创建具有 O 和 X 的交替矩形的矩阵 · GeeksForGeeks 中文系列教程

# 创建具有 O 和 X 的交替矩形的矩阵 > 原文： [https://www.geeksforgeeks.org/create-a-matrix-with-alternating-rectangles-of-0-and-x/](https://www.geeksforgeeks.org/create-a-matrix-with-alternating-rectangles-of-0-and-x/) 编写代码，输入两个数字 m 和 n，并创建一个大小为 mxn（m 行和 n 列）的矩阵，其中每个元素均为 X 或 0。 Xs 和 0s 必须交替填充，矩阵应具有 Xs 的最外面的矩形，然后是 0s 的矩形，然后是 Xs 的矩形，依此类推。 **示例**： ``` Input: m = 3, n = 3 Output: Following matrix X X X X 0 X X X X Input: m = 4, n = 5 Output: Following matrix X X X X X X 0 0 0 X X 0 0 0 X X X X X X Input: m = 5, n = 5 Output: Following matrix X X X X X X 0 0 0 X X 0 X 0 X X 0 0 0 X X X X X X Input: m = 6, n = 7 Output: Following matrix X X X X X X X X 0 0 0 0 0 X X 0 X X X 0 X X 0 X X X 0 X X 0 0 0 0 0 X X X X X X X X ``` ***我们强烈建议最小化浏览器，然后自己尝试。*** 在 Shreepartners Gurgaon 的校园招聘中提出了这个问题。我遵循以下方法。 **1）**使用[代码以螺旋形式](https://www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/)打印矩阵。 **2）**而不是打印数组，而是在数组中插入元素“ X”或“ 0”。以下是上述方法的实现。 ## C++ ```cpp #include <stdio.h> // Function to print alternating rectangles of 0 and X void fill0X(int m, int n) { /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ int i, k = 0, l = 0; // Store given number of rows and columns for later use int r = m, c = n; // A 2D array to store the output to be printed char a[m][n]; char x = 'X'; // Iniitialize the character to be stoed in a[][] // Fill characters in a[][] in spiral form. Every iteration fills // one rectangle of either Xs or Os while (k < m && l < n) { /* Fill the first row from the remaining rows */ for (i = l; i < n; ++i) a[k][i] = x; k++; /* Fill the last column from the remaining columns */ for (i = k; i < m; ++i) a[i][n-1] = x; n--; /* Fill the last row from the remaining rows */ if (k < m) { for (i = n-1; i >= l; --i) a[m-1][i] = x; m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m-1; i >= k; --i) a[i][l] = x; l++; } // Flip character for next iteration x = (x == '0')? 'X': '0'; } // Print the filled matrix for (i = 0; i < r; i++) { for (int j = 0; j < c; j++) printf("%c ", a[i][j]); printf("\n"); } } /* Driver program to test above functions */ int main() { puts("Output for m = 5, n = 6"); fill0X(5, 6); puts("\nOutput for m = 4, n = 4"); fill0X(4, 4); puts("\nOutput for m = 3, n = 4"); fill0X(3, 4); return 0; } ```