c++复数的四则运算

Question

实现两个复数的四则运算。

r1为第一个操作数的实部，i1为第一个操作数的虚部，op为操作符，r2为第二个操作数的实部，i2为第二个操作数的虚部。

输出两个复数的运算结果。

如果op为'+' '-' '*' '/'以外的字符，输出："error" 。op为'/'，且r2,i2均为0时，输出："Divisor cannot be 0"。(不带引号)
不会啊，难啊

Answer 1

我之前学C++基础的时候写过差不多的，基本考察的就是运算符的重载（在类中），我把代码贴上来，你看着改。

#include<iostream>
#include<iomanip>
#include<string>

using std::cin;
using std::cout;
using std::fixed;
using std::endl;
using std::setprecision;
using std::string;

class plural
{
public:
    plural(double a = 0.0, double b = 0.0);
    void show();
    void operator_show(double x, double y);
    void operator +(plural m);
    void operator -(plural m);
    void operator *(plural m);
    void operator /(plural m);

private:
    double a{ 0.0 };
    double b{ 0.0 };
};


plural::plural(double a, double b)
{
    this->a = a;
    this->b = b;
}

void plural::show()
{
    if (b >= 0)
        cout << fixed << setprecision(1) << '(' << a << '+' << b << "i)";
    else
        cout << fixed << setprecision(1) << '(' << a << b << "i)";
}

void plural::operator_show(double x, double y)
{
    if ((-0.1 < x&&x < 0.1) && !(-0.1 < y&&y < 0.1))
    {
        cout << fixed << setprecision(1) << y << 'i';
    }
    else if (!(-0.1 < x&&x < 0.1) && (-0.1 < y&&y < 0.1))
    {
        cout << fixed << setprecision(1) << x;
    }
    else if ((-0.1 < x&&x < 0.1) && (-0.1 < y&&y < 0.1))
    {
        cout << 0.0;
    }
    else if (y >= 0.1)
        cout << fixed << setprecision(1) << x << '+' << y << 'i';
    else
        cout << fixed << setprecision(1) << x << y << 'i';
}

void plural::operator +(plural m)
{
    double x = a + m.a;
    double y = b + m.b;
    this->show();
    cout << " + ";
    m.show();
    cout << " = ";
    operator_show(x, y);

}

void plural::operator -(plural m)
{
    double x = a - m.a;
    double y = b - m.b;
    this->show();
    cout << " - ";
    m.show();
    cout << " = ";
    operator_show(x, y);
}

void plural::operator *(plural m)
{
    double x = a * m.a - b * m.b;
    double y = a * m.b + b * m.a;
    this->show();
    cout << " * ";
    m.show();
    cout << " = ";
    operator_show(x, y);
}

void plural::operator /(plural m)
{
    double x = (a * m.a + b * m.b) / (m.a*m.a + m.b*m.b);
    double y = (b * m.a - a * m.b) / (m.a*m.a + m.b*m.b);
    this->show();
    cout << " / ";
    m.show();
    cout << " = ";
    operator_show(x, y);
}

int main()
{
    double a1, b1, a2, b2;
    cin >> a1 >> b1 >> a2 >> b2;
    plural p1{ a1,b1 };
    plural p2{ a2,b2 };
    p1 + p2;
    cout << endl;
    p1 - p2;
    cout << endl;
    p1*p2;
    cout << endl;
    p1 / p2;

}