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internbootcamp/bootcamp/cbracketsubsequence/cbracketsubsequence.py

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+"""# 
+
+### 谜题描述
+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.
+
+Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
+
+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.
+
+It is guaranteed that such sequence always exists.
+
+Input
+
+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.
+
+The second line is a string s — regular bracket sequence of length n.
+
+Output
+
+Print a single string — a regular bracket sequence of length exactly k such that it is also a subsequence of s.
+
+It is guaranteed that such sequence always exists.
+
+Examples
+
+Input
+
+6 4
+()(())
+
+
+Output
+
+()()
+
+
+Input
+
+8 8
+(()(()))
+
+
+Output
+
+(()(()))
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+[n,k] = map(int,raw_input().split())
+bracket = raw_input()
+result = []
+i = 0
+for i in bracket:
+    if i == \")\" and n>k:
+        result.pop()
+        n-=2
+    else:
+        result.append(i)
+print \"\".join(result)
+```
+
+
+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。
+"""
+
+from bootcamp import Basebootcamp
+import re
+import random
+from bootcamp import Basebootcamp
+
+class Cbracketsubsequencebootcamp(Basebootcamp):
+    def __init__(self, max_insertions=3):
+        self.max_insertions = max_insertions
+    
+    def case_generator(self):
+        # 生成目标k长度序列
+        k = 2 * random.randint(1, 6)  # 允许更大范围k值
+        t = self.generate_regular_sequence(k)
+        
+        # 插入多层括号结构
+        insertions = random.randint(0, self.max_insertions)
+        s = t
+        for _ in range(insertions):
+            # 随机插入合法括号对
+            pos = random.randint(0, len(s))
+            s = s[:pos] + '()' + s[pos:]
+        
+        assert self.is_regular(s), f"Invalid regular sequence: {s}"
+        return {'n': len(s), 'k': k, 's': s}
+    
+    @staticmethod
+    def generate_regular_sequence(length):
+        """生成更复杂的嵌套结构"""
+        if length == 0:
+            return ""
+        if length == 2:
+            return "()"
+        choice = random.random()
+        if choice < 0.5:
+            return "(" + Cbracketsubsequencebootcamp.generate_regular_sequence(length-2) + ")"
+        else:
+            split = random.randint(1, length//2 - 1) * 2
+            return (Cbracketsubsequencebootcamp.generate_regular_sequence(split) +
+                    Cbracketsubsequencebootcamp.generate_regular_sequence(length-split))
+    
+    @staticmethod
+    def prompt_func(question_case):
+        # 保持原prompt结构，增加示例多样性
+        return f"""给定长度为{question_case['n']}的正则括号序列：
+{question_case['s']}
+
+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。"""
+
+    @staticmethod
+    def extract_output(output):
+        # 强化格式校验
+        matches = re.findall(r'\[answer\]\s*([()]+)\s*\[/answer\]', output)
+        return matches[-1] if matches else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 添加更严格的校验
+        return (
+            len(solution) == identity['k'] and
+            solution.count('(') == solution.count(')') and
+            cls.is_regular(solution) and
+            cls.is_subsequence(solution, identity['s'])
+        )
+    
+    @staticmethod
+    def is_regular(s):
+        balance = 0
+        for c in s:
+            balance += 1 if c == '(' else -1
+            if balance < 0:
+                return False
+        return balance == 0
+    
+    @staticmethod
+    def is_subsequence(sub, s):
+        it = iter(s)
+        return all(c in it for c in sub)