好的，这是一个相当精细的代码高尔夫挑战：实施Reversi（奥赛罗）游戏。
以尽可能少的字符执行此操作。
会话应该是这样的：
abcdefgh
1
2
3
4 wb
5 bw
6
7
8
b>d3
abcdefgh
1
2
3 b
4 bb
5 bw
6
7
8
答案 0 :(得分：7)
Haskell， 821 个字符†，包括换行符。
import Array
import List
import Maybe
import IO
a=((1,0),(8,7))
z=zip
x=range
y=repeat
o=putStr
i=['a'..'h']
r=[0,1,-1]
d b=' ':i++concatMap(\i->'\n':show(i+1)++map g(take 8$drop(i*8)$elems b))[0..7]++"\n"
e=array a(z(x a)$y 0)//z(x((4,3),(5,4)))[1,-1,-1,1]
f b p i|b!i/=0=Nothing|True=if b==v then Nothing else Just v where v=foldr u b(map(takeWhile(inRange a).($i).iterate.(\(i,j)(k,l)->(i+k,j+l)))$tail[(i,j)|i<-r,j<-r]);u r m=m//case(group$map(m!)r)of([_]:c@(m:_):(l:_):_)|m/=0&&l==p->z(take(1+length c)$r)(y p);_->[]
g=fromJust.(`lookup`z[-1..1]"b w")
h b p=do o$g p:">";hFlush stdout;c:r<-getLine;case f b p(read r,head$elemIndices c i)of;Nothing->h b p;Just m->o(d m)>>case filter(\s->mapMaybe(f m s)(x a)/=[])[-1*p,p]of(n:_)->h m n;_->print$if w==' 'then 'd'else w where w=g$signum$sum$elems m
main=o(d e)>>h e 1
以上内容基于此版本：
import Data.Array
import Data.List
import Data.Maybe
import System.IO
type Index = (Int, Int)
type Player = Int -- (-1, 0, 1) for white, unclaimed and black, resp.
type Board = Array Index Player
-- Infix function which add two vectors
(|+|) :: Index -> Index -> Index
(i, j) |+| (k, l) = (i + k, j + l)
-- The functions dim, pieces and board2Str must be updated if one wishes to
-- alter the board's dimensions.
-- Board dimensions
dim :: (Index, Index)
dim = ((0, 0), (7, 7))
-- The pieces that will initially be on the board
pieces :: [(Index, Player)]
pieces = zip (range ((3, 3), (4, 4))) [1, -1, -1, 1]
-- Return a textual representation of the given board
board2Str :: Board -> String
board2Str b = ' ' : ['a'..'h'] ++ -- column names
concatMap (\i -> '\n' : show (i + 1) ++ -- row names
map player2Str (take 8 $ drop (i * 8) $ elems b)) [0..7] ++ "\n" -- rows
-- The initial board, including the initial pieces
initial :: Board
initial = array dim (zip (range dim) $ repeat 0) // pieces -- the empty board
-- The directions in which pieces can be flipped.
deltas :: [(Int, Int)]
deltas = tail [(i, j) | i <- [0,1,-1], j <- [0,1,-1]]
-- All 'lines' in any direction starting from the given index
dirs :: Index -> [[Index]]
dirs c = map (takeWhile (inRange dim) . ($ c) . iterate . (|+|)) deltas
-- Attempt to perform a move
move :: Board -> Player -> Index -> Maybe Board
move b p i | b ! i /= 0 = Nothing
| otherwise = if b == updateAll then Nothing else Just updateAll
where
-- Attempt to swap pieces in all directions
updateAll = foldr update b (dirs i)
-- Attempt to swap pieces in the given direction
update r b' = b' // case (group $ map (b' !) r) of
([_]:c@(m:_):(l:_):_) | m /= 0 && l == p -> -- all conditions are met
zip (take (1 + length c) $ r) (repeat p) -- so swap the pieces
_ -> [] -- nothing to swap
player2Str :: Player -> Char
player2Str = fromJust . (`lookup` zip [-1..1] "b w")
-- Test whether the given player can make a move
possible :: Board -> Player -> Bool
possible b p = mapMaybe (move b p) (range dim) /= []
-- Ask the players to make a move in turns, and update the board when required
mainLoop :: Board -> Player -> IO ()
mainLoop b p = do
putStr $ player2Str p : "> "
hFlush stdout
c:r <- getLine -- get user input (this is very fragile!)
-- Attempt to perform the suggested move...
case move b p (read r - 1, head $ elemIndices c ['a'..]) of
Nothing -> mainLoop b p -- ...try again if not possible
Just b' -> do -- ...accept if possible
putStr $ board2Str b' -- print the new board state
case filter (possible b') [-1 * p, p] of -- select next player to move
(p':_) -> mainLoop b' p' -- move if possible, otherwise game over
_ -> print $ if winner == ' ' then 'd' else winner
where winner = player2Str $ signum $ sum $ elems b'
-- Let the games begin!
main :: IO()
main = putStr (board2Str initial) >> mainLoop initial 1
†：是的，我是756个字符，但注意到代码没有正确应用Reversi的规则。长号。
答案 1 :(得分：4)
编辑2：
更多的调整和技巧，我设法削减了另外124个字符，直到 752 字符：
using C=System.Console;class O{static int[]b=new int[100];static int c=1;static void Main(){b[44]=b[55]=-1;b[45]=b[54]=1;var g=true;while(g){C.WriteLine(" abcdefgh");for(int i=10;i<90;i++){switch(i%10){case 0:C.Write(i/10);break;case 9:C.WriteLine();break;default:C.Write("w b"[b[i]+1]);break;}}C.Write("w b"[c+1]+">");var l=C.ReadLine();if(l.Length>1){int x=l[0]-96,y=l[1]-48;if(x>0&x<9&y>0&y<9&&b[y*10+x]<1)c*=f(y*10+x,1);}g=false;for(int y=10;y<90;y+=10)for(int x=y;x<y+8;)g|=b[++x]==0&&f(x,0)<0;}int s=0;foreach(int p in b)s+=p;C.WriteLine(s==0?"d":s<0?"w":"b");}static int f(int ofs,int y){var x=1;foreach(int d in new int[]{-11,-10,-9,-1,1,9,10,11}){int l=1,o=ofs;while(b[o+=d]==-c)l++;if(b[o]==c&l>1){x=-1;while(y>0&l-->0)b[o-=d]=c;}}return x;}}
压缩之前：
using Con = System.Console;
class O {
// Initialise game board, made of ints - 0 is empty, 1 is a black piece, -1 a white.
// Using a 10x10 board, with the outer ring of empties acting as sentinels.
static int[] board = new int[100];
// Color of the current player.
static int currentColor = 1;
static void Main() {
// Set up the four pieces in the middle.
board[44] = board[55] = -1;
board[45] = board[54] = 1;
var legal = true;
while (legal) {
// Print game board.
Con.WriteLine(" abcdefgh");
for (int i = 10; i < 90; i++) {
switch (i % 10) {
case 0: Con.Write(i / 10); break;
case 9: Con.WriteLine(); break;
default: Con.Write("w b"[board[i] + 1]); break;
}
}
// Print input, indicating which color's turn it is.
Con.Write("w b"[currentColor+1] + ">");
// Parse input.
var line = Con.ReadLine();
if (line.Length > 1) {
int x = line[0] - 96, y = line[1] - 48;
// Discard malformed input.
if (x > 0 & x < 9 & y > 0 & y < 9 && board[y * 10 + x] < 1)
// Check if valid move and if so flip and switch players
currentColor *= flip(y * 10 + x, 1);
}
// See if there are any legal moves by considering all possible ones.
legal = false;
for (int y = 10; y < 90; y += 10)
for (int x = y; x < y + 8;)
legal |= board[++x] == 0 && flip(x, 0) < 0;
}
// Calculate final score: negative is a win for white, positive one for black.
int score = 0;
foreach (int piece in board) score += piece;
Con.WriteLine(score == 0 ? "d" : score < 0 ? "w" : "b");
}
// Flip pieces, or explore whether putting down a piece would cause any flips.
static int flip(int ofs, int commitPutDown) {
var causesFlips = 1;
// Explore all straight and diagonal directions from the piece put down.
foreach (int d in new int[] { -11, -10, -9, -1, 1, 9, 10, 11 }) {
// Move along that direction - if there is at least one piece of the opposite color next
// in line, and the pieces of the opposite color are followed by a piece of the same
// color, do a flip.
int distance = 1, o = ofs;
while (board[o += d] == - currentColor) distance++;
if (board[o] == currentColor & distance > 1) {
causesFlips = -1;
while (commitPutDown > 0 & distance-- > 0) board[o -= d] = currentColor;
}
}
return causesFlips;
}
}
876个字符C＃版本：
using C=System.Console;class O{static void Main(){int[]b=new int[100];b[44]=b[55]=2;b[45]=b[54]=1;int c=1;while(true){C.WriteLine(" abcdefgh");for(int i=10;i<90;i++){switch(i%10){case 0:C.Write(i/10);break;case 9:C.WriteLine();break;default:C.Write(" bw"[b[i]]);break;}}bool g=false;for(int y=10;y<90;y+=10)for(int x=1;x<9;x++){g|=(b[x+y]==0&&f(x+y,b,c,false));}if(!g)break;C.Write(" bw"[c]+">");var l=C.ReadLine();if(l.Length>1){int x=l[0]-96,y=l[1]-48,ofs;if(x>0&x<9&y>0&y<9&&b[ofs=y*10+x]==0)if(f(ofs,b,c,true))b[ofs]=c;c=3-c;}}int s=0;for(int y=10;y<90;y+=10)for(int x=1;x<9;x++)switch(b[y+x]){case 1:s++;break;case 2:s--;break;}C.WriteLine(s==0?"d":s<0?"w":"b");}static bool f(int ofs,int[]b,int p,bool c){var x=false;foreach(int d in new int[]{-11,-10,-9,-1,1,9,10,11}){int l=1,o=ofs;while(b[o+=d]==3-p)l++;if(b[o]==p&l>1){x=true;if(c)while(l-->1)b[o-=d]=p;}}return x;}}
压缩之前：
using Con = System.Console;
class Othello {
static void Main() {
// Initialise game board, made of ints - 0 is empty, 1 is a black piece, 2 a white.
// Using a 10x10 board, with the outer ring of empties acting as sentinels.
int[] board = new int[100];
// Set up the four pieces in the middle.
board[44] = board[55] = 2;
board[45] = board[54] = 1;
// Color of the current player.
int currentColor = 1;
while (true) {
// Print game board.
Con.WriteLine(" abcdefgh");
for (int i = 10; i < 90; i++) {
switch (i % 10) {
case 0: Con.Write(i / 10); break;
case 9: Con.WriteLine(); break;
default: Con.Write(" bw"[board[i]]); break;
}
}
// See if there are any legal moves by considering all possible ones.
bool legal = false;
for (int y = 10; y < 90; y += 10) for (int x = 1; x < 9; x++) {
legal |= (board[x + y] == 0 && flip(x + y, board, currentColor, false));
}
if (!legal) break;
// Print input, indicating which color's turn it is.
Con.Write(" bw"[currentColor] + ">");
// Parse input.
string l = Con.ReadLine();
if (l.Length > 1) {
int x = l[0] - 96, y = l[1] - 48;
int ofs;
// Discard malformed input.
if (x > 0 & x < 9 & y > 0 & y < 9 && board[ofs = y * 10 + x] == 0)
// Check if valid move & flip if it is - if not, continue.
if (flip(ofs, board, currentColor, true))
// Put down the piece itself.
board[ofs] = currentColor;
// Switch players.
currentColor = 3 - currentColor;
}
}
// Calculate final score: negative is a win for white, positive one for black.
int score = 0;
for (int y = 10; y < 90; y += 10) for (int x = 1; x < 9; x++)
switch (board[y + x]) {
case 1: score++; break;
case 2: score--; break;
}
Con.WriteLine(score == 0 ? "d" : score < 0 ? "w" : "b");
}
/** Flip pieces, or explore whether putting down a piece would cause any flips. */
static bool flip(int ofs, int[] board, int playerColor, bool commitPutDown) {
bool causesFlips = false;
// Explore all straight and diagonal directions from the piece put down.
foreach (int d in new int[] { -11, -10, -9, -1, 1, 9, 10, 11 }) {
// Move along that direction - if there is at least one piece of the opposite color next
// in line, and the pieces of the opposite color are followed by a piece of the same
// color, do a flip.
int distance = 1, o = ofs;
while (board[o += d] == 3 - playerColor) distance++;
if (board[o] == playerColor & distance > 1) {
causesFlips = true;
if (commitPutDown) while (distance-- > 1) board[o -= d] = playerColor;
}
}
return causesFlips;
}
}
我使用int [100]而不是char [10,10]，使一些循环和比较更简单。那里有很多小技巧，但除此之外它与原始Java代码基本相同。
编辑：
编辑器显示了一些奇怪的“Col”值，但是你应该看看“Ch”值...它不是943个字符，它是876 ...
答案 2 :(得分：4)
这是我的版本。它的重量为672个字符，包括换行符：
#include <iostream>
#define F for
#define G F(i p=9;p<90;++p)
#define R return
#define C std::cout
typedef int i;i b[100];i f(i p, i s, i d){i t=0;i q[]={-11,-10,-9,-1,1,9,10,11};F(i o=0;o<8;++o){i c=p+q[o];F(;b[c]==-s;c+=q[o]);if(b[c]==s)F(--t;c!=p;c-=q[o],++t)if(d)b[c]=s;}R t;}i h(i s){G if(!b[p]&&f(p,s,0))R 1;R 0;}i main(){F(i p=0;p<100;++p)b[p]=p<10||p>89||!(p%10%9)?2:0;b[44]=b[55]=-1;b[45]=b[54]=1;i s=1;F(;;){C<<" abcdefgh";G C<<(char)(p%10?"w b\n"[b[p]+1]:'0'+p/10);s=h(s)?s:-s;if(!h(s)){s=-34;G s+=b[p];C<<(s<0?'s':s?'b':'d')<<'\n';R 0;}i p=0;F(;p<9||p>90||b[p]||!f(p,s,1);){C<<"w b"[s+1]<<">";std::string m;std::cin>>m;p=m[0]-'`'+(m[1]-'0')*10;}b[p]=s;s=-s;}}
答案 3 :(得分：3)
这是Perl，现在低至408个字符。可以使用更简单的方法切断另外25个字符来声明获胜者（比如说"B"而不是"Winner: Black\n"）。前两个换行很重要;包含其他内容是为了便于阅读。
sub O{2&$B[$q+=$_]*$%}sub A{grep{$q=$=;$"=''while&O;$B[$q]*O$q=$=}$B[$=]?():@d}
sub F{$B[$q]=$%;O&&&F}sub D{print'
',a..h,$/,(map{($e="@B[$_*9+1..$_*9+8]
")=~y/012/ bw/;$_,$e}1..8),@_}
@d=map{$_,-$_}1,8..10;@B=(@z=(0)x40,$%=2,1,(0)x7,1,2,@z);
for$!(%!){$%^=3;for$=(9..80){$=%9*A&&do{{D$%-2?b:w,"> ";
$_=<>;$==(/./g)[1]*9-96+ord;A||redo}F$q=$=for A;last}}}
$X+=/1/-/2/for@B;D"Winner: ",$X<0?White:$X?Black:None,$/
@B持有游戏板。 $B[$i*9+$j]指行$ i（1..8）和列$ j（1..8）
@d是8个有效路线的清单
O是一种便捷方法。它将$q增加$_（当前方向），如果$B[$q]处的棋子属于当前玩家的对手，则返回非零值
F处理当前方向$_
A检查当前玩家是否可以在$B[$=]进行合法移动并返回可以翻转件的方向集
D(@_)绘制棋盘并打印@_
主循环切换$%（当前玩家）并遍历棋盘上的位置以找到该玩家的合法移动。
如果发现任何合法移动，请从标准输入读取移动（重复输入有效移动）并更新电路板。
答案 4 :(得分：2)
我可以用Java的1143个字符来完成：
import java.io.*;public class R{public static void main(String[]args)throws Exception{char[][]b=new char[10][10];for(inty=0;y<10;y++)for(int x=0;x<10;x++)b[y][x]=' ';b[4][4]='w';b[4][5]='b';b[5][4]='b';b[5][5]='w';char c='b';BufferedReader r=new BufferedReader(new InputStreamReader(System.in));while(true){System.out.println(" abcdefgh");boolean m=false;for(int y=1;y<9;y++){System.out.print(y);for(int x=1;x<9;x++){System.out.print(b[y][x]);m=m||(b[y][x]==' '&&f(x,y,b,c,false));}System.out.println();}if(!m)break;System.out.print(c+">");String l = r.readLine();if(l.length()<2)continue;int x=l.charAt(0)-'a'+1;int y=l.charAt(1)-'0';if(x<1||x>8||y<1||y>8||b[y][x]!=' '||!f(x, y, b, c, true))continue;b[y][x]=c;c=c=='b'?'w':'b';}int s=0;for(int y=1;y<9;y++)for(int x=1;x<9;x++)s+=b[y][x]=='b'?1:b[y][x]=='w'?-1:0;System.out.println(s==0?"d":s<0?"w":"b");}static boolean f(int x,int y,char[][]b,char c,boolean o){boolean p=false;for(int u=-1;u<2;u++)for(int v=-1;v<2;v++){if(u==0&&v==0)continue;int d=0;do d++;while(b[y+d*u][x+d*v]==(c=='b'?'w':'b'));if(b[y+d*u][x+d*v]==c&&d>1){p=true;if(o)for(int e=1;e<d;e++)b[y+e*u][x+e*v]=c;}}return p;}}
这是（几乎）相同Java的简写版本：
import java.io.*;
public class Reversi {
public static void main(String[] args) throws Exception {
// Initialise game board, made of chars - ' ' is empty, 'b' is a black piece, 'w' a white.
// Using a 10x10 board, with the outer ring of empties acting as sentinels.
char[][] board = new char[10][10];
for (int y = 0; y < 10; y++) { for (int x = 0; x < 10; x++) { board[y][x] = ' '; } }
// Set up the four pieces in the middle.
board[4][4] = 'w'; board[4][5] = 'b'; board[5][4] = 'b'; board[5][5] = 'w';
// Color of the current player.
char currentColor = 'b';
BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
while (true) {
// Print game board.
System.out.println(" abcdefgh");
for (int y = 1; y < 9; y++) {
System.out.print(y);
for (int x = 1; x < 9; x++) { System.out.print(board[y][x]); }
System.out.println();
}
// See if there are any legal moves by considering all possible ones.
boolean legalMovesAvailable = false;
for (int y = 1; y < 9; y++) { for (int x = 1; x < 9; x++) {
legalMovesAvailable = legalMovesAvailable ||
(board[y][x] == ' ' && flip(x, y, board, currentColor, false));
}}
if (!legalMovesAvailable) { break; }
// Print input, indicating which color's turn it is.
System.out.print(currentColor + ">");
// Parse input.
String l = r.readLine();
if (l.length() < 2) { continue; }
int x = l.charAt(0) - 'a' + 1;
int y = l.charAt(1) - '0';
// Discard malformed input.
if (x < 1 || x > 8 || y < 1 || y > 8 || board[y][x] != ' ') { continue; }
// Check if valid move & flip if it is - if not, continue.
if (!flip(x, y, board, currentColor, true)) { continue; }
// Put down the piece itself.
board[y][x] = currentColor;
// Switch players.
currentColor = currentColor == 'b' ? 'w' : 'b';
}
// Calculate final score: negative is a win for white, positive one for black.
int score = 0;
for (int y = 1; y < 9; y++) { for (int x = 1; x < 9; x++) {
score += board[y][x] == 'b' ? 1 : board[y][x] == 'w' ? -1 : 0;
}}
System.out.println(score == 0 ? "d" : score < 0 ? "w" : "b");
}
/** Flip pieces, or explore whether putting down a piece would cause any flips. */
static boolean flip(int pieceX, int pieceY, char[][] board, char playerColor, boolean commitPutDown) {
boolean causesFlips = false;
// Explore all straight and diagonal directions from the piece put down.
for (int dY = -1; dY < 2; dY++) { for (int dX = -1; dX < 2; dX++) {
if (dY == 0 && dX == 0) { continue; }
// Move along that direction - if there is at least one piece of the opposite color next
// in line, and the pieces of the opposite color are followed by a piece of the same
// color, do a flip.
int distance = 0;
do {
distance++;
} while (board[pieceY + distance * dY][pieceX + distance * dX] == (playerColor == 'b' ? 'w' : 'b'));
if (board[pieceY + distance * dY][pieceX + distance * dX] == playerColor && distance > 1) {
causesFlips = true;
if (commitPutDown) {
for (int distance2 = 1; distance2 < distance; distance2++) {
board[pieceY + distance2 * dY][pieceX + distance2 * dX] = playerColor;
}
}
}
}}
return causesFlips;
}
}
答案 5 :(得分：2)
这并没有完全解决所述的问题，但这个实现实际上在相当强的水平上播放奥赛罗（黑白棋）。在我看来，这是有史以来最好的IOCCC条目之一。您可以在Previous IOCCC Winners页面的“lievaart”下找到有关它的更多信息。
（我已经将实现复制为已发布，但GCC不接受#define D define的全部概念。使用s/D/define/g，程序编译并运行正常。）
#define D define
#D Y return
#D R for
#D e while
#D I printf
#D l int
#D W if
#D C y=v+111;H(x,v)*y++= *x
#D H(a,b)R(a=b+11;a<b+89;a++)
#D s(a)t=scanf("%d",&a)
#D U Z I
#D Z I("123\
45678\n");H(x,V){putchar(".XO"[*x]);W((x-V)%10==8){x+=2;I("%d\n",(x-V)/10-1);}}
l V[1600],u,r[]={-1,-11,-10,-9,1,11,10,9},h[]={11,18,81,88},ih[]={22,27,72,77},
bz,lv=60,*x,*y,m,t;S(d,v,f,_,a,b)l*v;{l c=0,*n=v+100,j=d<u-1?a:-9000,w,z,i,g,q=
3-f;W(d>u){R(w=i=0;i<4;i++)w+=(m=v[h[i]])==f?300:m==q?-300:(t=v[ih[i]])==f?-50:
t==q?50:0;Y w;}H(z,0){W(E(v,z,f,100)){c++;w= -S(d+1,n,q,0,-b,-j);W(w>j){g=bz=z;
j=w;W(w>=b||w>=8003)Y w;}}}W(!c){g=0;W(_){H(x,v)c+= *x==f?1:*x==3-f?-1:0;Y c>0?
8000+c:c-8000;}C;j= -S(d+1,n,q,1,-b,-j);}bz=g;Y d>=u-1?j+(c<<3):j;}main(){R(;t<
1600;t+=100)R(m=0;m<100;m++)V[t+m]=m<11||m>88||(m+1)%10<2?3:0;I("Level:");V[44]
=V[55]=1;V[45]=V[54]=2;s(u);e(lv>0){Z do{I("You:");s(m);}e(!E(V,m,2,0)&&m!=99);
W(m!=99)lv--;W(lv<15&&u<10)u+=2;U("Wait\n");I("Value:%d\n",S(0,V,1,0,-9000,9000
));I("move: %d\n",(lv-=E(V,bz,1,0),bz));}}E(v,z,f,o)l*v;{l*j,q=3-f,g=0,i,w,*k=v
+z;W(*k==0)R(i=7;i>=0;i--){j=k+(w=r[i]);e(*j==q)j+=w;W(*j==f&&j-w!=k){W(!g){g=1
;C;}e(j!=k)*((j-=w)+o)=f;}}Y g;}
您可以在我的Othello页面找到此版本的现代重新实现。
答案 6 :(得分：1)
假设您的代码在此处使用优化版本。
1081个字符
import java.util.*;import java.io.*;class R{public static void main(String[]a)throws Exception{PrintStream h=System.out;char[][]b=new char[10][10];for(int y=0;y<10;y++)for(int x=0;x<10;x++)b[y][x]=' ';b[4][4]='w';b[4][5]='b';b[5][4]='b';b[5][5]='w';char c='b';Scanner r=new Scanner(System.in);while(true){h.println(" abcdefgh");boolean m=false;for(int y=1;y<9;y++){h.print(y);for(int x=1;x<9;x++){h.print(b[y][x]);m=m||(b[y][x]==' '&&f(x,y,b,c,false));}h.println();}if(!m)break;h.print(c+">");String l=r.nextLine();if(l.length()<2)continue;int x=l.charAt(0)-'a'+1;int y=l.charAt(1)-'0';if(x<1||x>8||y<1||y>8||b[y][x]!=' '||!f(x,y,b,c,true))continue;b[y][x]=c;c=c=='b'?'w':'b';}int s=0;for(int y=1;y<9;y++)for(int x=1;x<9;x++)s+=b[y][x]=='b'?1:b[y][x]=='w'?-1:0;h.println(s==0?"d":s<0?"w":"b");}static boolean f(int x,int y,char[][]b,char c,boolean o){boolean p=false;for(int u=-1;u<2;u++)for(int v=-1;v<2;v++){if(u==0&&v==0)continue;int d=0;do d++;while(b[y+d*u][x+d*v]==(c=='b'?'w':'b'));if(b[y+d*u][x+d*v]==c&&d>1){p=true;if(o)for(int e=1;e<d;e++)b[y+e*u][x+e*v]=c;}}return p;}}
答案 7 :(得分：1)
JavaScript / HTML 1419个字符（不完整的条目，放弃）
Kinopiko正在放弃此条目。在提交条目之后，在对ungolfed版本进行了更多工作之后，我意识到这个问题比看起来要复杂得多。例如，如果两个玩家都处于无法玩的位置，则有必要检查无法玩的位置并宣布获胜者。我不知道其他答案是否真的这样做，但逻辑很复杂，这不再是一个有趣的问题。如果其他人想要接管并完成此任务，请随意这样做。
尚未完成（不保留分数）。我点了点击而不是阅读输入。
<html>
<head>
<style>
.s{background-color:green;width:50;height:50;font-size:40}
.w{color:white;font-size:40}
.b{color:black;font-size:40}
.h{width:50;height:50;font-size:30}
</style>
<script>
b=new Array()
for(i=0;i<10;i++)b[i]=new Array(0,0,0,0,0,0,0,0,0);
function j(x,y,c){a=k("s"+x+y)
a.innerHTML=String.fromCharCode(parseInt("25CF",16))
a.className="s "+c}
z="b"
function v(x,y,c){
d=0
for(g=(x>1?x-1:1);g<=(x<8?x+1:8);g++){
for(h=(y>1?y-1:1);h<=(y<8?y+1:8);h++){
if(g==x&&h==y)continue
n=b[g][h]
if(n!=0&&n!=c){e=g-x
f=h-y
for(t=1;;t++){p=x+t*e
q=y+t*f
if(p<1||q<1||p>8||q>8)break
if(!b[p][q])break
if(b[p][q]==c){d=1
for(s=1;s<t;s++){r=x+s*e
u=y+s*f
b[r][u]=c
a=k("s"+r+u)
a.className="s "+c}break}}}}}return d}
function w(x,y){if(b[x][y])return
if(!v(x,y,z))return
b[x][y]=z
j(x,y,z)
z=(z=="b"?"w":"b")}
function create_b(){
b[4][4]=b[5][5]="w";b[4][5]=b[5][4]="b"
t=k("b");
r=o("tr",t)
for(x=0;x<9;x++){
if(x){
h=o("th",r)
h.innerHTML=x}
else
h=o("th",r)
h.className="h"}
for(y=1;y<9;y++){
r=o("tr",t)
for(x=0;x<9;x++){
if(x){td=o("td",r)
td.className="s a"
td.id="s"+x+y;(function(x,y){td.onclick=function(){w(x,y)}}(x,y))
if(b[x][y])j(x,y,b[x][y])
}else{
h=o("th",r)
h.innerHTML=y
h.className="h"}}}}
function o(p,q){n=document.createElement(p);q.appendChild(n);return n}
function k(i){return document.getElementById(i)}
</script>
</head>
<body onload="create_b()">
<table id="b"></table>
</body>
</html>
答案 8 :(得分：0)
Here's a 1266 byte WinXP .COM file。这可能看起来有点大，但确实超出了要求。好的，我被带走了一点。
控制：
w,s,a,d = move cursor space = place piece enter = exit program
玩起来应该相当直观。
P.S。 Source code is here!