"""# ### 谜题描述 A two dimensional array is called a bracket array if each grid contains one of the two possible brackets — \"(\" or \")\". A path through the two dimensional array cells is called monotonous if any two consecutive cells in the path are side-adjacent and each cell of the path is located below or to the right from the previous one. A two dimensional array whose size equals n × m is called a correct bracket array, if any string formed by writing out the brackets on some monotonous way from cell (1, 1) to cell (n, m) forms a correct bracket sequence. Let's define the operation of comparing two correct bracket arrays of equal size (a and b) like that. Let's consider a given two dimensional array of priorities (c) — a two dimensional array of same size, containing different integers from 1 to nm. Let's find such position (i, j) in the two dimensional array, that ai, j ≠ bi, j. If there are several such positions, let's choose the one where number ci, j is minimum. If ai, j = \"(\", then a < b, otherwise a > b. If the position (i, j) is not found, then the arrays are considered equal. Your task is to find a k-th two dimensional correct bracket array. It is guaranteed that for the given sizes of n and m there will be no less than k two dimensional correct bracket arrays. Input The first line contains integers n, m and k — the sizes of the array and the number of the sought correct bracket array (1 ≤ n, m ≤ 100, 1 ≤ k ≤ 1018). Then an array of priorities is given, n lines each containing m numbers, number pi, j shows the priority of character j in line i (1 ≤ pi, j ≤ nm, all pi, j are different). Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. Output Print the k-th two dimensional correct bracket array. Examples Input 1 2 1 1 2 Output () Input 2 3 1 1 2 3 4 5 6 Output (() ()) Input 3 2 2 3 6 1 4 2 5 Output () )( () Note In the first sample exists only one correct two-dimensional bracket array. In the second and in the third samples two arrays exist. A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python #include int main() { int N, M; long long int what; scanf(\"%d %d %I64i\", &N, &M, &what); static int data[205][205]; static int p[205 * 205]; int i, j; for (i = 0; i < N; i++) { for (j = 0; j < M; j++) { scanf(\"%d\", &(data[i][j])); (data[i][j])--; p[data[i][j]] = i + j; } } static char res[205]; memset(res, '.', sizeof(res)); static long long int a[205][205]; int x, y; int K = N + M - 1; for (i = 0; i < (N * M); i++) { if (res[p[i]] == '.') { res[p[i]] = '('; memset(a, 0, sizeof(a)); a[0][0] = 1LL; for (x = 1; x <= K; x++) { for (y = 0; y <= K; y++) { if ((y > 0) && (res[x - 1] != ')')) a[x][y] += a[x - 1][y - 1]; if (res[x - 1] != '(') a[x][y] += a[x - 1][y + 1]; if (a[x][y] > 2000000000000000000LL) a[x][y] = 2000000000000000000LL; } } if (a[K][0] < what) { res[p[i]] = ')'; what -= a[K][0]; } } } for (i = 0; i < N; i++) { for (j = 0; j < M; j++) { printf(\"%c\", res[i + j]); } printf(\"\n\"); } return 0; } ``` 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 def generate_correct_array(n, m, k, priority): size = n * m p = [0] * size for i in range(n): for j in range(m): val = priority[i][j] - 1 # Convert to 0-based index p[val] = i + j res = ['.'] * (n + m - 1) K = n + m - 1 for i in range(size): s = p[i] if res[s] == '.': res[s] = '(' a = [[0] * (K + 2) for _ in range(K + 2)] a[0][0] = 1 for x in range(1, K + 1): for y in range(K + 1): a[x][y] = 0 current_char = res[x-1] if (x-1) < len(res) else '.' if y > 0 and current_char != ')': a[x][y] += a[x-1][y-1] if current_char != '(': a[x][y] += a[x-1][y+1] if a[x][y] > 1e18: a[x][y] = 1e18 total = a[K][0] if total < k: res[s] = ')' k -= total result = [] for i in range(n): row = [] for j in range(m): s = i + j row.append(res[s]) result.append(''.join(row)) return result class Cbracketsbootcamp(Basebootcamp): def __init__(self, n=2, m=3, k=1, priority=None): super().__init__() self.n = n self.m = m self.k = k if priority is None: # Generate a random priority matrix with 1..n*m elements = list(range(1, n * m + 1)) random.shuffle(elements) self.priority = [] idx = 0 for i in range(n): row = elements[idx:idx + m] self.priority.append(row) idx += m else: self.priority = [row.copy() for row in priority] self.validate_priority() def validate_priority(self): elements = [] for row in self.priority: elements.extend(row) if len(elements) != self.n * self.m: raise ValueError(f"Priority matrix size does not match n={self.n} and m={self.m}.") if len(set(elements)) != len(elements): raise ValueError("Priority matrix contains duplicate values.") expected = set(range(1, self.n * self.m + 1)) if set(elements) != expected: raise ValueError(f"Priority matrix elements must be unique and cover 1 to {self.n*self.m}.") def case_generator(self): # Generate a new random priority matrix each time to ensure diversity elements = list(range(1, self.n * self.m + 1)) random.shuffle(elements) priority = [] idx = 0 for i in range(self.n): row = elements[idx:idx + self.m] priority.append(row) idx += self.m # Ensure k is valid by setting to 1 (guaranteed by problem constraints) return { 'n': self.n, 'm': self.m, 'k': 1, 'priority': priority } @staticmethod def prompt_func(question_case): n = question_case['n'] m = question_case['m'] k = question_case['k'] priority = question_case['priority'] prompt = f"""You are to solve the k-th correct two-dimensional bracket array problem. The task is to find the k-th smallest correct bracket array based on the given priority matrix. Input: {n} {m} {k} Priority matrix: """ for row in priority: prompt += ' '.join(map(str, row)) + '\n' prompt += f""" A correct bracket array satisfies that every possible monotonous path from (1,1) to ({n},{m}) forms a valid bracket sequence. The k-th array is determined by comparing arrays using the priority matrix to find the earliest differing cell with the smallest priority. Output the correct {n}x{m} bracket array. Each row must contain exactly {m} parentheses. Place your answer between [answer] and [/answer], with each row on a separate line. Example format: [answer] () [/answer] Now, provide the answer for the given input:""" return prompt @staticmethod def extract_output(output): matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL) if not matches: return None content = matches[-1].strip() lines = [line.strip() for line in content.split('\n') if line.strip()] return lines @classmethod def _verify_correction(cls, solution, identity): try: if not solution: return False # Validate solution dimensions n = identity['n'] m = identity['m'] if len(solution) != n: return False for row in solution: if len(row) != m: return False # Generate correct answer correct = generate_correct_array( identity['n'], identity['m'], identity['k'], identity['priority'] ) return solution == correct except Exception as e: return False