shj_11.cpp

来自「此软件可以用元函数实现复数的加减乘除」· C++ 代码 · 共 51 行

#include <iostream.h>
class Complicate
{
public:
	Complicate(double shibu,double xubu){x=shibu;y=xubu;}
	void Print();
	friend void Plus(Complicate &a,Complicate &b);
	friend void Substract(Complicate &a,Complicate &b);
	friend void Multiply(Complicate &a,Complicate &b);
	friend void Division(Complicate &a,Complicate &b);
private:
	double x,y;
};
void Complicate::Print()
{
	cout<<"("<<x<<")"<<"+"<<"i"<<"("<<y<<")"<<endl;
}
void Plus(Complicate &a,Complicate &b)
{
	double newx=a.x+b.x;
	double newy=a.y+b.y;
	cout<<"Plus="<<"("<<newx<<")"<<"+"<<"i"<<"("<<newy<<")"<<endl;
}
void Substract(Complicate &a,Complicate &b)
{
	double newx=a.x-b.x;
	double newy=a.y-b.y;
	cout<<"Substract="<<"("<<newx<<")"<<"+"<<"i"<<"("<<newy<<")"<<endl;
}
void Multiply(Complicate &a,Complicate &b)
{
	double newx=a.x*b.x-a.y*b.y;
	double newy=a.x*b.y+b.x*a.y;
	cout<<"Multiply="<<"("<<newx<<")"<<"+"<<"i"<<"("<<newy<<")"<<endl;
}
void Division(Complicate &a,Complicate &b)
{
	double newx=(a.x*b.x+a.y*b.y)/(b.x*b.x+b.y*b.y);
	double newy=(b.x*a.y-a.x*b.y)/(b.x*b.x+b.y*b.y);
	cout<<"Division="<<"("<<newx<<")"<<"+"<<"i"<<"("<<newy<<")"<<endl;
}
void main()
{
	Complicate c1(5.0,-5.0),c2(-3.0,4.0);
	c1.Print();
	c2.Print();
	Plus(c1,c2);
	Substract(c1,c2);
	Multiply(c1,c2);
	Division(c1,c2);
}