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142 lines
4.3 KiB
Python
Executable file
142 lines
4.3 KiB
Python
Executable file
"""#
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### 谜题描述
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A bracket sequence is a string containing only characters \"(\" and \")\". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters \"1\" and \"+\" between the original characters of the sequence. For example, bracket sequences \"()()\" and \"(())\" are regular (the resulting expressions are: \"(1)+(1)\" and \"((1+1)+1)\"), and \")(\", \"(\" and \")\" are not.
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Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
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You are given a regular bracket sequence s and an integer number k. Your task is to find a regular bracket sequence of length exactly k such that it is also a subsequence of s.
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It is guaranteed that such sequence always exists.
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Input
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The first line contains two integers n and k (2 ≤ k ≤ n ≤ 2 ⋅ 10^5, both n and k are even) — the length of s and the length of the sequence you are asked to find.
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The second line is a string s — regular bracket sequence of length n.
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Output
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Print a single string — a regular bracket sequence of length exactly k such that it is also a subsequence of s.
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It is guaranteed that such sequence always exists.
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Examples
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Input
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6 4
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()(())
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Output
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()()
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Input
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8 8
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(()(()))
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Output
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(()(()))
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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[n,k] = map(int,raw_input().split())
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bracket = raw_input()
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result = []
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i = 0
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for i in bracket:
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if i == \")\" and n>k:
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result.pop()
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n-=2
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else:
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result.append(i)
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print \"\".join(result)
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import re
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import random
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from bootcamp import Basebootcamp
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class Cbracketsubsequencebootcamp(Basebootcamp):
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def __init__(self, max_insertions=3):
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self.max_insertions = max_insertions
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def case_generator(self):
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# 生成目标k长度序列
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k = 2 * random.randint(1, 6) # 允许更大范围k值
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t = self.generate_regular_sequence(k)
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# 插入多层括号结构
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insertions = random.randint(0, self.max_insertions)
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s = t
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for _ in range(insertions):
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# 随机插入合法括号对
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pos = random.randint(0, len(s))
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s = s[:pos] + '()' + s[pos:]
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assert self.is_regular(s), f"Invalid regular sequence: {s}"
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return {'n': len(s), 'k': k, 's': s}
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@staticmethod
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def generate_regular_sequence(length):
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"""生成更复杂的嵌套结构"""
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if length == 0:
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return ""
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if length == 2:
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return "()"
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choice = random.random()
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if choice < 0.5:
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return "(" + Cbracketsubsequencebootcamp.generate_regular_sequence(length-2) + ")"
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else:
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split = random.randint(1, length//2 - 1) * 2
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return (Cbracketsubsequencebootcamp.generate_regular_sequence(split) +
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Cbracketsubsequencebootcamp.generate_regular_sequence(length-split))
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@staticmethod
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def prompt_func(question_case):
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# 保持原prompt结构，增加示例多样性
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return f"""给定长度为{question_case['n']}的正则括号序列：
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{question_case['s']}
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请找出长度为{question_case['k']}的正则子序列。将答案置于[answer][/answer]标签内。"""
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@staticmethod
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def extract_output(output):
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# 强化格式校验
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matches = re.findall(r'\[answer\]\s*([()]+)\s*\[/answer\]', output)
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return matches[-1] if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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# 添加更严格的校验
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return (
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len(solution) == identity['k'] and
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solution.count('(') == solution.count(')') and
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cls.is_regular(solution) and
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cls.is_subsequence(solution, identity['s'])
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)
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@staticmethod
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def is_regular(s):
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balance = 0
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for c in s:
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balance += 1 if c == '(' else -1
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if balance < 0:
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return False
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return balance == 0
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@staticmethod
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def is_subsequence(sub, s):
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it = iter(s)
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return all(c in it for c in sub)
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