結果
| 問題 |
No.2076 Concon Substrings (ConVersion)
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 20:47:57 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,279 bytes |
| コンパイル時間 | 248 ms |
| コンパイル使用メモリ | 82,784 KB |
| 実行使用メモリ | 76,704 KB |
| 最終ジャッジ日時 | 2025-03-20 20:48:09 |
| 合計ジャッジ時間 | 3,180 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 22 WA * 6 |
ソースコード
import bisect
def main():
import sys
input = sys.stdin.read().split()
n = int(input[0])
A = int(input[1])
B = int(input[2])
s = input[3]
# Preprocess to find all 'con' runs
runs = []
i = 0
while i <= len(s) - 3:
if s[i:i+3] == 'con':
count = 0
j = i
while j <= len(s) - 3 and s[j:j+3] == 'con':
count += 1
j += 3
runs.append(count)
i = j
else:
i += 1
# We need to process the runs dynamically considering possible overlaps after operations
# But we'll use a greedy approach with the sorted list
available = sorted(runs)
steps = 0
current_step = 1 # 1-based steps
while True:
required = A if current_step % 2 == 1 else B
if not available:
break
# Find the smallest run >= required
idx = bisect.bisect_left(available, required)
if idx == len(available):
break # no such run
selected = available.pop(idx)
remaining = selected - required
if remaining > 0:
bisect.insort(available, remaining)
steps += 1
current_step += 1
print(steps)
if __name__ == '__main__':
main()
lam6er