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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
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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.

@Las Vegas

not hard though

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