結果
| 問題 |
No.1446 ハンバーグと納豆ごはん
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 18:55:02 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 38 ms / 2,000 ms |
| コード長 | 1,690 bytes |
| コンパイル時間 | 200 ms |
| コンパイル使用メモリ | 82,712 KB |
| 実行使用メモリ | 54,216 KB |
| 最終ジャッジ日時 | 2025-03-20 18:56:58 |
| 合計ジャッジ時間 | 2,776 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 33 |
ソースコード
A, B, N, M = map(int, input().split())
max_t = min(A, B)
# Function to compute best T when converting hamburgers (case 2: y=0)
def compute_case2():
best = 0
denominator = N + 1
if denominator == 0:
return 0 # This case should not happen as N >=1
x_opt = (A - B) // denominator
candidates = [x_opt - 2, x_opt -1, x_opt, x_opt +1, x_opt +2, 0, A//N]
seen = set()
for x in candidates:
x = max(0, x)
if N == 0:
x = 0
else:
x = min(x, A // N)
if x in seen:
continue
seen.add(x)
a_new = A - x * N
if a_new < 0:
continue
b_new = B + x
current = min(a_new, b_new)
if current > best:
best = current
return best
current_case2 = compute_case2()
if current_case2 > max_t:
max_t = current_case2
# Function to compute best T when converting natto (case 3: x=0)
def compute_case3():
best = 0
denominator = M + 1
if denominator == 0:
return 0 # This case should not happen as M >=1
y_opt = (B - A) // denominator
candidates = [y_opt - 2, y_opt -1, y_opt, y_opt +1, y_opt +2, 0, B//M]
seen = set()
for y in candidates:
y = max(0, y)
if M == 0:
y = 0
else:
y = min(y, B // M)
if y in seen:
continue
seen.add(y)
b_new = B - y * M
if b_new < 0:
continue
a_new = A + y
current = min(a_new, b_new)
if current > best:
best = current
return best
current_case3 = compute_case3()
if current_case3 > max_t:
max_t = current_case3
print(max_t)
lam6er