"""# ### 谜题描述 You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible. Input Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100). The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n). It is guaranteed that the sum of n by all test cases doesn't exceed 100. Output For each test case, if there is no appropriate permutation, print one number -1. Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n. Example Input 5 1 1 2 4 1 3 4 1 3 4 2 3 4 5 5 1 5 7 2 8 Output 1 2 -1 4 5 1 2 3 6 -1 1 3 5 6 7 9 2 4 8 10 Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python from bisect import bisect for _ in range(input()): n = input() a = map(int,raw_input().split()) if 1 not in a or 2*n in a: print -1 continue l = [] for i in range(1,2*n+1): if i not in a: l.append(i) d = {} f = 0 for i in range(n): p = bisect(l,a[i]) if p==len(l): f = 1 break d[a[i]] = l[p] l.remove(l[p]) if f==1: print -1 continue for i in range(n): print a[i],d[a[i]], print ``` 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 random from bisect import bisect_right as bisect import re from bootcamp import Basebootcamp class Crestoringpermutationbootcamp(Basebootcamp): def __init__(self, min_n=1, max_n=5, unsolvable_prob=0.2): self.min_n = min_n self.max_n = max_n self.unsolvable_prob = unsolvable_prob def case_generator(self): if random.random() < self.unsolvable_prob: n = random.randint(self.min_n, self.max_n) case_type = random.choice([1, 2]) b = [] if case_type == 1: b = [2 * n] remaining = list(range(1, 2 * n)) if n > 1: others = random.sample(remaining, n-1) b.extend(others) else: possible = list(range(2, 2 * n + 1)) b = random.sample(possible, k=n) random.shuffle(b) return { 'n': n, 'b': b, 'expected': -1 } else: while True: n = random.randint(self.min_n, self.max_n) possible_values = list(range(1, 2 * n + 1)) if 2 * n in possible_values: possible_values.remove(2 * n) if 1 not in possible_values: continue b = [1] if n > 1: remaining = possible_values.copy() remaining.remove(1) others = random.sample(remaining, n-1) b.extend(others) if len(set(b)) != n or 2 * n in b or 1 not in b: continue sorted_b = sorted(b) l = sorted([num for num in range(1, 2 * n + 1) if num not in sorted_b]) d = {} f = 0 for bi in sorted_b: pos = bisect(l, bi) if pos >= len(l): f = 1 break selected = l[pos] d[bi] = selected del l[pos] if f: continue a = [] for num in sorted_b: a.append(num) a.append(d[num]) return { 'n': n, 'b': b, 'expected': a } @staticmethod def prompt_func(question_case): n = question_case['n'] b = question_case['b'] prompt = f"""You are given a sequence b of length {n}. Find the lexicographically smallest permutation a of 1 to {2*n} such that for each i (1 ≤ i ≤ {n}), b[i] is the minimum of a[2i-1] and a[2i]. If impossible, output -1. Input: n = {n} b = {b} Format your answer as space-separated numbers within [answer] tags. Example: [answer]1 2 3 4[/answer] or [answer]-1[/answer]""" return prompt @staticmethod def extract_output(output): pattern = r'\[answer\](.*?)\[/answer\]' matches = re.findall(pattern, output, re.DOTALL) if not matches: return None last_match = matches[-1].strip() if last_match == '-1': return -1 try: return list(map(int, last_match.split())) except: return None @classmethod def _verify_correction(cls, solution, identity): n = identity['n'] b = identity['b'] expected = identity.get('expected') if solution == -1: return expected == -1 if not isinstance(solution, list) or len(solution) != 2 * n: return False if set(solution) != set(range(1, 2 * n + 1)): return False for i in range(n): if min(solution[2*i], solution[2*i+1]) != b[i]: return False if expected != -1: return solution == expected return False