迷宫，80字节
?::`}:("(!@
perfect:
{:{:;%"}
+puts; "
}zero: "
}else{(:
"negI" _~
""""""{{{"!@
拉丁字符perfect puts zero else neg I实际上只是注释*。
即，如果输入是完美的，0则打印a，否则打印-1。
在线尝试！
*所以这个或这个工作也...
?::`}:("(!@ ?::`}:("(!@
: BEWARE :
{:{:;%"} {:{:;%"}
+ ; " +LAIR; "
} : " } OF : "
} {(: }MINO{(:
" " _~ "TAUR" _~
""""""{{{"!@ """"""{{{"!@
怎么样？
将一个正整数作为输入n并将一个累加器变量-n放在辅助堆栈上，然后对每个整数执行从除n-1到（包括）与1除法运算的整数的除法测试n。如果累加器变量非零-1，则一旦完成，将输出a，否则输出a 0。
在?::`}:(执行开始时，仅执行一次：
?::`}:( Main,Aux
? - take an integer from STDIN and place it onto Main [[n],[]]
: - duplicate top of Main [[n,n],[]]
: - duplicate top of Main [[n,n,n],[]]
` - negate top of Main [[n,n,-n],[]]
} - place top of Main onto Aux [[n,n],[-n]]
: - duplicate top of Main [[n,n,n],[-n]]
( - decrement top of Main [[n,n,n-1],[-n]]
下一条指令"No是空操作，但我们有3条相邻指令，因此我们根据Main顶部的值进行分支，零将我们向前，而非零将我们向右。
如果输入是1我们，则前进，因为Main的顶部为零：
(!@ Main,Aux
( - decrement top of Main [[1,1,-1],[-1]]
! - print top of Main, a -1
@ - exit the labyrinth
但是，如果输入大于1我们右转，因为Main的顶部非零：
:} Main,Aux
: - duplicate top of Main [[n,n,n-1,n-1],[-n]]
} - place top of Main onto Aux [[n,n,n-1],[-n,n-1]]
此时，我们有一个三邻分支，但我们知道n-1它不是零，所以我们右转...
"% Main,Aux
" - no-op [[n,n,n-1],[-n,n-1]]
% - place modulo result onto Main [[n,n%(n-1)],[-n,n-1]]
- ...i.e we've got our first divisibility indicator n%(n-1), an
- accumulator, a=-n, and our potential divisor p=n-1:
- [[n,n%(n-1)],[a,p]]
我们现在在的另一个三邻分支机构%。
如果的结果为%非零，我们将减除我们的潜在除数p=p-1，然后将累加器保留a为：
;:{(:""}" Main,Aux
; - drop top of Main [[n],[a,p]]
: - duplicate top of Main [[n,n],[a,p]]
{ - place top of Aux onto Main [[n,n,p],[a]]
- three-neighbour branch but n-1 is non-zero so we turn left
( - decrement top of Main [[n,n,p-1],[a]]
: - duplicate top of Main [[n,n,p-1,p-1],[a]]
"" - no-ops [[n,n,p-1,p-1],[a]]
} - place top of Main onto Aux [[n,n,p-1],[a,p-1]]
" - no-op [[n,n,p-1],[a,p-1]]
% - place modulo result onto Main [[n,n%(p-1)],[a,p-1]]
- ...and we branch again according to the divisibility
- of n by our new potential divisor, p-1
...但是，如果的结果%为零（仅当时为第一次通过n=2），我们直接继续将除数添加到累加器中，然后a=a+p递减潜在的除数p=p-1：
;:{:{+}}""""""""{(:""} Main,Aux
; - drop top of Main [[n],[a,p]]
: - duplicate top of Main [[n,n],[a,p]]
{ - place top of Aux onto Main [[n,n,p],[a]]
: - duplicate top of Main [[n,n,p,p],[a]]
{ - place top of Aux onto Main [[n,n,p,p,a],[]]
+ - perform addition [[n,n,p,a+p],[]]
} - place top of Main onto Aux [[n,n,p],[a+p]]
} - place top of Main onto Aux [[n,n],[a+p,p]]
""""""" - no-ops [[n,n],[a+p,p]]
- a branch, but n is non-zero so we turn left
" - no-op [[n,n],[a+p,p]]
{ - place top of Aux onto Main [[n,n,p],[a+p]]
- we branch, but p is non-zero so we turn right
( - decrement top of Main [[n,n,p-1],[a+p]]
: - duplicate top of Main [[n,n,p-1,p-1],[a+p]]
"" - no-ops [[n,n,p-1,p-1],[a+p]]
} - place top of Main onto Aux [[n,n,p-1],[a+p,p-1]]
在这一点上，如果p-1仍然非零，我们向左转：
"% Main,Aux
" - no-op [[n,n,p-1],[a+p,p-1]]
% - modulo [[n,n%(p-1)],[a+p,p-1]]
- ...and we branch again according to the divisibility
- of n by our new potential divisor, p-1
...但是如果p-1为零，我们会直接:进入迷宫的第二行（您之前已经看过所有说明，所以我不再赘述它们，只是给出效果）：
:":}"":({):""}"%;:{:{+}}"""""""{{{ Main,Aux
: - [[n,n,0,0],[a,0]]
" - [[n,n,0,0],[a,0]]
- top of Main is zero so we go straight
- ...but we hit the wall and so turn around
: - [[n,n,0,0,0],[a,0]]
} - [[n,n,0,0],[a,0,0]]
- top of Main is zero so we go straight
"" - [[n,n,0,0],[a,0,0]]
: - [[n,n,0,0,0],[a,0,0]]
( - [[n,n,0,0,-1],[a,0,0]]
{ - [[n,n,0,0,-1,0],[a,0]]
- top of Main is zero so we go straight
- ...but we hit the wall and so turn around
( - [[n,n,0,0,-1,-1],[a,0]]
: - [[n,n,0,0,-1,-1,-1],[a,0]]
"" - [[n,n,0,0,-1,-1,-1],[a,0]]
} - [[n,n,0,0,-1,-1],[a,0,-1]]
- top of Main is non-zero so we turn left
" - [[n,n,0,0,-1,-1],[a,0,-1]]
% - (-1)%(-1)=0 [[n,n,0,0,0],[a,0,-1]]
; - [[n,n,0,0],[a,0,-1]]
: - [[n,n,0,0,0],[a,0,-1]]
{ - [[n,n,0,0,0,-1],[a,0]]
: - [[n,n,0,0,0,-1,-1],[a,0]]
{ - [[n,n,0,0,0,-1,-1,0],[a]]
+ - [[n,n,0,0,0,-1,-1],[a]]
} - [[n,n,0,0,0,-1],[a,-1]]
} - [[n,n,0,0,0],[a,-1,-1]]
""""""" - [[n,n,0,0,0],[a,-1,-1]]
- top of Main is zero so we go straight
{ - [[n,n,0,0,0,-1],[a,-1]]
{ - [[n,n,0,0,0,-1,-1],[a]]
{ - [[n,n,0,0,0,-1,-1,a],[]]
现在，它{具有三个相邻的指令，所以...
...如果a为零，那将是完美的n，那么我们继续：
"!@ Main,Aux
" - [[n,n,0,0,0,-1,-1,a],[]]
- top of Main is a, which is zero, so we go straight
! - print top of Main, which is a, which is a 0
@ - exit the labyrinth
...如果a为非零值（对于非完美值）n，那么我们向左转：
_~"!@ Main,Aux
_ - place a zero onto Main [[n,n,0,0,0,-1,-1,a,0],[]]
~ - bitwise NOT top of Main (=-1-x) [[n,n,0,0,0,-1,-1,a,-1],[]]
" - [[n,n,0,0,0,-1,-1,a,-1],[]]
- top of Main is NEGATIVE so we turn left
! - print top of Main, which is -1
@ - exit the labyrinth