"""# ### 谜题描述 Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions. * All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once. * [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column. You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n. Input The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4. Output Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid. If there are multiple answers, print any. We can show that an answer always exists. Examples Input 4 Output 8 9 1 13 3 12 7 5 0 2 4 11 6 10 15 14 Input 8 Output 19 55 11 39 32 36 4 52 51 7 35 31 12 48 28 20 43 23 59 15 0 8 16 44 3 47 27 63 24 40 60 56 34 38 6 54 17 53 9 37 14 50 30 22 49 5 33 29 2 10 18 46 41 21 57 13 26 42 62 58 1 45 25 61 Note In the first example, XOR of each row and each column is 13. In the second example, XOR of each row and each column is 60. Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python from __future__ import division, print_function def main(): n=int(input()) res=[[0]*n for _ in range(n)] x=0 for k in range(n//4): for i in range(n): for j in range(4): res[i][j+k*4]=x x+=1 for item in res: print(*item,sep=\" \") ######## Python 2 and 3 footer by Pajenegod and c1729 # Note because cf runs old PyPy3 version which doesn't have the sped up # unicode strings, PyPy3 strings will many times be slower than pypy2. # There is a way to get around this by using binary strings in PyPy3 # but its syntax is different which makes it kind of a mess to use. # So on cf, use PyPy2 for best string performance. py2 = round(0.5) if py2: from future_builtins import ascii, filter, hex, map, oct, zip range = xrange import os, sys from io import IOBase, BytesIO BUFSIZE = 8192 class FastIO(BytesIO): newlines = 0 def __init__(self, file): self._file = file self._fd = file.fileno() self.writable = \"x\" in file.mode or \"w\" in file.mode self.write = super(FastIO, self).write if self.writable else None def _fill(self): s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0]) return s def read(self): while self._fill(): pass return super(FastIO,self).read() def readline(self): while self.newlines == 0: s = self._fill(); self.newlines = s.count(b\"\n\") + (not s) self.newlines -= 1 return super(FastIO, self).readline() def flush(self): if self.writable: os.write(self._fd, self.getvalue()) self.truncate(0), self.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable if py2: self.write = self.buffer.write self.read = self.buffer.read self.readline = self.buffer.readline else: self.write = lambda s:self.buffer.write(s.encode('ascii')) self.read = lambda:self.buffer.read().decode('ascii') self.readline = lambda:self.buffer.readline().decode('ascii') sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip('\r\n') # Cout implemented in Python import sys class ostream: def __lshift__(self,a): sys.stdout.write(str(a)) return self cout = ostream() endl = '\n' # Read all remaining integers in stdin, type is given by optional argument, this is fast def readnumbers(zero = 0): conv = ord if py2 else lambda x:x A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read() try: while True: if s[i] >= b'0' [0]: numb = 10 * numb + conv(s[i]) - 48 elif s[i] == b'-' [0]: sign = -1 elif s[i] != b'\r' [0]: A.append(sign*numb) numb = zero; sign = 1 i += 1 except:pass if s and s[-1] >= b'0' [0]: A.append(sign*numb) return A if __name__== \"__main__\": main() ``` 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 Cmagicgridbootcamp(Basebootcamp): def __init__(self, **params): """ 支持两种模式： 1. 固定n模式：直接指定n参数 2. 随机模式：通过min_n/max_n指定范围（默认4-1000） """ if 'n' in params: # 固定n模式 self.n = params['n'] if not (self.n %4 ==0 and 4<=self.n<=1000): raise ValueError("n must be multiple of 4 (4 <= n <= 1000)") self.mode = 'fixed' else: # 随机模式 self.min_n = max(4, ((params.get('min_n',4)+3)//4)*4) self.max_n = min(1000, (params.get('max_n',1000)//4)*4) if self.min_n > self.max_n: raise ValueError("Invalid range for n generation") self.mode = 'random' def case_generator(self): """动态生成不同n值的案例""" if self.mode == 'fixed': n = self.n else: possible_n = list(range(self.min_n, self.max_n+1, 4)) n = random.choice(possible_n) return {'n': n} @staticmethod def prompt_func(question_case) -> str: n = question_case['n'] prompt = f"""You need to construct a {n}x{n} magic grid where: 1. All integers 0-{n*n-1} appear exactly once 2. All rows have the same XOR 3. All columns have the same XOR as rows Generate the grid with {n} rows of {n} space-separated integers. Put your answer between [answer] and [/answer] tags. Example for n=4: [answer] 8 9 1 13 3 12 7 5 0 2 4 11 6 10 15 14 [/answer]""" return prompt @staticmethod def extract_output(output): # 提取最后一个answer块并转换为二维列表 answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL) if not answer_blocks: return None last_answer = answer_blocks[-1].strip() try: return [list(map(int, line.split())) for line in last_answer.split('\n') if line.strip()] except: return None @classmethod def _verify_correction(cls, solution, identity): n = identity['n'] # 结构验证 if len(solution)!=n or any(len(row)!=n for row in solution): return False # 数值范围验证 flat = [num for row in solution for num in row] if set(flat) != set(range(n*n)): return False # 异或一致性验证 row_xors = [0]*n col_xors = [0]*n for i in range(n): for j in range(n): row_xors[i] ^= solution[i][j] col_xors[j] ^= solution[i][j] return len(set(row_xors+col_xors)) == 1 # 使用示例（随机模式） # bootcamp = Cmagicgridbootcamp(min_n=8, max_n=16) # 生成8/12/16的案例 # print(bootcamp.case_generator()) # e.g. {'n': 12} # print(bootcamp.case_generator()) # e.g. {'n': 8} # # 使用示例（固定模式） # bootcamp_fixed = Cmagicgridbootcamp(n=4) # print(bootcamp_fixed.case_generator()) # 总是返回 {'n': 4}