矩阵操作(类,复数)

发布于 7 天前  19 次阅读


#include <iostream>
#include <stdio.h>
using namespace std;

class Complex
{
public:
    Complex() { a[1][1][0] = 0; n = 0; m = 0; }
    friend Complex operator +(Complex& c1, Complex& c2);
    friend Complex operator -(Complex& c1, Complex& c2);
    friend Complex operator *(Complex& c1, Complex& c2);
    friend Complex operator !(Complex& c1);
    friend ostream& operator <<(ostream&, Complex&);
    friend istream& operator >>(istream&, Complex&);
private:
    double a[10][10][2]; //定义矩阵,矩阵中a[i][j][0]表示复数的实部,a[i][j][1]表示复数的虚部
    int n, m; //标识矩阵的大小,由用户输入
};

Complex operator +(Complex& c1, Complex& c2) //矩阵加法的重载实现
{
    Complex c3;
    if (c1.n != c2.n || c1.m != c2.m) //判断是否能够进行矩阵加法
    {
        printf("这两个矩阵不能相加\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c3.a[i][j][0] = c1.a[i][j][0] + c2.a[i][j][0];
            c3.a[i][j][1] = c1.a[i][j][1] + c2.a[i][j][1];
        }
    return c3;
}
Complex operator -(Complex& c1, Complex& c2) //矩阵减法的重载实现
{
    Complex c3;
    if (c1.n != c2.n || c1.m != c2.m) //判断是否能够进行矩阵减法
    {
        printf("这两个矩阵不能相减\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c3.a[i][j][0] = c1.a[i][j][0] - c2.a[i][j][0];
            c3.a[i][j][1] = c1.a[i][j][1] - c2.a[i][j][1];
        }
    return c3;
}

Complex operator *(Complex& c1, Complex& c2) //矩阵乘法的实现
{
    Complex c3;
    if (c1.m != c2.n) //判断能否进行矩阵乘法
    {
        printf("这两个矩阵不能相乘\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c2.m;
    double s1 = 0, s2 = 0;
    for (int i = 1; i <= c1.n; i++)
    {        
        for (int m = 1; m <= c2.m; m++)
        {
            for (int j = 1; j <= c1.m; j++)
            {
                s1 += c1.a[i][j][0] * c2.a[j][m][0] - c1.a[i][j][1] * c2.a[j][m][1];
                s2 += c1.a[i][j][0] * c2.a[j][m][1] + c1.a[i][j][1] * c2.a[j][m][0];
            }
            c3.a[i][m][0] = s1;
            c3.a[i][m][1] = s2;
            s1 = 0;
            s2 = 0;
        }
    }
    return c3;
}
Complex operator !(Complex& c1) //矩阵转置的实现
{
    Complex c2;
    c2.m = c1.n;
    c2.n = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c2.a[j][i][0] = c1.a[i][j][0];
            c2.a[j][i][1] = -c1.a[i][j][1];
        }        
    return c2;
}

ostream & operator << (ostream & output, Complex &c) //矩阵输出
{
    for (int i = 1; i <= c.n; i++)
    {
        for (int j = 1; j <= c.m; j++)
            if (c.a[i][j][1] >= 0)
                output << c.a[i][j][0] << '+' << c.a[i][j][1] << "i ";
            else output << c.a[i][j][0] << c.a[i][j][1] << "i ";
        output << endl;
    }
    return output;
}
istream & operator >> (istream & input, Complex &c) //矩阵输入
{
    printf("请输入矩阵的大小:\n"); 
    input >> c.n >> c.m; //先输入该矩阵的大小
    printf("请输入矩阵:\n");
    for (int i = 1; i <= c.n; i++)
    {
        for (int j = 1; j <= c.m; j++)
            input >> c.a[i][j][0] >> c.a[i][j][1];
    }
    return input;
}


int main()
{
    Complex c1, c2;
    cin >> c1;
    cin >> c2;
    printf("矩阵1为:\n");
    cout << c1;
    printf("矩阵2为:\n");
    cout << c2;
    Complex c3;
    printf("矩阵相加结果为:\n");
    c3 = c1 + c2;
    cout << c3;
    printf("矩阵相减结果为:\n");
    c3 = c1 - c2;
    cout << c3;
    printf("矩阵相乘结果为:\n");
    c3 = c1 * c2;
    cout << c3;
    printf("矩阵1的共轭转置矩阵为:\n");
    c3 = !c1;
    cout << c3;
    printf("矩阵2的共轭转置矩阵为:\n");
    c3 = !c2;
    cout << c3;
}

实数域

#include <iostream>
#include <stdio.h>
using namespace std;

class Complex
{
public:
    Complex() { a[1][1][0] = 0; n = 0; m = 0; }
    friend Complex operator +(Complex& c1, Complex& c2);
    friend Complex operator -(Complex& c1, Complex& c2);
    friend Complex operator *(Complex& c1, Complex& c2);
    friend Complex operator !(Complex& c1);
    friend ostream& operator <<(ostream&, Complex&);
    friend istream& operator >>(istream&, Complex&);
private:
    double a[10][10][2]; //定义矩阵,矩阵中a[i][j][0]表示复数的实部,a[i][j][1]表示复数的虚部
    int n, m; //标识矩阵的大小,由用户输入
};

Complex operator +(Complex& c1, Complex& c2) //矩阵加法的重载实现
{
    Complex c3;
    if (c1.n != c2.n || c1.m != c2.m) //判断是否能够进行矩阵加法
    {
        printf("这两个矩阵不能相加\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c3.a[i][j][0] = c1.a[i][j][0] + c2.a[i][j][0];
           // c3.a[i][j][1] = c1.a[i][j][1] + c2.a[i][j][1];
        }
    return c3;
}
Complex operator -(Complex& c1, Complex& c2) //矩阵减法的重载实现
{
    Complex c3;
    if (c1.n != c2.n || c1.m != c2.m) //判断是否能够进行矩阵减法
    {
        printf("这两个矩阵不能相减\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c3.a[i][j][0] = c1.a[i][j][0] - c2.a[i][j][0];
         //   c3.a[i][j][1] = c1.a[i][j][1] - c2.a[i][j][1];
        }
    return c3;
}

Complex operator *(Complex& c1, Complex& c2) //矩阵乘法的实现
{
    Complex c3;
    if (c1.m != c2.n) //判断能否进行矩阵乘法
    {
        printf("这两个矩阵不能相乘\n");
        return c3;
    }
    c3.n = c1.n;
    c3.m = c2.m;
    double s1 = 0, s2 = 0;
    for (int i = 1; i <= c1.n; i++)
    {        
        for (int m = 1; m <= c2.m; m++)
        {
            for (int j = 1; j <= c1.m; j++)
            {
                s1 += c1.a[i][j][0] * c2.a[j][m][0];
               // s2 += c1.a[i][j][0] * c2.a[j][m][1] + c1.a[i][j][1] * c2.a[j][m][0];
            }
            c3.a[i][m][0] = s1;
          //  c3.a[i][m][1] = s2;
            s1 = 0;
          //  s2 = 0;
        }
    }
    return c3;
}
Complex operator !(Complex& c1) //矩阵转置的实现
{
    Complex c2;
    c2.m = c1.n;
    c2.n = c1.m;
    for (int i = 1; i <= c1.n; i++)
        for (int j = 1; j <= c1.m; j++)
        {
            c2.a[j][i][0] = c1.a[i][j][0];
          //  c2.a[j][i][1] = -c1.a[i][j][1];
        }        
    return c2;
}

ostream & operator << (ostream & output, Complex &c) //矩阵输出
{
    for (int i = 1; i <= c.n; i++)
    {
        for (int j = 1; j <= c.m; j++)
            if (c.a[i][j][1] >= 0)
                output << c.a[i][j][0] << ' ';
            else output << c.a[i][j][0] << ' ';
        output << endl;
    }
    return output;
}
istream & operator >> (istream & input, Complex &c) //矩阵输入
{
    printf("请输入矩阵的大小:\n"); 
    input >> c.n >> c.m; //先输入该矩阵的大小
    printf("请输入矩阵:\n");
    for (int i = 1; i <= c.n; i++)
    {
        for (int j = 1; j <= c.m; j++)
            input >> c.a[i][j][0];
    }
    return input;
}


int main()
{
    Complex c1, c2;
    cin >> c1;
    cin >> c2;
    printf("矩阵1为:\n");
    cout << c1;
    printf("矩阵2为:\n");
    cout << c2;
    Complex c3;
    printf("矩阵相加结果为:\n");
    c3 = c1 + c2;
    cout << c3;
    printf("矩阵相减结果为:\n");
    c3 = c1 - c2;
    cout << c3;
    printf("矩阵相乘结果为:\n");
    c3 = c1 * c2;
    cout << c3;
    printf("矩阵1的转置矩阵为:\n");
    c3 = !c1;
    cout << c3;
    printf("矩阵2的转置矩阵为:\n");
    c3 = !c2;
    cout << c3;
}

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