結果
| 問題 |
No.489 株に挑戦
|
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-16 01:06:36 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,313 bytes |
| コンパイル時間 | 263 ms |
| コンパイル使用メモリ | 82,480 KB |
| 実行使用メモリ | 77,348 KB |
| 最終ジャッジ日時 | 2025-04-16 01:08:10 |
| 合計ジャッジ時間 | 3,553 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 33 WA * 2 |
ソースコード
import sys
from collections import deque
def main():
N, D, K = map(int, sys.stdin.readline().split())
x = [int(sys.stdin.readline()) for _ in range(N)]
dq = deque()
max_profit = 0
best_j = -1
best_k = -1
for k in range(1, N):
window_left = max(0, k - D)
# Remove elements out of the window from the front
while dq and dq[0] < window_left:
dq.popleft()
current_j = k - 1
# Remove elements from the back that are >= current_j's x
while dq and x[dq[-1]] >= x[current_j]:
dq.pop()
dq.append(current_j)
# The front is the index of the minimum value in the current window
min_j = dq[0]
current_profit = x[k] - x[min_j]
total_profit = current_profit * K
if total_profit > max_profit:
max_profit = total_profit
best_j = min_j
best_k = k
elif total_profit == max_profit and total_profit > 0:
# Check lexicographical order
if (min_j < best_j) or (min_j == best_j and k < best_k):
best_j = min_j
best_k = k
if max_profit > 0:
print(max_profit)
print(best_j, best_k)
else:
print(0)
if __name__ == "__main__":
main()
lam6er