結果

問題 No.2103 ±1s Game
ユーザー lam6er
提出日時 2025-03-20 20:38:21
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,735 bytes
コンパイル時間 214 ms
コンパイル使用メモリ 82,280 KB
実行使用メモリ 54,552 KB
最終ジャッジ日時 2025-03-20 20:38:46
合計ジャッジ時間 2,735 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
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 * 24 WA * 10
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 #
プレゼンテーションモードにする

X, Y, K, P = map(int, input().split())
d = 0 if P == 1 else 1
s_min = max(0, K - X)
s_max = min(Y, K)
# Check if there's any s in [s_min, s_max] with parity d
has_d_parity = False
# The minimum and maximum possible s values. If any of them or any in between has parity d.
low = s_min
high = s_max
if low > high:
has_d_parity = False
else:
if low % 2 == d or high % 2 == d:
has_d_parity = True
else:
# Check if there are elements with parity d between low and high
if low < high:
has_d_parity = True
if not has_d_parity:
print("Bob")
exit()
r = (Y - d) % 2
M = X + Y - K
a_min = max(0, M - X)
a_max = min(Y, M)
# Check if there exists a in [a_min, a_max] with parity r
valid_r = False
if a_min > a_max:
valid_r = False
else:
# Check a_min and a_max first
if a_min % 2 == r or a_max % 2 == r:
valid_r = True
else:
# If there's at least one a between a_min and a_max with parity r
# The difference between a_min and a_max >=1 and they have different parities
# then there's a a with parity r in between.
if a_min < a_max:
valid_r = True
if not valid_r:
print("Bob")
exit()
if M % 2 == 1:
print("Alice")
else:
# Check if there exists a in [a_min, a_max] with parity (r ^ 1)
valid_opposite = False
if a_min <= a_max:
# Check a_min and a_max first
if a_min % 2 == (r ^ 1) or a_max % 2 == (r ^ 1):
valid_opposite = True
else:
# Check if there are any elements in between
if a_min < a_max:
valid_opposite = (a_min % 2 != a_max % 2)
if valid_opposite:
print("Bob")
else:
print("Alice")
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