老师让用VB写个坐标反算方位角的小程序，然后我写了，被同学拿去转抄了，我就不想用写的那个了，正好自己在学java就写来试试，代码很烂，新手，不过可以给跟我一样的新手参考下嘿嘿。顺便今天在github建了个项目，以后写的小程序会在github更新代码。项目地址：github https://github.com/shuiqingliu/survey

代码实现：

import java.util.Scanner; //导入Scanner类
class Test{ //定义计算类
public double x1; //定义坐标变量
public double y1;
public double x2;
public double y2;
public double r;
public double s;
public double d;
public int e;
public double a;
public int b;
public double c;
public int h;
public void calculate(){ //定义计算方法
if(x2 - x1 > 0 && y2-y1 >0){ //判断x,y的坐标增量
getInfo();
s = r;
change(s);
}else if(x2-x1<0 && y2-y1>0){
getInfo();
s = r + 180;
change(s);

}else if(x2-x1 <0 && y2-y1 <0){
getInfo();
s = r+ 180;
change(s);
}else if (x2-x1 >0 && y2-y1 <0){
getInfo();
s = r + 360;
change(s);
}
if(x2-x1 >0 && y2-y1 ==0){ //判断是否位于坐标轴上
s = 0;
change(s);
}else if (x2-x1 == 0 && y2-y1 >0){
s = 90;
change(s);

}else if (x2-x1 <0 && y2-y1 ==0){
s = 180;
change(s);

}else if (x2-x1 == 0 && y2-y1 <0){
s =270;
change(s);

}
System.out.println(“坐标方位角是：” + e + “°” + b + “‘“ + h + “‘’”);
}
public double getInfo(){ //定义计算R的方法
r = Math.atan((y2-y1)/(x2-x1)) 206265/3600;
return this.r;
}
public void change(double d){ //将度数转换为度分秒
e =(int)Math.floor(d);
a = (d - e) 60;
b = (int)Math.floor(a);
c = (a - b) * 60;
h = (int)Math.floor(c);
}
}
public class Postion{
public static void main(String[] args){
Test t1 = new Test(); //实例化Test
Scanner sc = new Scanner(System.in); //调用Scanner类
System.out.print(“请输入坐标A 的X坐标x1:”);
t1.x1 = sc.nextDouble();
System.out.print(“请输入坐标A 的Y坐标y1:”);
t1.y1 = sc.nextDouble();
System.out.print(“请输入坐标B 的x坐标x2:”);
t1.x2 = sc.nextDouble();
System.out.print(“请输入坐标B 的Y坐标y2:”);
t1.y2 = sc.nextDouble();
t1.calculate(); //调用Test 类中的calculate方法
}
}