even on A's turn if it starts even.
    // A wins if and only if the initial count of 1-candy boxes is odd.
    
    // Check if the count of boxes with 1 candy is odd.
    if (count_ones % 2 != 0) {
        // If count_ones is odd, Player A (the first player) wins.
        std::cout << "A" << std::endl; // Output 'A' followed by a newline.
    } else {
        // If count_ones is even (including 0), Player B (the second player) wins.
        std::cout << "B" << std::endl; // Output 'B' followed by a newline.
    }
    
    return 0; // Return 0 to indicate successful execution of the program.
}
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