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java - 使用抽象形状类的三角形类

转载 作者:行者123 更新时间:2023-12-01 14:38:34 25 4
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所以我需要使用抽象类创建这个三角形类。它也将由测试人员类绘制。我正在完成它的一部分,但我在数学部分遇到了严重的问题。我已经在测试器类中设置了坐标,我不知道如何让笔转动一定角度来绘制三角形的下一条边。附件是我迄今为止所学的所有类(class)。任何帮助将不胜感激。

测试员类

    import TurtleGraphics.*;

public class TestShapes1 {

public static void main (String[] args) {

// Declare and instantiate a pen, a circle and a wheel
Pen p = new StandardPen();
//Shape s1 = new Circle1 (20, 20, 20);
//Shape s2 = new Wheel1 (-20, -20, 20, 6);
Shape1 t2 = new Triangle1 (0, 0, 50, 0, 0, 30);

// Draw the circle and wheel
//s1.draw (p);
t2.draw (p);
}

}

形状类

    import TurtleGraphics.Pen;

public interface Shape1 {
public double area();
public void draw (Pen p);
public double getXPos();
public double getYPos();
public void move (double xLoc, double yLoc);
public void stretchBy (double factor);
public String toString();
}

三角形类

    import TurtleGraphics.Pen;

public class Triangle1 implements Shape1 {


private double x1, y1, x2, y2, x3, y3;
private double s1, s2, s3;
private double d1, d2;
//private double height, width;

public Triangle1() {
x1 = 0;
y1 = 0;
x2 = 1;
y2 = 0;
x3 = 0;
y3 = 1;
//height = 1;
//width = 1;
}

public Triangle1 (double xLoc1, double yLoc1, double xLoc2, double yLoc2, double xLoc3, double yLoc3) {
x1 = xLoc1;
y1 = yLoc1;
x2 = xLoc2;
y2 = yLoc2;
x3 = xLoc3;
y3 = yLoc3;
//height = h;
//width = w;
}

public double area() {
return (Math.abs(x1*y2-x2*y1+x2*y3-x3*y2+x3*y1-x1*y3))/2.0;
}

public void draw (Pen p) {
s1 = Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
s2 = Math.sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
s3 = Math.sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
p.up();
p.move (x1, y1);
p.down();
p.setDirection (0);
p.move (s1);
d1 = (Math.acos((s2*s2+s3*s3-s1*s1)/(2.0*s2*s3)))*180/Math.PI;
p.turn (180 - d1);
p.move (s2);
d2 = (Math.acos((s3*s3+s1*s1-s2*s2)/(2.0*s3*s1)))*180/Math.PI;
p.turn (180 - d2);
p.move (s3);
p.turn (-90);
//p.move ();
}

public double getXPos() {
return x1;
}

public double getYPos() {
return y1;
}

public void move (double xLoc, double yLoc) {
x1 = x1 + xLoc;
y1 = y1 + yLoc;
x2 = x2 + xLoc;
y2 = y2 + yLoc;
x3 = x3 + xLoc;
y3 = y3 + yLoc;
}

public void stretchBy (double factor) {
x1 *= factor;
y1 *= factor;
}

public String toString() {
String str = "TRIANGLE\n";
// + "Width & Height: " + width + " & " + height +"\n"
// + "(X,Y) Position: (" + xPos + "," + yPos + ")\n"
// + "Area: " + area();
return str;
}
}

最佳答案

你不需要任何数学。只需将度数传递给 p.turn() 即可。所以使用

p.turn(180); 

而不是

d1 = (Math.acos((s2*s2+s3*s3-s1*s1)/(2.0*s2*s3)))*180/Math.PI;
p.turn (180 - d1);

请参阅documentation供引用:

The degrees can be an integer or floating-point number. Example: pen.turn(-45); Rotate the pen 45 degrees clockwise.

关于java - 使用抽象形状类的三角形类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16229841/

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