結果
| 問題 |
No.227 簡単ポーカー
|
| ユーザー |
|
| 提出日時 | 2020-08-20 10:12:58 |
| 言語 | Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 844 bytes |
| コンパイル時間 | 91 ms |
| コンパイル使用メモリ | 12,800 KB |
| 実行使用メモリ | 10,752 KB |
| 最終ジャッジ日時 | 2024-10-13 03:00:45 |
| 合計ジャッジ時間 | 1,051 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 11 WA * 3 |
ソースコード
all_number = input().split()
number = []
for i in all_number:
number.append(i)
number.sort()
if number[0] == number[1] == number[2] and number[3] == number[4] or number[0] == number[1] and number[2] == number[3] == number[4]:
print('FULL HOUSE')
elif number[0] == number[1] == number[2] and number[3] != number[4] or number[1] == number[2] == number[3] and number[0] != number[4] or number[2] == number[3] == number[4] and number[0] != number[1]:
print('THREE CARD')
elif number[0] == number[1] == number[2] == number[3] == number[4] or number[0] != number[1] != number[2] != number[3] != number[4]:
print('NO HAND')
elif number[0] == number[1] and number[2] == number[3] or number[0] == number[1] and number[3] == number[4] or number[1] == number[2] and number[3] == number[4]:
print('TWO PAIR')
else:
print('ONE PAIR')