由于本周没有人发布过代码高尔夫挑战，我会试一试。我这样做是为了在漫长的编译周期中除了玩剑之外你还能做些什么。
绘制ASCII艺术星，在标准输入上给出三个数字（尖峰的数量，星的类型（通过连接n个顶点的顶点绘制星形）和星的直径）。例子：
Input: Input: Input:
5 2 20 7 2 20 7 3 20
Output: Output: Output:
x x x
xx xx x
xx x xx xx xx x
x x x xxxx x xx xx
xx x x xxxxx x x x x xxx
xxxx x x xxxx x x x x x x
x xxx x xx x xx xxx x x x
x x xxxx x x xx x xxxxx xx x
x x xxx x x x x xx xxxxxx
x x x xxx x x x x x xx xxxxxx
x x xx x x x x x x xxx
x x x xxx x x x x xx xxxxxx
x x xxx x x x x xx xxxxxx
x x xxxx x x xx x xxxxx xx x
x xxx x xx x xx xxx x x x
xxxx x x xxxx x x x x x x
xx x x xxxxx x x x x xxx
x x x xxxx x xx xx
xx x xx xx xx x
xx x x x
x x x
由于正确光栅化线条可以成为代码高尔夫挑战的PITA，我会留下一些余地，但不会太多。更多例子：
足够好：
x x x x
xx xx x x
x x x x
x xx xx x x x
x x x
x x x
xxx xxx x x
x x xxxxxxxxxxxxxxxxxxxxx
x x xx x x xx
xx x x xx xx x x xx
x x xxx xxx
xxxxxxxxxxxxxxxxxxxxx xxxxx
x x x xx xx x
x xx xx x
xxx xxx
x x xx x x xx
x
x
不削减它：
x xx xx
xx x x
xx x x
x x xx xx
xx x x x x
xxxx x x x xx xx x
x xxxx x x
x xxxx x x x
x x xxx xxx xxx
xx x xxxxxx x x
xx x xxxxxx x x
xx xxx xx x x xx
x x xxx x x x
x xxx x xxxxxxxxxxxxxxxxxxxxx
xxx x x x x
xx x x
x x
x x x x
x x
xx
x
Lack of precission Lack of clipping
玩得开心！
答案 0 :(得分：6)
a,b,c=gets.split.map &:to_i
o=(0..c).map{' '*c}
m=(1..a).map{|i|r=2*h=Math::PI*i/a
[(1+Math.sin(r))*c/2+0.5,(1+Math.cos(r))*c/2]}
a.times{|n|j,i,x,y=*m[n],*m[n-b]
j,x,i,y=i,y,j,x if k=((i-y)/(j-x))**2>1
s,z=[x,j].sort
s.to_i.upto(z){|m|t=i+(m-j)*(i-y)/(j-x)
k ?(o[t][m]='x'):(o[m][t]='x')}}
puts o
x xx x xx
x x x x x x
xx x x xx xx x x x
x x x xxxx x xx xx xxxxxxxxxxxxxxxx
xxx x x xxxxx xx x x x xx x x x x
x xxx x x xxxxx x x x x x x x x x x x
x xxx x xx x xx xxxx x xxx x x x xx
x x xxxx x x x x xxxxxx x xx x
x x xxx x x xxx x x xxxxx x xx
x x x xxx xx x x x xx x xxxxx xx x x
xx x xx xx x x xx x xxxx x x x
xx x xxx xx x x x xx x xxxx xx x x
x x xxx xx xxx x x xxxxx x xx
x x xxxx x x x x xxxxxxx x xx x
x xxx x xx x xx xxxx x xxx x x x xx
x xxx x x xxxxx x x x x x xx x x x x
xxx x x x xxxx xx x x x xx x x x x
x x x xxxx x x x x xxxxxxxxxxxxxxxx
x x x x xx xx x x x
xx x x xx x x
x xx x x
答案 1 :(得分：5)
using System.Drawing;class P{static void Main(string[]a){int p=int.Parse(a[0]),i=int.Parse(a[1]),l=int.Parse(a[2]),n;var o=System.Console.Out;var b=new Bitmap(l*4/3,l*4/3);var g=Graphics.FromImage(b);g.TranslateTransform(l/8,l/3);for(n=0;n<p;n++){g.DrawLine(Pens.Red,0,0,l,0);g.TranslateTransform(l,0);g.RotateTransform(360*i/p);}for(i=0;i<b.Height;i++,o.WriteLine())for(p=0;p<b.Width;p++)o.Write(b.GetPixel(p,i).A>0?"#":" ");}}
明星5 2 20
#
# #
# #
# #
# #
# #
#####################
## # # ##
# # # #
# # # #
## ##
## ##
# # # #
# ### #
# # # #
# ## ## #
# # # #
## ##
# #
答案 2 :(得分：4)
OCaml，659个字符（wc -c）。
使用经典的Bresenham算法绘制线条，使其具有递归效果。
保存为stars.ml并使用echo 5 2 20 | ocaml stars.ml运行。
let pl b x y=b.(x).[y]<-'x'
let ln b x0 y0 x1 y1 =
let rec h p x0 y0 x1 y1 =
if x0>x1 then h p x1 y1 x0 y0
else(
let dx,dy,ys=x1-x0,abs(y1-y0),if y0<y1 then 1 else -1 in
let rec k x y e=
if x<=x1 then(
p y x;
if e<dy then k(x+1)(y+ys)(e-dy+dx)
else k(x+1)y(e-dy)
)in
k x0 y0(dx/2)
)in
if abs(y1-y0)>abs(x1-x0)then
h(fun x y->pl b y x)y0 x0 y1 x1
else h(pl b)x0 y0 x1 y1
let f=float
let g n s d=
let b=Array.init d(fun _->String.make d ' ')in
let r=f(d/2)-.0.1 in
let k h i=int_of_float(r+.r*.h(6.2831853*.(f(i mod n)/.(f n))))in
let x,y=k cos,k sin in
for i=0 to n do ln b(x i)(y i)(x(i+s))(y(i+s))done;Array.iter print_endline b
let()=Scanf.scanf"%d %d %d"g
输出
x x x
xx xxx xx
xx x x xxx xx x
x x x xxxxx x x x x
xx x x xxxxxx xx x x x xxx
xxxxx x x xxx x x x xxx x x x x
x xxxx x x x xx xxxxxxx xx x
x x xxx x x xx x xxxxx x
x x x xxxx x x x x x xx xxxxxx
x x xx xx x x xxx x xxx
x x xxx x x x xxx x xxxx
x x xxx xx xx x xx xxxx
x x xxxx x x xx x xxxxxx x
x xxx x x x x x xxxxx x xx x
xxxx x x xxx x x x xx x x x x
xx x x xxxxxx x x x x xxx
x x x xxxxx x x x x
xx x xx xxx xx x
xx x x xx
x x x
答案 3 :(得分：4)
我第一次尝试代码高尔夫......
另存为k.py。以python k.py 5 2 20
import sys,math as m
A=abs;R=range;X=round
p,t,w=map(int,sys.argv[1:])
G=[[' ']*-~w for i in R(w+1)]
def L(a,b,c,d):
s=A(d-b)>A(c-a)
if s:a,b,c,d=b,a,d,c
if a>c:a,b,c,d=c,d,a,b
f=c-a;e=f/2;y=int(X(b));x=int(X(a))
while x<X(c):
e-=A(d-b);p,q=((x,y),(y,x))[s];G[p][q]='x';x+=1
if e<0:y+=(-1,1)[y<d];e+=f
r=w/2;k=m.pi*2/p
s=[(m.sin(k*i)*r+r,m.cos(k*i)*r+r)for i in R(p+t)]
for i in R(p):L(*(s[i]+s[i+t]))
for i in G:print''.join(i)
（通过用标签代替第二级缩进，可以节省2个字符）。
此处提供的扩展版本：http://gist.github.com/591485
x x x
xx xx x
xx x xx x xx x
x x x xxxxx xx xx
xx x x x xxxx x x x x xxx
xxxx x x xxxx x x x x x x
x xxx x xxx x xx xxx x x x
x x xxxx x x xx x xxxxx xx x
x x xxx x x x x xx xxxxxx
x x xxx x x x x x xx xxxxx
x x x x x x x xx xxx
x x x xxx x x x x xx xxxxx
x x xxx x x x x xx xxxx x
x xxxxx x x xx x xxxxx xx x
x xxx x xxx x xx xxx x xx x
x xxx x x xxxx x x x x x x
xx x x x xxxx x x xx xx x
x x x xxxxx xx xx
xx x xx x
x x x x
x
答案 4 :(得分：3)
我还将Bresenham算法用于线条。
g=Split(WScript.StdIn.ReadAll):t=g(1):d=g(2):ReDim r(d*d)
For i=0 To d*d:r(i)=" ":Next
c=1.57:u=(d-1)/2:b=c*4/g(0):For i=1 To g(0)
Z n:x=g:Z n+c:y=g:Z n-b*t:f=g:Z n-b*t+c:n=n+b
s=Abs(g-y)>Abs(f-x):If s Then W x,y:W f,g
If x>f Then W x,f:W y,g
p=f-x:q=Abs(g-y):h=p\2:v=(y>g)*2+1
For x=x To f:r((s+1)*(y*d+x)-s*(x*d+y))="x"
h=h-q:If h<0 Then y=y+v:h=h+p
Next:Next:For i=0 To d:WScript.Echo Mid(Join(r,""),i*d+1,d):Next
Sub W(a,b):e=a:a=b:b=e:End Sub
Sub Z(e):g=Int(u*Cos(e)+u):End Sub
输出：
x x x
xx xx x
xx x xx xxx xx x
x x x xxxx x xx xx
xx x x xxxxx x x x x xx
xxxx x x xxx x x x xx x x xx x
x xxx x x x xx xxxxxx x x
x x xxx x x xx x xxxx x
x x xxx x x x x xx xx xxx
xx x xxx x x x xx xx xxxx
xx x xx x x x xx xx xxx
x x x xxxx x x x x xx xx xxxxxx
x x xxx x x xxx x xxxxxxx
x xxx x x x x x xxxx x x
xxxxx x x xxx x xx xxx x x xx x
xx x x xxxxx xx x x x xx
x x x xxxx x xx xx
xx x x xxx xx x
xx xxx x
x x x
扩展代码：
Dim PI, args, output, i
PI = 4 * Atn(1)
args = Split(WScript.StdIn.ReadAll, " ")
output = Join(Star(args(0), args(1), args(2)), vbNullString)
For i = 1 To Len(output) Step args(2)
WScript.Echo Mid(output, i, args(2))
Next
Function Star(spikes, star_type, diameter)
Dim result(), i, vertexes(), angle, radius, p1, p2
ReDim result(diameter * diameter - 1)
ReDim vertexes(spikes - 1)
For i = 0 To UBound(result)
result(i) = " "
Next
radius = (diameter - 1) / 2
For i = 0 To UBound(vertexes)
vertexes(i) = Array(CLng(radius * Cos(angle) + radius), _
CLng(radius * Sin(angle) + radius))
angle = angle - (2 * PI / spikes)
Next
For i = 0 To UBound(vertexes)
p1 = vertexes(i)
p2 = vertexes((i + star_type) Mod spikes)
DrawLine p1(0), p2(0), p1(1), p2(1), result, diameter
Next
Star = result
End Function
Sub DrawLine(ByVal x0, ByVal x1, ByVal y0, ByVal y1, arr, diameter)
Dim steep, deltax, deltay, error, ystep
steep = Abs(y1 - y0) > Abs(x1 - x0)
If steep Then
Swap x0, y0
Swap x1, y1
End If
If x0 > x1 Then
Swap x0, x1
Swap y0, y1
End If
deltax = x1 - x0
deltay = Abs(y1 - y0)
error = deltax \ 2
If y0 < y1 Then ystep = 1 Else ystep = -1
For x0 = x0 To x1
If steep Then
arr(x0 * diameter + y0) = "x"
Else
arr(y0 * diameter + x0) = "x"
End If
error = error - deltay
If error < 0 Then
y0 = y0 + ystep
error = error + deltax
End If
Next
End Sub
Function Swap(a, b)
Dim temp
temp = a
a = b
b = temp
End Function
答案 5 :(得分：1)
Javascript（基于Isc的python解决方案）， 598 591 586 559字节（与rhino shell一起使用：另存为'stars.js '，与'rhino stars.js 7 2 20'一起运行：
for(var a=Math,b=parseInt,c=a.floor,d=a.abs,e=arguments,f=b(e[0]),g=b(e[1]),h=b(e[2]),i=[],j=0;j<(h+1)*h;j++)i.push((j+1)%(h+1)?" ":"\n");function k(w,x){i[x*(h+1)+w]="x"}var l=(h-1)/2,m=[];for(j=0;j<f;j++)m.push([c(l*a.cos(a.PI*2/f*j)+h/2),c(l*a.sin(a.PI*2/f*j)+h/2)]);
for(j=0;j<f;j++){var n=m[j][0],o=m[j][1],p=m[(j+g)%f][0],q=m[(j+g)%f][1],r=void 0,s=void 0;if(r=d(q-o)>d(p-n)){s=n;n=o;o=s;s=p;p=q;q=s}if(n>p){s=n;n=p;p=s;s=o;o=q;q=s}for(var t=p-n,u=0,v=d(q-o)/t,y=o<q?1:-1,z=o,A=n;A<=p;A++){r?k(z,A):k(A,z);u+=v;if(u>=0.5){z+=y;u-=1}}}print(i.join(""));
输出：
x
xx
xx
x x
xx x x
xxxx x x
x xxx x
x x xxx
x x xxx
xx x xxx
xx x xx
x x x xxxx
x x xxx
x xxx x
xxxxx x x
xx x x
x x
xx
xx
x
x
xx
x x xxx
x xxxx x
xxxxx x x
xxx x x x
x x xx
x x x
x x xxx
x x x
x x x
x x x xx
x x xx
x x x
xxx x xxx
xxxxx x x
x xxxx x
x xx xxx
xx
x
x
x
xx x
xx xx
x x xx
xx x x xx x
xxxxxx x x
x xxxx x
xx xx xxxx
xx xx xxxx
xx xx xxx
xx xx xxxxxx
x xxxxxxx
x xxxx x x
xxx x x xx x
x x xx
xx xx
xx x
x
x
扩展代码：
var p = parseInt(arguments[0]),
t = parseInt(arguments[1]),
w = parseInt(arguments[2]);
var g = [];
for(var i = 0; i < (w + 1) * w; i++)
g.push((i + 1) % (w + 1) ? ' ' : '\n');
function plot(x, y) { g[y * (w + 1) + x] = 'x';}
function line(x0, y0, x1, y1) {
var steep = Math.abs(y1 - y0) > Math.abs(x1 - x0), temp;
if(steep) {
temp = x0; x0 = y0; y0 = temp;
temp = x1; x1 = y1; y1 = temp;
}
if(x0 > x1) {
temp = x0; x0 = x1; x1 = temp;
temp = y0; y0 = y1; y1 = temp;
}
var deltax = x1 - x0;
var deltay = Math.abs(y1 - y0);
var error = 0;
var deltaerr = deltay / deltax;
var ystep
var y = y0;
if(y0 < y1) ystep = 1; else ystep = -1;
for(var x = x0; x <= x1; x++) {
if(steep) plot(y,x); else plot(x,y);
error = error + deltaerr;
if(error >= 0.5) {
y = y + ystep;
error = error - 1.0;
}
}
}
var r = (w-1)/2, points = [];
for(var i = 0; i < p; i++) {
points.push([
Math.floor(r * Math.cos(Math.PI * 2 / p * i) + w/2),
Math.floor(r * Math.sin(Math.PI * 2 / p * i) + w/2)
]);
}
for(var i = 0; i < p; i++) {
line(points[i][0], points[i][1], points[((i + t) % p)][0], points[((i + t) % p)][1]);
}
print(g.join(''));