結果
| 問題 | 
                            No.2254 Reverse Only
                             | 
                    
| コンテスト | |
| ユーザー | 
                             gew1fw
                         | 
                    
| 提出日時 | 2025-06-12 14:47:15 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                WA
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 1,494 bytes | 
| コンパイル時間 | 186 ms | 
| コンパイル使用メモリ | 82,700 KB | 
| 実行使用メモリ | 174,032 KB | 
| 最終ジャッジ日時 | 2025-06-12 14:49:53 | 
| 合計ジャッジ時間 | 8,911 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge4 / judge5 | 
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| ファイルパターン | 結果 | 
|---|---|
| other | AC * 32 WA * 15 | 
ソースコード
from collections import defaultdict, Counter
def main():
    import sys
    input = sys.stdin.read().split()
    ptr = 0
    N = int(input[ptr])
    ptr += 1
    k = int(input[ptr])
    ptr += 1
    A = list(map(int, input[ptr:ptr+N]))
    ptr += N
    B = list(map(int, input[ptr:ptr+N]))
    ptr += N
    
    # Check if the multisets of A and B are the same
    if Counter(A) != Counter(B):
        print("No")
        return
    
    # If k is even or 1, output Yes
    if k % 2 == 0 or k == 1:
        print("Yes")
        return
    
    # For odd k > 1, check parity of positions for each element
    def get_parity_counts(arr):
        counts = defaultdict(lambda: [0, 0])  # [odd_count, even_count]
        for idx, num in enumerate(arr):
            pos = idx + 1  # 1-based position
            if pos % 2 == 1:
                counts[num][0] += 1
            else:
                counts[num][1] += 1
        return counts
    
    a_parity = get_parity_counts(A)
    b_parity = get_parity_counts(B)
    
    # Check if all elements have the same parity counts in A and B
    # First, collect all unique elements from both (though Counter is same)
    all_elements = set(a_parity.keys()).union(set(b_parity.keys()))
    for elem in all_elements:
        a_odd, a_even = a_parity[elem]
        b_odd, b_even = b_parity.get(elem, [0, 0])
        if a_odd != b_odd or a_even != b_even:
            print("No")
            return
    print("Yes")
if __name__ == "__main__":
    main()
            
            
            
        
            
gew1fw