在此示例中,您将学习在C编程中查找矩阵的转置。
在此程序中,要求用户输入行r和列c的数量 。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( r*c的顺序)。
然后,下面的程序计算矩阵的转置并将其打印在屏幕上。
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; }
结果
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
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