結果
問題 |
No.2121 帰属関係と充足可能性
|
ユーザー |
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提出日時 | 2025-06-12 18:39:58 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 1,144 bytes |
コンパイル時間 | 319 ms |
コンパイル使用メモリ | 82,072 KB |
実行使用メモリ | 113,264 KB |
最終ジャッジ日時 | 2025-06-12 18:40:21 |
合計ジャッジ時間 | 4,265 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 5 TLE * 1 -- * 43 |
ソースコード
import sys from itertools import product def generate_v(n): if n == 0: return [frozenset()] prev_v = generate_v(n-1) current_v = set() for bits in range(1 << len(prev_v)): subset = set() for i in range(len(prev_v)): if (bits >> i) & 1: subset.add(prev_v[i]) current_v.add(frozenset(subset)) return list(current_v) def main(): N = int(sys.stdin.readline()) A = list(map(int, sys.stdin.readline().split())) A0, A1, A2, A3, A4, A5 = A v = generate_v(N) for m0, m1, m2 in product(v, repeat=3): mA0 = [m0, m1, m2][A0] mA1 = [m0, m1, m2][A1] mA2 = [m0, m1, m2][A2] mA3 = [m0, m1, m2][A3] mA4 = [m0, m1, m2][A4] mA5 = [m0, m1, m2][A5] condition1 = all(elem in mA1 for elem in mA0) condition2 = mA1 in mA2 condition3 = mA3 in mA2 condition4 = mA4 in mA2 condition5 = mA5 in mA0 if condition1 and condition2 and condition3 and condition4 and condition5: print("YES") return print("NO") if __name__ == "__main__": main()