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java - 打印出一个 ASCII 圆和轴

转载 作者:行者123 更新时间:2023-12-04 17:53:25 25 4
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此程序在笛卡尔平面中打印一个圆。

输入是:半径、圆心坐标 (cx,cy) 以及我们要打印圆的字符。

如果圆上的点与轴重叠,则点优先。我在方法 drawCircle 中写了打印轴的条件,但图像失真了...

有些东西在躲避我...有人可以帮我找出我的错误吗?

这是我的整个程序(有问题的方法是最后一个,drawCircle):

public class Circle {
public static void main(String[] args) {
System.out.println(onCircle(1, 2, 3, 4, 5));

drawCircle(1, 3, 3, '*');
drawCircle(3, 3, 3, '*');
drawCircle(5, 10, 12, '*');
}

//Question 1A
public static boolean onCircle(int radius, int cx, int cy, int x, int y) {
//default answer is false, but if the
//inequality holds then it is set to true
boolean isDrawn = false;
if (Math.pow(radius,2)<=(Math.pow((x-cx),2)+Math.pow((y-cy),2))
&& (Math.pow((x-cx),2)+Math.pow((y-cy),2))<=(Math.pow(radius,2)+1)) {
isDrawn = true;
}
return isDrawn;
}

//Question 1B
public static void verifyInput(int radius, int cx, int cy) {
//if radius is negative, display error message
if (radius <= 0) {
throw new IllegalArgumentException(
"The radius of the circle must be a positive number.");
}
//if the center of the circle with radius 'radius' causes the circle
//to 'overflow' into other quadrants, display error message
if ((cx - radius) < 0 || (cy - radius) < 0) {
throw new IllegalArgumentException(
"the circle with requested parameters does not fit " +
"in the quadrant. Consider moving the center of the " +
"circle further from the axes.");
}
}

//Question 1C
public static void drawCircle(int radius, int cx, int cy, char symbol) {
verifyInput(radius, cx, cy);
//set the values for extension of the axes (aka how long are they)
int xMax = cx + radius + 1;
int yMax = cy + radius + 1;
for (int j = yMax; j >= 0; j--) {
for (int i = 0; i <= xMax; i++) {
//set of if-block to print the axes
if (i == 0 && j == 0) {
System.out.print('+');
} else if (i == 0) {
if (j == yMax) {
System.out.print('^');
}
if (j != yMax && onCircle(radius, cx, cy, i, j) == false) {
System.out.print('|');
}
} else if (j == 0) {
if (i == xMax) {
System.out.print('>');
}
if (i != xMax && onCircle(radius, cx, cy, i, j) == false) {
System.out.print('-');
}
}

//if block to print the circle
//verify for each coordinate (i,j)
//in the quadrant if they are on circle
//if =true print symbol, if =false print empty character
if (onCircle(radius, cx, cy, i, j) == true) {
System.out.print(symbol);
} else {
System.out.print(' ');
}
}
System.out.println();
}
}
}

这是我得到的:

false
^
| ***
| * *
| ***
|
+ - - - - >
^
| ***
|
* *
* *
* *
|
+ - ***- - >
^
| ***
| * *
| * *
|
| * *
| * *
| * *
|
| * *
| * *
| ***
|
|
|
|
|
|
+ - - - - - - - - - - - - - - - >

如您所见,第 1 个和第 3 个圆很好,但与轴重叠的圆变形了。

最佳答案

以原点为中心的一般圆方程:

Equation of a circle

Java 中可以这样实现:

i*i + j*j == r*r

但是在整数坐标系的情况下,您必须以某种方式四舍五入这个方程,以便圆的所有点都反射(reflect)在这个坐标系中:

(int) Math.sqrt(i*i + j*j) == r

如果 r=8,则圆和轴如下所示:

r=8
* * * * * * * * *
* * * * *
* * *
* * *
* * * * *
* * *
* * *
* * *
* * * * * * * * * * * * * * * * *
* * *
* * *
* * *
* * * * *
* * *
* * *
* * * * *
* * * * * * * * *

Try it online!

int r = 8;
System.out.println("r=" + r);
IntStream.rangeClosed(-r, r)
.peek(i -> IntStream.rangeClosed(-r, r)
.mapToObj(j -> i == 0 || j == 0 ||
(int) Math.sqrt(i*i + j*j) == r ?
"* " : " ")
.forEach(System.out::print))
.forEach(i -> System.out.println());

另请参阅:Print out an ASCII star in java

关于java - 打印出一个 ASCII 圆和轴,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42447764/

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