定义并实现一个复数类Complex，适当添加函数，使得下面的代码能够工作。

Question

定义并实现一个复数类Complex，适当添加函数，使得下面的代码能够工作。
int main()
{
complex c1(3.4,5.0);
complex c2=c1;
complex c3;
c1.add(5.0); //复数c1的实部增加5.0
c2.sub(5.0); //复数c2的实部减少5.0
c3=c1.add(c2);// 复数c1和c2相加求和
c3.display(); // 显示输出复数
c3=c1.sub(c2); // 复数c1和c2相减求差
c3.display();
c3=add(c1,c2); // 复数c1和c2相加求和
c3.display();
c3=10; //复数c3的实部为10.0，虚部为0.0
c3.display();
return 0;
}

Answer 1

有两个参数的构造函数，拷贝构造函数，无参构造函数，加减函数及显示函数

#include <stdio.h>
class complex
{
    double a,b;
public:
    complex() {}
    complex(double a,double b) {this->a = a;this->b = b;}
    complex(const complex &x) {a = x.a;b = x.b;}
    complex(double a) {this->a = a;b = 0;}
    void add(double k) {a+=k;}
    void sub(double k) {a-=k;}
    complex add(const complex &x) {a+=x.a;b+=x.b;return *this;}
    void display() 
    {
        if(a == 0)
        {
            cout<<b<<"i"<<endl;
        }
        else if(b==0)
        {
            cout<<a<<endl;
        }
        else if(b<0)
        {
            cout<<a<<b<<"i"<<endl;
        }
        else
            cout<<a<<"+"<<b<<"i"<<endl;
    }
    complex sub(const complex &x) {a-=x.a;b-=x.b;return *this;}
    friend complex add(const complex &x1,const complex &x2) ;
};

complex add(const complex &x1,const complex &x2) {complex x;x.a = x1.a+x2.a;x.b = x1.b+x2.b;return x;}

int main()
{
    complex c1(3.4,5.0);
    complex c2=c1;
    complex c3;
    c1.add(5.0); //复数c1的实部增加5.0
    c2.sub(5.0); //复数c2的实部减少5.0
    c3=c1.add(c2);// 复数c1和c2相加求和
    c3.display(); // 显示输出复数
    c3=c1.sub(c2); // 复数c1和c2相减求差
    c3.display();
    c3=add(c1,c2); // 复数c1和c2相加求和
    c3.display();
    c3=10; //复数c3的实部为10.0，虚部为0.0
    c3.display();
    return 0;
}