25

由于本周没有人发布代码高尔夫挑战，我会尝试一下。我这样做是为了在那些漫长的编译周期中你可以做一些除了玩剑之外的事情。

挑战：

绘制 ASCII 艺术星，给定标准输入上的三个数字（尖峰的数量、星的类型（通过连接相隔 n 个顶点的顶点来绘制星）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能是代码高尔夫挑战的 PITA，我会留下一些余地，但不会太多。更多示例：

够好了：

    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

不剪：

            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！

4

6 回答 6

6

红宝石 — 323 276 304 297

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
于 2010-09-21T23:55:19.397 回答
5

C#：555 428 个字符

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?"#":" ");}}

星星 5 2 20

            #
           # #
           # #
           # #
          #   #
          #   #
  #####################
   ##    #     #    ##
     #   #     #   #
      #  #     #  #
       ##       ##
        ##     ##
        # #   # #
       #   ###   #
       #   # #   #
       # ##   ## #
      # #       # #
      ##         ##
      #           #
于 2010-09-22T13:08:46.227 回答
4

OCaml, 659 characters (wc -c).

Uses the classical Bresenham algorithm to draw the lines, made recursive for fun. Save as stars.ml and run with echo 5 2 20 | ocaml stars.ml.

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

Outputs

            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
于 2010-09-21T14:52:36.053 回答
4

我第一次尝试代码高尔夫......

Python - 550 533 507 484 451 437 个字符

另存为k.py. 运行方式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)

（通过用制表符替换二级缩进，可能会再节省 2 个字符......）

此处提供扩展版本：http: //gist.github.com/591485

示例输出：

             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
于 2010-09-21T17:25:10.573 回答
3

VBScript，584 511 482 字节

我还对线条使用了 Bresenham 算法。

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
于 2010-09-21T17:36:20.417 回答
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'，使用 'rhino stars.js 7 2 20' 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       
          xx       
  xx xx       
   xxxxx       
   x xxx x       
    xxxxx       
     xxxxx     
     xx x xxx  
     xx x xx
     xxxxxx  
    xxxxx      
   x xxx x       
   xxxxx       
  xx xx       
          xx       
           xx       
           xx       
            X       

       X            
       xx           
      xxxxx    
      x xxxx x    
    xxxxx    
 xxxxxx    
 xx xx    
  xxx    
  xxxxx  
   xxx
   xxx
  某某某某
  xx xx   
 xxx    
 xxx x xxx    
    xxxxx                                                                                                                                              
      x xxxx x                                                                                                                                              
      x xx xxx                                                                                                                                              
       xx                                                                                                                                                     
       X            

       X            
       X            
       某某    
       xx xx    
       xx xx     
 xx xx xx xx     
  xxxxxx xx      
   x xxxx x      
    xx xx xxxx      
      xx xx xxxx  
      xx xx xxx
    xx xx xxxxxx   
   x xxxxxxx      
  x xxxx xx      
 xxx xx xx x     
       xx xx     
       xx xx    
       某某    
       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(''));
于 2010-09-22T12:50:31.300 回答