Mamy wiele poziomych osi dla liczb, ale szczerze uważam, że są one trochę nudne. Twoim zadaniem dzisiaj jest zbudowanie części osi ukośnej między dwiema różnymi liczbami całkowitymi nieujemnymi podanymi jako dane wejściowe.

Jak zbudować oś przekątną?

• Weźmy przykład z danymi wejściowymi 0, 5 . Nasza oś powinna wyglądać następująco:

  0
   1
    2)
     3)
      4
       5
  

• Nasza oś powinna jednak dobrze wyglądać w przypadku liczb, które mają więcej cyfr! Jeśli na przykład wejściem jest, 0, 14nowa oś powinna być:

  0
   1
    2)
     3)
      4
       5
        6
         7
          8
           9
            10
              11
                12
                  13
                    14
  

• Chodzi o to, że pierwsza cyfra następnego numeru na osi zawsze musi być umieszczona dokładnie za ostatnią cyfrą poprzedniego numeru. Aby jeszcze lepiej zrozumieć ten pomysł, oto kolejny przykład 997, 1004:

  997
     998
        999
           1000
               1001
                   1002
                       1003
                           1004
  

Zasady

• Możesz założyć, że dane wejściowe są w porządku rosnącym lub malejącym (możesz wybierać między 5,3i 3,5).

• Możesz również założyć, że różnica między dwiema liczbami całkowitymi jest mniejsza niż 100.

• Możesz mieć wiodącą nową linię lub spójną spację wiodącą (w każdej linii). Spacje końcowe / znaki nowej linii również są w porządku.

• Domyślne luki są zabronione.

• Możesz przyjmować dane wejściowe i dostarczać dane wyjściowe dowolnym standardowym środkiem .

• To jest golf golfowy , więc wygrywa najkrótszy kod w bajtach w każdym języku!

Inne przypadki testowe

• 1, 10:

  1
   2)
    3)
     4
      5
       6
        7
         8
          9
           10
  

• 95, 103:

  95
    96
      97
        98
          99
            100
               101
                  102
                     103
  

• 999999, 1000009:

  999999
        1000000
               1000001
                      1000002
                             1000003
                                    1000004
                                           1000005
                                                  1000006
                                                         1000007
                                                                1000008
                                                                       1000009
  

05AB1E, 8 7 6 bytes

Thanks to Magic Octopus Urn for saving a byte!

It somehow works, but honestly I have no idea why.

Code

Ÿvy.O=

Uses the 05AB1E encoding. Try it online!

Explanation

Ÿ          # Create the range [a, .., b] from the input array
 vy        # For each element
   .O      #   Push the connected overlapped version of that string using the
                 previous version of that string. The previous version initially
                 is the input repeated again. Somehow, when the input array is
                 repeated again, this command sees it as 1 character, which gives
                 the leading space before each line outputted. After the first
                 iteration, it reuses on what is left on the stack from the
                 previous iteration and basically attaches (or overlaps) itself 
                 onto the previous string, whereas the previous string is replaced 
                 by spaces and merged into the initial string. The previous string
                 is then discarded. We do not have to worry about numbers overlapping 
                 other numbers, since the incremented version of a number never
                 overlaps entirely on the previous number. An example of 123 and 456:

                 123
                    456

                 Which leaves us "   456" on the stack.
     =     #   Print with a newline without popping

Python 2, 43 bytes

lambda a,b:'\v'.join(map(str,range(a,b+1)))

Makes use of vertical tab to make the ladder effect. The way thet \v is rendered is console dependent, so it may not work everywhere (like TIO).

Charcoal, 9 8 bytes

Ｆ…·ＮＮ⁺¶ι

Try it online!

Link is to the verbose version of the code. Input in ascending order.

• 1 byte saved thanks to ASCII-only!

R, 70 69 61 bytes

function(a,b)for(i in a:b){cat(rep('',F),i,'
');F=F+nchar(i)}

Function that takes the start and end variable as arguments. Loops over the sequence, and prints each element, prepended with enough spaces. F starts as FALSE=0, and during each iteration, the amount of characters for that value is added to it. F decides the amount of spaces printed.

Try it online!

-8 bytes thanks to @Giuseppe

C#, 90 89 85 bytes

s=>e=>{var r="";for(int g=0;e>s;g+=(s+++"").Length)r+="".PadLeft(g)+s+"\n";return r;}

Saved 1 byte thanks to @LiefdeWen.
Saved 4 bytes thanks to @auhmaan.

Try it online!

Full/Formatted version:

namespace System
{
    class P
    {
        static void Main()
        {
            Func<int, Func<int, string>> f = s => e =>
            {
                var r = "";
                for (int g = 0; e > s; g += (s++ + "").Length)
                    r += "".PadLeft(g) + s + "\n";

                return r;
            };

            Console.WriteLine(f(0)(5));
            Console.WriteLine(f(0)(14));
            Console.WriteLine(f(997)(1004));
            Console.WriteLine(f(1)(10));
            Console.WriteLine(f(95)(103));
            Console.WriteLine(f(999999)(1000009));

            Console.ReadLine();
        }
    }
}

Python 2, 58 54 bytes

def f(a,b,s=''):print s;b<a or f(a+1,b,' '*len(s)+`a`)

Try it online!

Mathematica, 59, bytes

Grid[(DiagonalMatrix@Range[1+##]/. 0->""+1)-1,Spacings->0]&

input

[10,15]

-3 bytes @JungHwanMin
problem with 0 fixed (see comments for details)
thanx to @ngenisis

Jelly, 9 bytes

rD⁶ṁ$;¥\Y

Try it online!

Mathematica, 48 bytes

Rotate[""<>Table[ToString@i<>" ",{i,##}],-Pi/4]&

since there are so many answers, I thought this one should be included

input

[0,10]

output

C, 166 134 95 82 Bytes

New Answer

Just as a function not as a whole program.

f(a,b){int s=0,i;while(a<=b){i=s;while(i--)printf(" ");s+=printf("%i\n",a++)-1;}}

Thanks to Falken for helping knock off 13 Bytes (and fix a glitch)!

Thanks to Steph Hen for helping knock off 12 Bytes!

Thanks to Zacharý for help knock off 1 Byte!

Old Answers

Got rid of the int before main and changed const char*v[] to char**v and got rid of return 0;

main(int c,char**v){int s=0;for(int a=atoi(v[1]);a<=atoi(v[2]);a++){for(int i=0;i<s;i++)printf(" ");printf("%i\n",a);s+=log10(a)+1;}}


int main(int c,const char*v[]){int s=0;for(int a=atoi(v[1]);a<=atoi(v[2]);a++){for(int i=0;i<s;i++)printf(" ");printf("%i\n",a);s+=log10(a)+1;}return 0;}

This is my first time golfing and I wanted to try something in C. Not sure if I formatted this correctly, but I had fun making it!

int main(int c, const char * v[]) {
    int s = 0;
    for(int a=atoi(v[1]); a<=atoi(v[2]); a++) {
        for(int i=0; i<s; i++) printf(" ");
        printf("%i\n",a);
        s += log10(a)+1;
    }
    return 0;
}

Explanation

int s = 0; // Number of spaces for each line

for(int a=atoi(argv[1]); a<=atoi(argv[2]); a++) { // Loop thru numbers

for(int i=0; i<s; i++) printf(" "); // Add leading spaces

printf("%i\n",a); // Print number

s += log10(a)+1; // Update leading spaces

Usage

C++, 167 165 bytes

-2 bytes thanks to Zacharý

#include<string>
#define S std::string
S d(int l,int h){S r;for(int m=0,i=l,j;i<=h;){for(j=0;j<m;++j)r+=32;S t=std::to_string(i++);r+=t;r+=10;m+=t.size();}return r;}

APL (Dyalog), 25 24 bytes

-1 thanks to Zacharý.

Assumes ⎕IO←0 for zero based counting. Takes the lower bound as left argument and the upper bound as right argument.

{↑⍵↑⍨¨-+\≢¨⍵}(⍕¨⊣+∘⍳1--)

Try it online!

(…) apply the following tacit function between the arguments:

 - subtract the upper lower from the upper bound

 1- subtract that from one (i.e. 1 + ∆)

 ⊣+∘⍳ left lower bound plus the integers 0 through that

 ⍕¨ format (stringify) each

{…} apply the following anonymous on that (represented by ⍵):

 ≢¨ length of each (number)

 +\ cumulative sum

 - negate

 ⍵↑⍨¨ for each stringified number, take that many characters from the end (pads with spaces)

 ↑ mix list of strings into character matrix

Retina, 81 78 bytes

.+
$*
+`\b(1+)¶11\1
$1¶1$&
1+
$.& $.&
 (.+)
$.1$* 
+1`( *)(.+?)( +)¶
$1$2¶$1$3

Try it online! Takes input as a newline-separated list of two integers. Edit: Saved 3 bytes by stealing the range-expansion code from my answer to Do we share the prime cluster? Explanation:

.+
$*

Convert both inputs to unary.

+`\b(1+)¶11\1
$1¶1$&

While the last two elements (a, b) of the list differ by more than 1, replace them with (a, a+1, b). This expands the list from a tuple into a range.

1+
$.& $.&

Convert back to decimal in duplicate.

 (.+)
$.1$* 

Convert the duplicate copy to spaces.

+1`( *)(.+?)( +)¶
$1$2¶$1$3

Cumulatively sum the spaces from each line to the next.

Pyth, 14 13 bytes

V}FQ
=k+*dlkN

Try it online!

LOGO, 53 bytes

[for[i ? ?2][repeat ycor[type "\ ]pr :i fd count :i]]

There is no "Try it online!" link because all online LOGO interpreter does not support template-list.

That is a template-list (equivalent of lambda function in other languages).

Usage:

apply [for[i ? ?2][repeat ycor[type "\ ]pr :i fd count :i]] [997 1004]

(apply calls the function)

will print

997
   998
      999
         1000
             1001
                 1002
                     1003
                         1004

Note:

This uses turtle's ycor (Y-coordinate) to store the number of spaces needed to type, therefore:

• The turtle need to be set to home in its default position and heading (upwards) before each invocation.
• window should be executed if ycor gets too large that the turtle moves off the screen. Description of window command: if the turtle is asked to move past the boundary of the graphics window, it will move off screen., unlike the default setting wrap, which if the turtle is asked to move past the boundary of the FMSLogo screen window, it will "wrap around" and reappear at the opposite edge of the window.

Explanation:

for[i ? ?2]        Loop variable i in range [?, ?2], which is 2 input values
repeat ycor        That number of times
type "\            space character need to be escaped to be typed out.
pr :i              print the value of :i with a newline
fd count :i        increase turtle's y-coordinate by the length of the word :i. (Numbers in LOGO are stored as words)

05AB1E, 8 bytes

Ÿʒ¾ú=þv¼

Try it online!

-2 thanks to Adnan.

JavaScript (ES8), 69 67 62 bytes

Takes input as integers, in ascending order, using currying syntax. Returns an array of strings.

x=>y=>[...Array(++y-x)].map(_=>s="".padEnd(s.length)+x++,s="")

Try it

o.innerText=(f=

x=>y=>[...Array(++y-x)].map(_=>s="".padEnd(s.length)+x++,s="")

)(i.value=93)(j.value=105).join`\n`
oninput=_=>o.innerText=f(Math.min(i.value,j.value))(Math.max(i.value,j.value)).join`\n`

label,input{font-family:sans-serif}input{margin:0 5px 0 0;width:100px;}

<label for=i>x: </label><input id=i type=number><label for=j>y: </label><input id=j type=number><pre id=o>

Japt, 12 bytes

òV
£¯Y ¬ç +X

Takes input in either order and always returns the numbers in ascending order, as an array of lines.

Try it online! with the -R flag to join the array with newlines.

Explanation

Implicit input of U and V.

òV
£

Create inclusive range [U, V] and map each value to...

¯Y ¬ç

The values before the current (¯Y), joined to a string (¬) and filled with spaces (ç).

+X

Plus the current number. Resulting array is implicitly output.

Python 2, 65 63 62 61 bytes

-2 bytes Thanks to @Mr. Xcoder: exec doesn't need braces

-1 bye thanks to @Zacharý: print s*' ' as print' '*s

def f(m,n,s=0):exec(n-m+1)*"print' '*s+`m`;s+=len(`m`);m+=1;"

Try it online!

JavaScript, 57 bytes

f=(x,y,s='')=>y>=x?s+`
`+f(x+1,y,s.replace(/./g,' ')+x):s

Python 2, 60 59 bytes

-1 byte thanks to Mr.Xcoder for defining my s=0 as an optional variable in my function.

def f(l,u,s=0):
 while l<=u:print' '*s+`l`;s+=len(`l`);l+=1

Try it online!

I think it is possible to transfer this into a lambda version, but I do not know how. I also think that there is some sort of mapping between the spaces and the length of the current number, but this I also did not figure out yet. So I think there still is room for improvement.

What i did was creating a range from the lowerbound lto the upper bound u printing each line with a space multiplied with a number s. I am increasing the multiplier with the length of the current number.

Python 2, 78 77 79 bytes

def f(a,b):
 for i in range(a,b+1):print sum(len(`j`)for j in range(i))*' '+`i`

Try it online!

f(A, B) will print the portion of the axis between A and B inclusive.

First time I answer a challenge!

Uses and abuses Python 2's backticks to count the number of spaces it has to add before the number.

-1 byte thanks to Mr.Xcoder

+2 because I forgot a +1

C (gcc), 41 38 bytes

-3 bytes Thanks to ASCII-only

t(x,v){while(x<=v)printf("%d\v",x++);}

Works on RedHat6, accessed via PuTTY

Try it online!

V, 16 bytes

ÀñÙywÒ $pça/jd

Try it online!

This would be way easier if I could take start end - start but I think that's changing the challenge a bit too much.

This takes the start number as input in the buffer and the end number as an argument. It actually creates the ladder from start to start + end and then deletes everything after the end number.

MATL, 11 bytes

vii&:"t~@Vh

Try it online!

Explanation

This works by generating a string for each number and concatenating it with a logically-negated copy of the previous string. Thus char 0 is prepended 0 as many times as the length of the previous string. Char 0 is displayed as a space, and each string is displayed on a different line

v       % Concatenate stack (which is empty): pushes []
ii      % Input two numbers
&:      % Range between the two numbers
"       % For each
  t     %   Duplicate
  ~     %   Logical negation. This gives a vector of zeros
  @     %   Push current number
  V     %   Convert to string
  h     %   Concatenate with the vector of zeros, which gets automatically 
        %   converted into chars.
        % End (implicit). Display stack (implicit), each string on a diferent
        % line, char 0 shown as space

Swift 4, 115 bytes

I think nobody would have posted a Swift solution anyway...

func f(l:Int,b:Int){for i in l...b{print(String(repeating:" ",count:(l..<i).map{String($0).count}.reduce(0,+)),i)}}

Try it online!

Perl, 19 bytes

Note: \x0b is counted as one byte.

Along with others, I thought using cursor movements would be the shortest route, this does mean it doesn't work on TIO:

print"$_\x0b"for<>..<>

Usage

perl -e 'print"$_\x0b"for<>..<>' <<< '5
10'
5
 6
  7
   8
    9
     10

Japt, 10 9 bytes

òV åÈç +Y

Test it online! Returns an array of lines; -R flag included to join on newlines for easier viewing.

Explanation

 òV åÈ   ç +Y
UòV åXY{Xç +Y}   Ungolfed
                 Implicit: U, V = inputs, P = empty string
UòV              Create the range [U, U+1, ..., V-1, V].
    åXY{     }   Cumulative reduce: Map each previous result X and current item Y to:
        Xç         Fill X with spaces.
           +Y      Append Y.
                 Implicit: output result of last expression

Old version, 10 bytes:

òV £P=ç +X

Test it online!

 òV £  P= ç +X
UòV mX{P=Pç +X}  Ungolfed
                 Implicit: U, V = inputs, P = empty string
UòV              Create the range [U, U+1, ..., V-1, V].
    mX{       }  Map each item X to:
         Pç        Fill P with spaces.
            +X     Append X.
       P=          Re-set P to the result.
                   Implicitly return the same.
                 Implicit: output result of last expression

D, 133 127 126 125 121 119 bytes

import std.conv,std.stdio;void f(T)(T a,T b,T s=0){for(T j;j++<s;)' '.write;a.writeln;if(a-b)f(a+1,b,s+a.text.length);}

Jelly and APL were taken.

Try it online!

If you're fine with console-dependent results (goes off the same principle as Giacomos's C answer) here's one for 72 71 bytes:

import std.stdio;void f(T)(T a,T b){while(a<=b){a++.write;'\v'.write;}}

How? (Only D specific tricks)

• f(T)(T a,T b,T s=0) D's template system can infer types
• for(T j;j++<s;) Integers default to 0.
• ' '.write;a.writeln D lets you call fun(arg) like arg.fun (one of the few golfy things D has)
• a.text.length Same as above, and D also allows you to call a method with no parameters as if it was a property (text is conversion to string)
• One thing that might be relevant (I didn't use this though) newlines can be in strings!

Java 8, 79 bytes

(a,b)->{for(String s="";a<=b;System.out.printf("%"+s.length()+"d\n",a++))s+=a;}

Try it online!