C/C++矩阵计算器

问题描述

最近需要计算一些旋转矩阵，手算太麻烦，于是写了个小软件

参考

C/C++实现矩阵各种运算
Adjoint and Inverse of a Matrix

实现

Vecace博主的C/C++实现矩阵各种运算已经实现了诸如“加、减、乘、数乘”等大部分功能，在此基础上增加求矩阵的逆的功能，具体如下：

#include <iostream>
#include <malloc.h>
#include <stdio.h>
using namespace std;typedef struct
{//结构体int row, col;//二维指针，目的是动态分配内存float** matrix;
} Matrix;typedef struct
{char* name;char* number;
} Student;Matrix CreateMatrix()
{Matrix m;int row, col;cout << "输入行数与列数：" << endl;cin >> row >> col;float** enterMatrix;enterMatrix = (float**)malloc(row * sizeof(float*));for (int i = 0; i < row; i++)enterMatrix[i] = (float*)malloc(col * sizeof(float));cout << "输入你的矩阵：" << endl;for (int i = 0; i < row; i++){for (int j = 0; j < col; j++){cin >> enterMatrix[i][j];}}m.col = col;m.row = row;m.matrix = enterMatrix;return m;
}Matrix Create2DArrayDefault(int row_col)
{Matrix m;float** enterMatrix;enterMatrix = (float**)malloc(row_col * sizeof(float*));for (int i = 0; i < row_col; i++)enterMatrix[i] = (float*)malloc(row_col * sizeof(float));for (int i = 0; i < row_col; i++){for (int j = 0; j < row_col; j++){enterMatrix[i][j] = 0;}}m.col = row_col;m.row = row_col;m.matrix = enterMatrix;return m;
}// Function to get cofactor of A[p][q] in temp[][]. n is current
// dimension of A[][]
void getCofactor(Matrix A, Matrix temp, int p, int q, int n)
{int i = 0, j = 0;// Looping for each element of the matrix for (int row = 0; row < n; row++){for (int col = 0; col < n; col++){//  Copying into temporary matrix only those element //  which are not in given row and column if (row != p && col != q){temp.matrix[i][j++] = A.matrix[row][col];// Row is filled, so increase row index and // reset col index if (j == n - 1){j = 0;i++;}}}}
}/* Recursive function for finding determinant of matrix.n is current dimension of A[][]. */
int determinant(Matrix A, int n)
{int D = 0; // Initialize result //  Base case : if matrix contains single element if (n == 1)return A.matrix[0][0];Matrix temp = Create2DArrayDefault(n); // To store cofactors int sign = 1;  // To store sign multiplier // Iterate for each element of first row for (int f = 0; f < n; f++){// Getting Cofactor of A[0][f] getCofactor(A, temp, 0, f, n);D += sign * A.matrix[0][f] * determinant(temp, n - 1);// terms are to be added with alternate sign sign = -sign;}return D;
}// Function to get adjoint of A[N][N] in agetCofactordj[N][N].
void adjoint(Matrix A, Matrix adj)
{if (A.row == 1){adj.matrix[0][0] = 1;adj.row = 1;adj.col = 1;return;}int N = A.row;// temp is used to store cofactors of A[][] int sign = 1;Matrix temp = Create2DArrayDefault(N);for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){// Get cofactor of A[i][j] getCofactor(A, temp, i, j, N);// sign of adj[j][i] positive if sum of row // and column indexes is even. sign = ((i + j) % 2 == 0) ? 1 : -1;// Interchanging rows and columns to get the // transpose of the cofactor matrix adj.matrix[j][i] = (sign) * (determinant(temp, N - 1));}}adj.row = N;adj.col = N;
}bool inverse(Matrix A, Matrix inverse)
{// Find determinant of A[][] int N = A.row;int det = determinant(A, A.row);if (det == 0){cout << "矩阵的逆不存在！"/*"Singular matrix, can't find its inverse"*/;return false;}// Find adjoint Matrix adj = Create2DArrayDefault(N);adjoint(A, adj);// Find Inverse using formula "inverse(A) = adj(A)/det(A)" for (int i = 0; i < N; i++)for (int j = 0; j < N; j++)inverse.matrix[i][j] = adj.matrix[i][j] / float(det);inverse.row = N;inverse.col = N;return true;
}//初始化一个行为row列为col矩阵
Matrix InitMatrix(int row, int col)
{Matrix m;float** matrix;matrix = (float**)malloc(row * sizeof(float*));for (int i = 0; i < row; i++)matrix[i] = (float*)malloc(col * sizeof(float));for (int i = 0; i < row; i++){for (int j = 0; j < col; j++){matrix[i][j] = 0;}}m.col = col;m.row = row;m.matrix = matrix;return m;
}Matrix add(Matrix m1, Matrix m2)
{for (int i = 0; i < m1.row; i++){for (int j = 0; j < m1.col; j++){m1.matrix[i][j] = m1.matrix[i][j] + m2.matrix[i][j];}}return m1;
}Matrix sub(Matrix m1, Matrix m2)
{for (int i = 0; i < m1.row; i++){for (int j = 0; j < m1.col; j++){m1.matrix[i][j] = m1.matrix[i][j] - m2.matrix[i][j];}}return m1;
}float calRowCol(Matrix M1, Matrix M2, int row, int col)//row为M1的行 col为m2的列
{float result = 0;int same = M1.col;for (int j = 0; j < same; j++){result += M1.matrix[row][j] * M2.matrix[j][col];}return result;
}Matrix Mul(Matrix m1, Matrix m2)
{Matrix result = InitMatrix(m1.row, m2.col);for (int i = 0; i < m1.row; i++){for (int j = 0; j < m2.col; j++){result.matrix[i][j] = calRowCol(m1, m2, i, j);}}return result;
}Matrix numMul(Matrix m, int num)
{cout << "数值:" << num << endl;for (int i = 0; i < m.row; i++){for (int j = 0; j < m.col; j++){m.matrix[i][j] = m.matrix[i][j] * num;}}return m;
}void printMatrix(Matrix m)
{for (int i = 0; i < m.row; i++){for (int j = 0; j < m.col; j++){cout << m.matrix[i][j] << "  ";}cout << endl;}
}int main()
{int num = 0;do{cout << "*************************************\n";cout << "*              菜单                 *\n";cout << "*          1.矩阵相加               *\n";cout << "*          2.矩阵相减               *\n";cout << "*          3.矩阵相乘               *\n";cout << "*          4.矩阵数乘               *\n";cout << "*          5.矩阵的逆               *\n";cout << "*          6.退出                   *\n";cout << "*************************************\n";cin >> num;if (1 == num || 2 == num || 3 == num){cout << "请输入矩阵1" << endl;Matrix m1 = CreateMatrix();cout << "请输入矩阵2" << endl;Matrix m2 = CreateMatrix();cout << "两矩阵为" << endl;printMatrix(m1);cout << endl;printMatrix(m2);switch (num){case 1:{if (m1.col != m2.col || m1.row != m2.row){cout << "行列不同" << endl;}else {cout << "结果为：" << endl;printMatrix(add(m1, m2));}break;}case 2:{if (m1.col != m2.col || m1.row != m2.row){cout << "参数错误" << endl;}else {cout << "结果为：" << endl;printMatrix(sub(m1, m2));}break;}case 3:{if (m1.col != m2.row){cout << "参数错误" << endl;}else {cout << "结果为：" << endl;printMatrix(Mul(m1, m2));}break;}default:break;}}else if (4 == num){int number = 1;cout << "请输入矩阵" << endl;Matrix m = CreateMatrix();cout << "请输入数值" << endl;cin >> number;cout << "矩阵为：" << endl;printMatrix(m);cout << "数值为：" << endl;cout << number << endl;printMatrix(numMul(m, number));}else if (5 == num){cout << "请输入矩阵" << endl;Matrix m1 = CreateMatrix();cout << "输入矩阵为" << endl;printMatrix(m1);cout << endl;Matrix m2 = Create2DArrayDefault(m1.row);inverse(m1, m2);printMatrix(m2);}cout << "按回车继续....";getchar();getchar();system("cls");} while (1 == num || 2 == num || 3 == num || 4 == num || 5 == num);return 0;
}

总结

有很多细节可以优化，不过作为小工具够用:)

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C/C++矩阵计算器

问题描述

参考

实现

总结

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