Python 实现：根据规则判定 2^n×2^n 方阵中作弊者的赦免命运

以下是使用 Python 语言解决此问题的代码： ```python def赦免矩阵(n): size = 2 n matrix = [[1] * size for _ in range(size)] 赦免(matrix, 0, 0, size) for row in matrix: for num in row: print(num, end=' ') print() def赦免(matrix, row_start, col_start, size): if size == 1: return half_size = size // 2 for i in range(row_start, row_start + half_size): for j in range(col_start, col_start + half_size): matrix[i][j] = 0 赦免(matrix, row_start, col_start + half_size, half_size) 赦免(matrix, row_start + half_size, col_start, half_size) 赦免(matrix, row_start + half_size, col_start + half_size, half_size) n = int(input()) 赦免矩阵(n) ``` 当您输入`3`时，它将输出与您期望相同的结果。 ######[AI写代码神器 | 247点数解答 | 2024-11-03 10:30:46]