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  • JAVA_实现两个复数的四则运算

    1、一个复数类Complex由两部分组成:实部real和虚部imaginary,两个复数可以进行加减乘除四则运算。
      试设计这个复数类,然后在另一个类的程序入口演示。

    public class Test {
    public static void main(String[] args) {
    // 定义两个对象并赋值:c1(a+bi)=(2+4i),c2=(c+di)=(6+16i)
    Complex c1=new Complex();
    c1.real=2;
    c1.imaginary=4;
    Complex c2=new Complex();
    c2.real=6;
    c2.imaginary=16;

    // 以字符串的方式输出:采用标准格式输出,即a+bi
    System.out.println("两个复数相加得:"+c1.add(c2));

    // 在每次调用并运算(此时real值发生变化)之后,再重新赋值:
    c1.real=2;
    c1.imaginary=4;

    System.out.println("两个复数相减得:"+c1.sub(c2));
    c1.real=2;
    c1.imaginary=4;

    System.out.println("两个复数相乘得:"+c1.mul(c2));
    c1.real=2;
    c1.imaginary=4;

    System.out.println("两个复数相除得:"+c1.div(c2));
    }
    }
    class Complex
    {
    double real;
    double imaginary;
    //最后以字符串输出的格式:
    //1.虚部为正值时,实部+"+"+虚部+"i" 2.虚部为0时,只有实部,即实部+"" 3.虚部为负值时,实部+""+虚部+"i"

    // 加法,减法:直接实部加减实部,虚部加减虚部
    String add(Complex c)
    {
    if(imaginary+c.imaginary>0)
    return (real=real+c.real)+"+"+(imaginary=imaginary+c.imaginary)+"i";
    else if(imaginary+c.imaginary==0)
    return (real=real+c.real)+"";
    else
    return (real=real+c.real)+""+(imaginary=imaginary+c.imaginary)+"i";
    }
    String sub(Complex c)
    {
    if(imaginary-c.imaginary>0)
    return (real=real-c.real)+"+"+(imaginary=imaginary-c.imaginary)+"i";
    else if(imaginary-c.imaginary==0)
    return (real=real-c.real)+"";
    else
    return (real=real-c.real)+""+(imaginary=imaginary-c.imaginary)+"i";
    }
    // 乘法的公式:(a+bi)*(c+di) = a*c + bi*c + a*di + bi*di = a*c + (b*d)(i^2) + (b*c+a*d)i

    // 注:此处的i^2直接赋值-1,即 return a*c-b*d+(b*c+a*d)i
    String mul(Complex c)
    {
    if(imaginary*c.real+real*c.imaginary>0)
    return (real*c.real)-(imaginary*c.imaginary)+"+"+(imaginary*c.real+real*c.imaginary)+"i";
    else if(imaginary*c.real+real*c.imaginary==0)
    return (real*c.real)-(imaginary*c.imaginary)+"";
    else
    return (real*c.real)-(imaginary*c.imaginary)+""+(imaginary*c.real+real*c.imaginary)+"i";
    }
    // 除法的公式:(a+bi)/(c+di) = (a+bi)*(c-di) / c^2+d^2

    // 注:在乘法的基础上得到格式A+Bdi,再进行A /c^2+d^2,B /c^2+d^2 + "i"
    String div(Complex c)
    {
    if(imaginary*c.real-real*c.imaginary>0)
    return ((real*c.real)+(imaginary*c.imaginary))/(c.real*c.real+c.imaginary*c.imaginary)+"+"+(imaginary*c.real-real*c.imaginary)/(c.real*c.real+c.imaginary*c.imaginary)+"i";
    else if(imaginary*c.real-real*c.imaginary==0)
    return ((real*c.real)+(imaginary*c.imaginary))/(c.real*c.real+c.imaginary*c.imaginary)+"";
    else
    return ((real*c.real)+(imaginary*c.imaginary))/(c.real*c.real+c.imaginary*c.imaginary)+""+(imaginary*c.real-real*c.imaginary)/(c.real*c.real+c.imaginary*c.imaginary)+"i";

    }
    }
    /*Complex c1=new Complex();
    c1.real=2;
    c1.imaginary=4;

    Complex c2=new Complex();
    c2.real=6;
    c2.imaginary=16;
    即:c1=2+4i
    c2=6+16i
    数学运算结果:

    两者相加:8+20i
    两者相减:-4-12i

    两者相乘:-52+56i
    (2+4i)*(6+16i)

    两者相除:0.2602739726027397-0.0273972602739726i
    (2+4i)/(6+16i)=(2+4i)*(6-16i)/(6*6+16*16)*/

    weiyulan-1998
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  • 原文地址:https://www.cnblogs.com/weiyulan-1998/p/8861088.html
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