# C 程序：查找矩阵的转置 > 原文： [https://www.programiz.com/c-programming/examples/matrix-transpose](https://www.programiz.com/c-programming/examples/matrix-transpose) #### 在此示例中，您将学习在 C 编程中查找矩阵的转置。 要理解此示例，您应该了解以下 [C 编程](/c-programming "C tutorial")主题： * [C 数组](/c-programming/c-arrays) * [C 多维数组](/c-programming/c-multi-dimensional-arrays) * * * 矩阵的转置是通过交换行和列而获得的新矩阵。 在该程序中，要求用户输入行数`r`和列数`c`。 在此程序中，它们的值应小于 10。 然后，要求用户输入矩阵的元素（顺序为`r*c`）。 然后，下面的程序计算矩阵的转置并将其打印在屏幕上。 * * * ## 查找矩阵转置的程序 ```c #include <stdio.h> int main() { int a[10][10], transpose[10][10], r, c, i, j; printf("Enter rows and columns: "); scanf("%d %d", &r, &c); // Assigning elements to the matrix printf("\nEnter matrix elements:\n"); for (i = 0; i < r; ++i) for (j = 0; j < c; ++j) { printf("Enter element a%d%d: ", i + 1, j + 1); scanf("%d", &a[i][j]); } // Displaying the matrix a[][] printf("\nEntered matrix: \n"); for (i = 0; i < r; ++i) for (j = 0; j < c; ++j) { printf("%d ", a[i][j]); if (j == c - 1) printf("\n"); } // Finding the transpose of matrix a for (i = 0; i < r; ++i) for (j = 0; j < c; ++j) { transpose[j][i] = a[i][j]; } // Displaying the transpose of matrix a printf("\nTranspose of the matrix:\n"); for (i = 0; i < c; ++i) for (j = 0; j < r; ++j) { printf("%d ", transpose[i][j]); if (j == r - 1) printf("\n"); } return 0; } ``` **输出** ```c Enter rows and columns: 2 3 Enter matrix elements: Enter element a11: 1 Enter element a12: 4 Enter element a13: 0 Enter element a21: -5 Enter element a22: 2 Enter element a23: 7 Entered matrix: 1 4 0 -5 2 7 Transpose of the matrix: 1 -5 4 2 0 7 ```