矩阵的转置是通过交换行和列得到的新矩阵。
在这个程序中，要求用户输入行数
r和列数
c。在本程序中，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)。
下面的程序然后计算矩阵的转置并将其打印在屏幕上。
求矩阵转置的程序
#include <stdio.h> int main() { int a[10][10], transpose[10][10], r, c; printf("Enter rows and columns: "); scanf("%d %d", &r, &c); // asssigning elements to the matrix printf("\nEnter matrix elements:\n"); for (int i = 0; i < r; ++i) for (int j = 0; j < c; ++j) { printf("Enter element a%d%d: ", i + 1, j + 1); scanf("%d", &a[i][j]); } // printing the matrix a[][] printf("\nEntered matrix: \n"); for (int i = 0; i < r; ++i) for (int j = 0; j < c; ++j) { printf("%d ", a[i][j]); if (j == c-1) printf("\n"); } // computing the transpose for (int i = 0; i < r; ++i) for (int j = 0; j < c; ++j) { transpose[j][i] = a[i][j]; } // printing the transpose printf("\nTranspose of the matrix:\n"); for (int i = 0; i < c; ++i) for (int j = 0; j < r; ++j) { printf("%d ", transpose[i][j]); if (j == r-1) printf("\n"); } return 0; }
输出
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
