C#编程：定义一个复数类,实现复数的简单加法运算,并能显示结果。

Posted 2023-04-14

求高手解答

参考技术A class fs

public double a;//实部
public double b;//虚部
//加法
public fs add(fs fs1)

fs fs_temp = new fs();
fs_temp.a = this.a + fs1.a;
fs_temp.b = this.b + fs1.b;
return fs_temp;

//减法
public fs jf(fs fs1)

fs fs_temp = new fs();
fs_temp.a = this.a - fs1.a;
fs_temp.b = this.b - fs1.b;
return fs_temp;

public string ToString()

if(b==0)
return "" + a;

else if (a == 0)

return b + "i";

else

return a + "+" + b + "i";

本回答被提问者采纳

用java定义一个复数类Complex，能够创建复数对象，并且实现复数之间的加、减运算

复数具有实部和虚部两部分,如2+5i就是一个复数,其中2为实部,5i为虚部;两复数相加时, 实部与实部相加, 虚部与虚部相加。两复数相减时，实部与实部相减, 虚部与虚部相减。
如(2+5i)+(3-4i)=5+1i
(2+5i)-(3-4i)=-1+9i
现在要求定义一个复数类Complex，能够创建复数对象，并且实现复数之间的加、减运算，在main()函数中测试该类。
我需要全部代码，急用，谢谢了~

public class ComplexDemo
// main方法
public static void main(String[] a)
Complex b = new Complex(2, 5);
Complex c = new Complex(3, -4);
System.out.println(b + "+" + c + "=" + b.add(c));
System.out.println(b + "-" + c + "=" + b.minus(c));
System.out.println(b + "*" + c + "=" + b.multiply(c));
System.out.println(b + "/" + c + "=" + b.divide(c));



// Complex类
class Complex
private double m;// 实部
private double n;// 虚部

public Complex(double m, double n)
this.m = m;
this.n = n;


// add
public Complex add(Complex c)
return new Complex(m + c.m, n + c.n);


// minus
public Complex minus(Complex c)
return new Complex(m - c.m, n - c.n);


// multiply
public Complex multiply(Complex c)
return new Complex(m * c.m - n * c.n, m * c.n + n * c.m);


// divide
public Complex divide(Complex c)
double d = Math.sqrt(c.m * c.m) + Math.sqrt(c.n * c.n);
return new Complex((m * c.m + n * c.n) / d, Math.round((m * c.n - n * c.m) / d));


public String toString()
String rtr_str = "";
if (n > 0)
rtr_str = "(" + m + "+" + n + "i" + ")";
if (n == 0)
rtr_str = "(" + m + ")";
if (n < 0)
rtr_str = "(" + m + n + "i" + ")";
return rtr_str;

参考技术A 定义2个成员变量，一个放实数，一个放虚数，再定义一个caculate方法，传进一个complex参数，计算后返回一个complex结果对象。
package test;

public class Complex
private int shishu;

private int xushu;

public Complex(int shishu, int xushu)
this.shishu = shishu;
this.xushu = xushu;


public Complex add(Complex complex)
return new Complex(this.shishu + complex.getShishu(), this.xushu + complex.getXushu() );


public Complex cut(Complex complex)
return new Complex(this.shishu - complex.getShishu(), this.xushu - complex.getXushu() );


public String toString()
if (this.xushu == 0)
return "" + this.shishu;
else if (this.xushu > 0)
return this.shishu + "+" + this.xushu + "i";
else
return this.shishu + "" + this.xushu + "i";



/**
* @param args
*/
public static void main(String[] args)
Complex a = new Complex(5,2);
Complex b = new Complex(4,3);
System.out.println(a);
System.out.println(b);
System.out.println(a.add(b));
System.out.println(a.cut(b));



public int getShishu()
return shishu;


public int getXushu()
return xushu;



运行结果：
5+2i
4+3i
9+5i
1-1i 参考技术B 用着这么麻烦吗？杀鸡你非得用拖拉机给它压死，至于吗