2015年大一下第6周项目1

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2015年大一下第6周项目1

/* 
*Copyright (c) 2014,烟台大学计算机学院 
*All rights reserved. 
*文件名称:Annpion.cpp 
*作者:王耀鹏 
*完成日期:2015年4月25日 
*版本号:v1.0 
* 
*问题描述:定义复数类重载运算符+、-、*、/，使之能用于复数的加减乘除。扩展+、-、*、/运算符的功能，使之能与double型数据进行运算。
*输入描述:无。 
*输出描述:复数之间的加减乘除，复数与小数之间的加减乘除。 
*/  
#include <iostream>
using namespace std;
class Complex
{
public:Complex(){real=0;imag=0;}Complex(double r,double i){real=r;imag=i;}friend Complex operator+(Complex &c1, Complex &c2);friend Complex operator+(double d1, Complex &c2);friend Complex operator+(Complex &c1, double d2);friend Complex operator-(Complex &c1, Complex &c2);friend Complex operator-(double d1, Complex &c2);friend Complex operator-(Complex &c1, double d2);friend Complex operator*(Complex &c1, Complex &c2);friend Complex operator*(double d1, Complex &c2);friend Complex operator*(Complex &c1, double d2);friend Complex operator/(Complex &c1, Complex &c2);friend Complex operator/(double d1, Complex &c2);friend Complex operator/(Complex &c1, double d2);void display();
private:double real;double imag;
};//复数相加：(a+bi)+(c+di)=(a+c)+(b+d)i.
Complex operator+(Complex &c1, Complex &c2)
{Complex c;c.real=c1.real+c2.real;c.imag=c1.imag+c2.imag;return c;
}
Complex operator+(double d1, Complex &c2)
{Complex c(d1,0);return c+c2; //按运算法则计算的确可以，但充分利用已经定义好的代码，既省人力，也避免引入新的错误，但可能机器的效率会不佳
}
Complex operator+(Complex &c1, double d2)
{Complex c(d2,0);return c1+c;
}
//复数相减：(a+bi)-(c+di)=(a-c)+(b-d)i.
Complex operator-(Complex &c1, Complex &c2)
{Complex c;c.real=c1.real-c2.real;c.imag=c1.imag-c2.imag;return c;
}
Complex operator-(double d1, Complex &c2)
{Complex c(d1,0);return c-c2;
}
Complex operator-(Complex &c1, double d2)
{Complex c(d2,0);return c1-c;
}//复数相乘：(a+bi)(c+di)=(ac－bd)+(bc+ad)i.
Complex operator*(Complex &c1, Complex &c2)
{Complex c;c.real=c1.real*c2.real-c1.imag*c2.imag;c.imag=c1.imag*c2.real+c1.real*c2.imag;return c;
}
Complex operator*(double d1, Complex &c2)
{Complex c(d1,0);return c*c2;
}
Complex operator*(Complex &c1, double d2)
{Complex c(d2,0);return c1*c;
}//复数相除：(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +(bc-ad)/(c^2+d^2)i
Complex operator/(Complex &c1, Complex &c2)
{Complex c;c.real=(c1.real*c2.real+c1.imag*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);c.imag=(c1.imag*c2.real-c1.real*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);return c;
}
Complex operator/(double d1, Complex &c2)
{Complex c(d1,0);return c/c2;
}
Complex operator/(Complex &c1, double d2)
{Complex c(d2,0);return c1/c;
}void Complex::display()
{cout<<"("<<real<<","<<imag<<"i)"<<endl;
}int main()
{Complex c1(3,4),c2(5,-10),c3;double d=11;cout<<"c1=";c1.display();cout<<"c2=";c2.display();cout<<"d="<<d<<endl<<endl;cout<<"下面是重载运算符的计算结果: "<<endl;c3=c1+c2;cout<<"c1+c2=";c3.display();cout<<"c1+d=";(c1+d).display();cout<<"d+c1=";(d+c1).display();c3=c1-c2;cout<<"c1-c2=";c3.display();cout<<"c1-d=";(c1-d).display();cout<<"d-c1=";(d-c1).display();c3=c1*c2;cout<<"c1*c2=";c3.display();cout<<"c1*d=";(c1*d).display();cout<<"d*c1=";(d*c1).display();c3=c1/c2;cout<<"c1/c2=";c3.display();cout<<"c1/d=";(c1/d).display();cout<<"d/c1=";(d/c1).display();return 0;
}

运行结果：