C语言实现：解决2ⁿ×2ⁿ方阵作弊者赦免问题

```c #include <stdio.h> #include <stdlib.h> void赦免(int n, int matrix, int row, int col, int size) { if (size == 1) { return; } for (int i = row + size / 2; i < row + size; i++) { for (int j = col + size / 2; j < col + size; j++) { matrix[i][j] = 1; } } 赦免(n, matrix, row, col, size / 2); 赦免(n, matrix, row + size / 2, col, size / 2); 赦免(n, matrix, row, col + size / 2, size / 2); } int main() { int n; scanf("%d", &n); int size = 1 << n; int matrix = (int )malloc(size * sizeof(int *)); for (int i = 0; i < size; i++) { matrix[i] = (int *)malloc(size * sizeof(int)); } 赦免(n, matrix, 0, 0, size); for (int i = 0; i < size; i++) { for (int j = 0; j < size; j++) { printf("%d ", matrix[i][j]); } printf("\n"); } for (int i = 0; i < size; i++) { free(matrix[i]); } free(matrix); return 0; } ``` ######[AI写代码神器 | 370点数解答 | 2024-11-03 10:31:31]