您的任务是编写一个打印ASCII三角形的程序或函数。他们看起来像这样：

|\
| \
|  \
----

您的程序将采用单个数字输入n，并带有约束0 <= n <= 1000。上面的三角形的值为n=3。

ASCII三角形将具有n反斜杠（\）和竖线（|），n+1线和破折号（-），并且每行除最终行外还将具有等于行号（从0开始，即第一行为行0）的空格。 。

例子：

输入：

4

输出：

|\
| \
|  \
|   \
-----

输入：

0

输出：

在此测试用例中，输出必须为空。没有空格。

输入：

1

输出：

|\
--

输入和输出必须完全是我指定的方式。

这是代码高尔夫球，因此请争取尽可能短的代码！

Jelly, 14 bytes

’⁶x⁾|\jṄµ€Ṫ”-ṁ

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How it works.

’⁶x⁾|\jṄµ€Ṫ”-ṁ  Main link. Argument: n

        µ       Combine the links to the left into a chain.
         €      Map the chain over [1, ..., n]; for each k:
’                 Decrement; yield k-1.
 ⁶x               Repeat the space character k-1 times, yielding a string.
   ⁾\j            Join the character array ['|', '\'], separating by those spaces.
      Ṅ           Print the result, followed by a linefeed.
         Ṫ      Tail; extract the last line.
                This will yield 0 if the array is empty.
          ⁾-ṁ   Mold the character '-' like that line (or 0), yielding a string
                of an equal amount of hyphen-minus characters.  

C, 58 bytes

i;f(n){for(i=2*n;~i--;printf(i<n?"-":"|%*c\n",2*n-i,92));}

--

Thanks to @Steadybox who's comments on this answer helped me shave a few bytes in my above solution

Javascript (ES6), 97 85 81 75 74 bytes

n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\
`+s):s)(n,"")+"-".repeat(n&&n+1)

Turns out I wasn't using nearly enough recursion

f=n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\
`+s):s)(n,"")+"-".repeat(n&&n+1)

console.log(f(0))
console.log(f(1))
console.log(f(2))
console.log(f(3))
console.log(f(4))

05AB1E, 16 15 16 bytes

Saved a byte thanks to Adnan

FðN×…|ÿ\}Dg'-×»?

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Explanation

F       }         # for N in range [0 ... input-1]
 ðN×              # push <space> repeated N times
    …|ÿ\          # to the middle of the string "|\"
         Dg       # get length of last string pushed
           '-×    # repeat "-" that many times
              »   # join strings by newline
               ?  # print without newline

V, 18 17 16 bytes

1 byte saved thanks to @nmjcman101 for using another way of outputting nothing if the input is 0

é\é|ÀñÙá ñÒ-xÀ«D

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Hexdump:

00000000: e95c e97c c0f1 d9e1 20f1 d22d 78c0 ab44  .\.|.... ..-x..D

Explanation (outdated)

We first have a loop to check if the argument is 0. If so, the code below executes (|\ is written). Otherwise, nothing is written and the buffer is empty.

Àñ     ñ            " Argument times do:
  é\é|              " Write |\
      h             " Exit loop by creating a breaking error

Now that we got the top of the triangle, we need to create its body.

Àñ   ñ              " Argument times do:
  Ù                 " Duplicate line, the cursor comes down
   à<SPACE>         " Append a space

Now we got one extra line at the bottom of the buffer. This has to be replaced with -s.

Ó-                  " Replace every character with a -
   x                " Delete the extra '-'

This answer would be shorter if we could whatever we want for input 0

V, 14 13 bytes

é\é|ÀñÙá ñÒ-x

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C#, 93 bytes

n=>{var s=n>0?new string('-',n+1):"";while(n-->0)s="|"+new string(' ',n)+"\\\n"+s;return s;};

Anonymous function which returns the ASCII triangle as a string.

Full program with ungolfed, commented function and test cases:

using System;

class ASCIITriangles
{
    static void Main()
    {
      Func<int, string> f =
      n =>
      {
          // creates the triangle's bottom, made of dashes
          // or an empty string if n == 0
          var s = n > 0 ? new string('-', n + 1) : "";

          // a bottom to top process
          while ( n-- > 0)
          // that creates each precedent line
            s = "|" + new string(' ', n) + "\\\n" + s;

          // and returns the resulting ASCII art
          return s;
      };

      // test cases:
      Console.WriteLine(f(4));
      Console.WriteLine(f(0));
      Console.WriteLine(f(1));
    }
}

Python 2, 69 bytes

lambda x:'\n'.join(['|'+' '*n+'\\'for n in range(x)]+['-'*-~x*(x>0)])

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CJam, 24 22 21 bytes

Saved 1 byte thanks to Martin Ender

ri_{S*'|\'\N}%\_g+'-*

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Explanation

ri                     e# Take an integer from input
  _                    e# Duplicate it
   {                   e# Map the following to the range from 0 to input-1
    S*                 e#   Put that many spaces
      '|               e#   Put a pipe
        \              e#   Swap the spaces and the pipe
         '\            e#   Put a backslash
           N           e#   Put a newline
            }%         e# (end of map block)
              \        e# Swap the top two stack elements (bring input to the top)
               _g+     e# Add the input's signum to itself. Effectively this increments any 
                       e#  non-zero number and leaves zero as zero.
                  '-*  e# Put that many dashes

SmileBASIC, 51 bytes

INPUT N
FOR I=0TO N-1?"|";" "*I;"\
NEXT?"-"*(N+!!N)

PowerShell, 51 67 bytes

param($n)if($n){1..$n|%{"|"+" "*--$_+"\"};write-host -n ('-'*++$n)}

Try it online!

(Byte increase to account for no trailing newline)

Takes input $n and verifies it is non-zero. Then loops to construct the triangle, and finishes with a line of -. Implicit Write-Output happens at program completion.

Retina, 39 bytes

.*
$*
*\`(?<=(.*)).
|$.1$* \¶
1
-
-$
--

Try it online

Convert decimal input to unary. Replace each 1 with |<N-1 spaces>\¶, print, and undo replace. Replace each 1 with a hyphen, and the last hyphen with 2 hyphens. Tadaa!

Common Lisp, 89 86 bytes

Creates an anonymous function that takes the n input and prints the triangle to *standard-output* (stdout, by default).

Golfed

(lambda(n)(when(< 0 n)(dotimes(i n)(format t"|~v@t\\~%"i))(format t"~v,,,'-<~>"(1+ n))))

Ungolfed

(lambda (n)
  (when (< 0 n)
    (dotimes (i n)
      (format t "|~v@t\\~%" i))
    (format t "~v,,,'-<~>" (1+ n))))

I'm sure I could make this shorter somehow.

C 101 93 75 bytes

f(n){i;for(i=0;i++<n;)printf("|%*c\n",i,92);for(;n--+1;)prin‌​tf("-");}

Ungolfed version

void f(int n)
{
  int i;

  for(i=0;i++<n;)
    printf("|%*c\n",i,92);

  for(;n--+1;)
    printf("-");

}

@Steadybox Thanks for pointing out, makes a lot of sense.

Charcoal, 15 bytes

Ｎβ¿β«↓β→⁺¹β↖↖β»

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Breakdown

Ｎβ¿β«↓β→⁺¹β↖↖β»
Ｎβ               assign input to variable β
   ¿β«         »  if β != 0:
      ↓β           draw vertical line β bars long
        →⁺¹β       draw horizontal line β+1 dashes long
            ↖      move cursor up one line and left one character
             ↖β    draw diagonal line β slashes long

Japt, 20 bytes

Saved 2 bytes thanks to @ETHproductions

o@'|+SpX +'\Ãp-pUÄ)·

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Explanation

o@'|+SpX +'\Ãp-pUÄ)·
o                       // Creates a range from 0 to input
 @                      // Iterate through the array
  '|+                   // "|" + 
     SpX +              // S (" ") repeated X (index) times +
          '\Ã            // "\" }
             p-pU       // "-" repeated U (input) +1 times
                 Ä)·    // Join with newlines

J, 20 bytes

-13 bytes thanks to bob

*#' \|-'{~3,~2,.=@i.

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original: 33 bytes

(#&'| \'@(1,1,~])"0 i.),('-'#~>:)

ungolfed

(#&'| \' @ (1,1,~])"0 i.) , ('-'#~>:)

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Pyke, 18 17 bytes

I Fd*\|R\\s)Qh\-*

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Python2, 73 bytes

n=input()
w=0
exec'print"|"+" "*w+"\\\\"+("\\n"+"-"*-~n)*(w>n-2);w+=1;'*n

A full program. I also tried string interpolation for the last line, but it turned out be a couple bytes longer :/

exec'print"|%s\\\\%s"%(" "*w,("\\n"+"-"*-~n)*(w>n-2));w+=1;'*n

Another solution at 73 bytes:

n=j=input()
exec'print"|"+" "*(n-j)+"\\\\"+("\\n"+"-"*-~n)*(j<2);j-=1;'*n

Test cases

0:

1:
|\
--

2:
|\
| \
---

3:
|\
| \
|  \
----

6:
|\
| \
|  \
|   \
|    \
|     \
-------

MATL, 19 bytes

?'\|- '2GXyYc!3Yc!)

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?         % Implicit input. If non-zero
  '\|- '  %   Push this string
  2       %   Push 2
  G       %   Push input
  Xy      %   Identity matrix of that size
  Yc      %   Prepend a column of 2's to that matrix
  !       %   Transpose
  3       %   Push 3
  Yc      %   Postpend a column of 3's to the matrix
  !       %   Transpose
  )       %   Index into string
          % Implicit end. Implicit display

QBIC, 41 bytes

:~a>0|[a|?@|`+space$(b-1)+@\`][a+1|Z=Z+@-

Explanation

:~a>0|  Gets a, and checks if a > 0
        If it isn't the program quits without printing anything
[a|     For b=1; b <= a; b++
?@|`+   Print "|"
space$  and a number of spaces
(b-1)   euqal to our current 1-based line - 1
+@\`    and a "\"
]       NEXT
[a+1|   FOR c=1; c <= a+1; c++
Z=Z+@-  Add a dash to Z$
        Z$ gets printed implicitly at the end of the program, if it holds anything
        The last string literal, IF and second FOR loop are closed implicitly.

R, 101 bytes

for(i in 1:(n=scan())){stopifnot(n>0);cat("|",rep(" ",i-1),"\\\n",sep="")};cat("-",rep("-",n),sep="")

This code complies with the n=0 test-case if you only consider STDOUT !
Indeed, the stopifnot(n>0) part stops the script execution, displays nothing to STDOUT but writes Error: n > 0 is not TRUE to SDTERR.

Ungolfed :

for(i in 1:(n=scan()))
    {
    stopifnot(n>0)
    cat("|", rep(" ", i-1), "\\\n", sep = "")
    }

cat("-", rep("-", n), sep = "")

Python 2, 62 bytes

n=input();s='\\'
exec"print'|'+s;s=' '+s;"*n
if n:print'-'*-~n

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Prints line by line, each time adding another space before the backslash. If a function that doesn't print would be allowed, that would likely be shorter.

JavaScript (ES6), 71 bytes

f=
n=>console.log(' '.repeat(n).replace(/./g,'|$`\\\n')+'-'.repeat(n+!!n))

<form onsubmit=f(+i.value);return!true><input id=i type=number><input type=submit value=Go!>

Outputs to the console. Save 6 bytes if printing to the SpiderMonkey JavaScript shell is acceptable. Save 13 bytes if returning the output is acceptable.

Python 2, 67 bytes

Another function in Python 2, using rjust.

lambda n:('|'.join(map('\\\n'.rjust,range(n+2)))+'-'*-~n)[4:]*(n>0)

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Python 3, 60 bytes

f=lambda n,k=0:k<n and'|'+' '*k+'\\\n'+f(n,k+1)or'-'[:n]*-~n

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Two more solutions with the same byte count.

f=lambda n,k=0:n and'|'+' '*k+'\\\n'+f(n-1,k+1)or-~k*'-'[:k]
f=lambda n,s='|':-~n*'-'[:n]if s[n:]else s+'\\\n'+f(n,s+' ')

Perl, 63 bytes

$n=shift;print'|',$"x--$_,"\\\n"for 1..$n;print'-'x++$n,$/if$n

Ungolfed:

$ perl -MO=Deparse triangle.pl
$n = shift @ARGV;
print '|', $" x --$_, "\\\n" foreach (1 .. $n);
print '-' x ++$n, $/ if $n;

$" is the list separator, which defaults to " ". $/ is the output record separator, which defaults to "\n". $_ is the implicit loop variable.

Haskell, 82 65 bytes

g 0=""
g n=((take n$iterate(' ':)"\\\n")>>=('|':))++([0..n]>>"-")

Try it online! Usage:

Prelude> g 4
"|\\\n| \\\n|  \\\n|   \\\n-----"

Or more nicely:

Prelude> putStr $ g 4
|\
| \
|  \
|   \
-----

Pyth, 23 18 bytes

VQ++\|*dN\\)IQ*\-h

Test suite available online.
Thanks to Ven for golfing off 5 bytes.

Explanation

VQ++\|*dN\\)IQ*\-h
 Q           Q    Q  [Q is implicitly appended, initializes to eval(input)]
       d             [d initializes to ' ' (space)]
VQ         )         For N in range(0, eval(input)):
      *dN             Repeat space N times
   +\|                Prepend |
  +      \\           Append \
                      Implicitly print on new line
            IQ       If (input): [0 is falsy, all other valid inputs are truthy]
                 hQ   Increment input by 1
              *\-     Repeat - that many times
                      Implicitly print on new line

Javascript 101(Full Program), 94(Function Output), 79(Return) bytes

Full Program

Will not run in Chrome (as process doesn't exist apparently)
Will not run in TIO (as prompt apparently isn't allowed)

x=prompt();s='';for(i=0;i<x;i++)s+='|'+' '.repeat(i)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))

Function with EXACT print

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))}

Try it Online

Function with return string

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;return s+'-'.repeat(x&&x+1)}

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Repeating characters in Javascript is dumb and so is suppressing newlines on output

Python 2, 82 bytes

def f(i,c=0):
 if c<i:print'|'+' '*c+'\\';f(i,c+1)
 print'-'*((c+1,c)[c<1]);exit()

Try it online!

Longer that the other Python answers but a recursive function just to be different.

It feels wasteful using two print statements but I can't find a shorter way round it. Also the exit() wastes 7 to stop it printing decreasing number of - under the triangle.