This website requires Javascript to function properly. Please go to the setting of your web browser and enable Javascript for this website.

Ad by

技多不压身，工到自然成：安省技工证书特训班，点击咨询报名！

Ad by

技多不压身，工到自然成：安省技工证书特训班，点击咨询报名！

not hard though

wangqingshui(忘情水)

1. write an integer function sqrt() (round to closest integer)
You do not need a accurate sqrt function. to calculate y = sqrt(r * r - x * x), you can loop y from r - x to r and compare y * y and r * r - x * x (of course it's better to save r * r - x * x into a variable) and get the closest y.

2.
for (x = 0; x <= r; ++x)
{
　　y = sqrt(r * r - x * x);
　　draw points (x, y), (x, -y), (-x, y), (-x, -y)
}

3. there might be some disconnected points when x is close to r. to improve this case, you can save the previous y to y1. if (y1 - y > 1), loop from y1 to y and for each step calculate x1 and draw the 4 points.

(#3617356@0)
Last Updated: 2007-4-16
This post has been archived. It cannot be replied.

Report

Replies, comments and Discussions:

工作学习 / 专业技术讨论 / 一道常见的编程面试题求解，多谢 -xm1223(andy); 2007-4-15 {144} (#3617306@0)
画圆：
Write a routine to draw a circle given a center coordiante (x,y) and a radius (r) without making use of any floating point computations.

not hard though -wangqingshui(忘情水); 2007-4-16 {631} (#3617356@0)
1. write an integer function sqrt() (round to closest integer)
You do not need a accurate sqrt function. to calculate y = sqrt(r * r - x * x), you can loop y from r - x to r and compare y * y and r * r - x * x (of course it's better to save r * r - x * x into a variable) and get the closest y.

2.
for (x = 0; x <= r; ++x)
{
　　y = sqrt(r * r - x * x);
　　draw points (x, y), (x, -y), (-x, y), (-x, -y)
}

3. there might be some disconnected points when x is close to r. to improve this case, you can save the previous y to y1. if (y1 - y > 1), loop from y1 to y and for each step calculate x1 and draw the 4 points.

好像我用过的所有计算机语言里sqrt都是floating 函数 -ccra(ccra); 2007-4-18 (#3622202@0)

for sure here we can not use the predefined function sqrt(). -wangqingshui(忘情水); 2007-4-19 (#3622983@0)

把屏幕上的每一点都计算(x-x0)^2+(y-y0)^2, 如果值等于r^2画点，如果不是不画。 -ccra(ccra); 2007-4-18 (#3622197@0)

You will only draw couple pixels. -wangqingshui(忘情水); 2007-4-19 (#3622986@0)

Not verified yet. -exception(违例); 2007-4-18 {295} (#3622375@0)
for (i = r; i >= 0; i--)
{
for (j = r - i; j <= r; j++)
{
if ((i * i + j * j <= r * r) && ( i * i + (j + 1) * (j + 1) >= r * r ))
{
drawpoint( x - i, y + j);
drawpoint( x + i, y - j);
drawpoint( x - i, y + j);
drawpoint( x + i, y - j);
}
}
}

This is a very good answer. -wangqingshui(忘情水); 2007-4-20 (#3625648@0)

You guys made me laugh. It is a very basic computer graphics concept. Even a non IT guy like me know the answer. -xtype(xtype); 2007-4-20 (#3625655@0)

hehe!! -arfeifei(老顽童); 2007-4-20 {24} (#3625672@0)
graphic.drawOval(x,y,r);
谁叫我今天心情好呢。写CODE我就不会，讲讲原理吧。先画第一象限的1/4。 -xtype(xtype); 2007-4-20 {439} (#3625703@0)
start from (x+r, y)

start from this point to draw the 1/4 circle, the first step should be up,

draw (x+r, y+1), decide this pixel is outside of the circle or inside, by
r*r + 1*1 is bigger than r*r

so the next step is left, draw (x+r-1, 1) decide it is in or out again, if it is in, go up, if it is out, go left until you reach (x, y+r)

now you got the 1/4 circle. Do the same thing four times, you now get your full circle.

@Calgary

not hard though

Replies, comments and Discussions:

More Topics