Ben n-üçlü üçlü n döndüren ve formu olan bir üçlü olarak tanımlar:

1 ? 2 ? 3 ? n - 1 ? n : 0 : 0 : 0  # n - 1 zeroes

N girdisi verildiğinde bir işlev veya tam program yazın ve n. Üçüncüyü döndürür. Kod-Golf.

testcases

0 #=> undefined behaviour
1 #=> 1
2 #=> 1 ? 2 : 0
3 #=> 1 ? 2 ? 3 : 0 : 0
10 #=> 1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ? 9 ? 10 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0

1000'inci üçlü, bence bir çeşit zen uyumu var.

Pyth - 19 18 17 bayt

Alanlar beni öldürüyor, onları idare etmenin daha iyi bir yolunu düşünüyor.

+j" ? "SQ*tQ" : 0

Sadece sayıları a " ? "ile birleştirir ve sonra ikinci kısmı birleştirir.

+              String concatenation
 j" ? "        Join by the string
  SQ           1-indexed inclusive range to input
 *             String repetition
  tQ           Input - 1
  " : 0        String implicitly closed by end of program

Burada çevrimiçi deneyin .

CJam, 18 18 bayt

ri,:)":?0"*2/ze_S*

Çevrimiçi deneyin .

açıklama

ri,:)          e# Generate the list 1..n.
":?0"*         e# Insert ":?0" between every two numbers.
2/             e# Split into pairs, e.g. 1:, ?0, 2:, ?0, ..., ?0, n.
z              e# First items in every pair before second items in every pair.
e_             e# Concatenate the two parts.
S*             e# Insert spaces.

Ruby, 31 bayt

f=->n{[*1..n]*' ? '+' : 0'*~-n}

Ölçek:

> f[1]
=> "1"
> f[7]
=> "1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 : 0 : 0 : 0 : 0 : 0 : 0"

CJam, 19 bayt

Sadece bir başlangıç ​​...

ri_,:)'?*\(":0"*+S*

Nasıl çalışır

ri_                       e# Read the number as integer and make a copy of it on stack
   ,:)                    e# Convert the copy to array [1 .. n]
      '?*                 e# Join the numbers with a '?'. So we have [1 '? 2 '? ... '? n]
         \(               e# Swap the stack to get original integer on top. Decrement it by 1
           ":0"*          e# Get n-1 repeated ":0" string
                +S*       e# Join the two strings and fill it with spaces. 

Buradan çevrimiçi deneyin

Brainfuck, 305

(STDIN'den giriş numarası olmadan, alttaki düzenlemeye bakın)

-[->+>+<<]>>>++++[>++++++++<-]>[<+>-]+++++++[>+++++++++<-]>[<+>-]++++++[>++++++++
<-]>[<+>-]++<<<<>>>+.-<<<[>.>.<.>>>>>++++++++++<<[->+>-[>+>>]>[+[-<+>]>+>>]<<<<<<
]>>[-]>>>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]>>[>++++++[-<++++++++>]<.<<
+>+>[-]]<[<[->-<]++++++[->++++++++<]>.[-]]<<++++++[-<++++++++>]<.[-]<<[-<+>]<+<<<
<-]>>-----<<<[->>.>.<.>>.<<<<]

Bu güzel algoritmayı, tüm programın 155 baytını alan bir sayı yazdırmak için kullandım .

32768'e kadar girişler için çalışır (algoritmanın 16 bit sınırlaması). Sondaki boşlukları üretmez ve girdi 1için de çalışır :

input    output
0        [infinite loop (til underflow)]
1        "1"
2        "1 ? 2 : 0"
4        "1 ? 2 ? 3 ? 4 : 0 : 0 : 0"
etc.

Hızlı gözden geçirme:

Kurulum (97 bayt)

-                                 Decrease input (position 0)
[->+>+<<]  >>>                    Copy input twice to the right and 
                                      shift 3 positions to the right
++++[>++++++++<-]>   [<+>-]       Precalculate number 32 (=" ") at position 3
+++++++[>+++++++++<-]>  [<+>-]    Precalculate number 63 (="?") at position 4
++++++[>++++++++<-]>    [<+>-]    Precalculate number 48 (="0") at position 5
++<<<<                            Precalculate number 2 for later use. This number
                                      will be printed in each iteration. (position 6)

İlk bölüm (181 bayt)

>>>+.-<<<                Go to the char "0" we saved, increase it, print it,
                             decrease it and go back (this prints "1" everytime)
[                        While our second copy of the number isn't zero
    >.>.<.>>>                Move to " ", print, move to "?", print,
                                 move to " " again, print, move to our
                                 number at the end which is initially 2

    [>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]
        ++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<
        [.[-]<]<             Algorithm to print the number at current position

    +<<<<                    Increase our number at the end and return to the beginning
-]                       Decrease the loop variable

İkinci kısım (27 bayt)

>>-----<<<        Move to our "?" char and decrease it by 5 to get ":"
[-                While our first copy of the number isn't zero decrease it
   >>.>.<.>>.<<<<     Print " ", print ":", print " ", print "0"
]

8 Brainfuck komutunu 3 bit ile eşleştirmeye izin verilirse, bu program 114 bayt ve 3 bit daha alacak

Tekli, ~ 4.08 * 10 ^ 275 bayt

Çok uzun burada olurdu, ama sadece 408452257862560239329948606295286361112603208650130608525040044700379331457759667646985586658469601803889628246410788572492437928714867190270708935427798983714797786123292750743771225096145575210320040188155473030775033228313350778616384531426430459309802833775612506568528463 sıfır olduğunu ve Brainfuck programı olarak aynı şekilde çalışır.

EDIT : Ben berbat, bu program gerçekten kullanıcı girişi almaz, sadece geçerli işaretçi değerini giriş olarak kullanır. Bir sayıyı ayrıştırabilmek için çok daha fazlası gerekirdi ve bunu yapmak beni rahatsız etmiyor.

Bu nedenle, doğrudan programa girilen bir değerle çalışır (programdan önce n kez "+" ekleyerek), ancak STDIN ile çalışmaz

JavaScript (ES6), 42 39 bayt

f=m=>(g=k=>k-m?k+` ? ${g(k+1)} : 0`:m)(1)

Dış işlev fgiriş değerini alır ve sonra temel işlevi gtest etmek için giriş değerini maksimum olarak kullanarak dizeyi ortadan kaldırmak için iç işlevi yinelemeli olarak çağırır .

Ungolfed:

function f(max) {
    function g(count) {
        if(count==max) {
            // base case: return max for the center
            return max;
        } else {
            // recursive case: build outer shell around center
            return count + " ? " + g(count+1) + " : 0";
        }
    }

    return g(1);
}

Python 56 55

t=lambda n:' ? '.join(map(str,range(1,n+1)))+' : 0'*~-n

C # - 76

Func<int,string>F=k=>{var r="";for(;k>1;)r=" ? "+k--+r+" : 0";return"1"+r;};

Haskell, 53 bayt

g n='1':foldr(\x s->" ? "++show x++s++" : 0")""[2..n]

Nasıl çalışır: boş bir dize ile başlayıp mevcut sayıyı ve a'yı ekleyerek ve a ekleyerek aşağıdan döngüye geçerek dizeyi içeriden dışarıya ndoğru oluşturun . Sonunda her şeyin önüne bir koy .2?: 01

Farklı bir yaklaşım (@Mauris sayesinde artık 9 bayt daha kısa):

Haskell, 60 51 bayt

 f n='1':((=<<[2..n])=<<[(" ? "++).show,\x->" : 0"])

Nasıl çalışır: bir değişmez 1ardından ? <x>her biri için <x>de [2..n]bir sabit ardından : 0her biri için <x>de [2..n].

Pyth, 17 bytes

jd.iSQs*RtQ,\?":0

Demonstration.

Julia, 44 31 bytes

n->join(1:n," ? ")*" : 0"^(n-1)

This creates an unnamed function that accepts an integer as input and returns a string. To call it, give it a name, e.g. f=n->....

First we join together the integers 1 to n, separating each with ? and spaces into a single string. Then we append to this the string " : 0" repeated n-1 times.

Examples:

julia> f(1)
"1"

julia> f(3)
"1 ? 2 ? 3 : 0 : 0"

julia> f(0)
can't repeat a string -1 times

JavaScript ES7, 62 bytes

n=>[for(i of Array(n).keys())i+1].join` ? `+' : 0'.repeat(n-1)

I don't know if I can golf this more. But it's a pretty straightforward solution

Firefox only:

var f=n=>[for(i of Array(n).keys())i+1].join` ? `+' : 0'.repeat(n-1)

alert(f(+prompt('Input: ')));

ES5 equivalent:

// Most browsers now support .repeat
String.prototype.repeat = String.prototype.repeat || function(n){var _n = '', i = 0; for (;i < n; i += 1){_n+=this};return _n}
                                                             //Function                         
function f(n){a=[];for(i of Array(n).keys()){a.push(i+1)};return a.join(' ? ')+' : 0'.repeat(n-1)}

alert(f(+prompt('Input: ')))

CoffeeScript, 52 bytes

f=(n)->s='';s=' ? '+n--+s+' : 0'while n;s.slice 3,-4

Explanation

f=(n)->
 s = ''                                # initialize string
 s = ' ? ' + n-- + s + ' : 0' while n  # prepend and append in decrementing loop
 s.slice 3,-4                          # chop off leading ?, trailing 0 and whitespace

SWI-Prolog, 90 bytes

a(X):-Y is X-1,\+ (between(1,Y,L),\+writef('%w ? ',[L])),write(X),writef('%r',[' : 0',Y]).

Definitely not going to win, but the \+ (between(1,TopBound,N),\+do_something(N)) construction is pretty interesting to repeat something on a sequence of integers.

Swift 145 (135 w/o whitespace)

func t(n:Int) -> String {
    let a = (1..<n).reverse().reduce("") {" ? \($1)\($0) : 0"}
    return a.substringFromIndex(advance(a.startIndex, 3))
}

Can you believe the part to substring is actually longer than the part to produce the expression.

Perl, 36 bytes

say join(" ? ",1..$_)." : 0"x($_-1)

35 characters +1 for -n.

Run with:

echo 10 | perl -nE'say join(" ? ",1..$_)." : 0"x($_-1)'

Java, 71

I couldn't help myself after commenting on RCB's answer. So here's another Java (71 like wow when is Java not the longest!)

String t(int n){String s=""+n;for(;--n>0;)s=n+" ? "+s+" : 0";return s;}

Java, 125 88 bytes

Original

String f(int n){if(n==1)return"1";String s="",e="";for(int i=1;i<n;i++){s+=i+" ? ";e+=i==n-1?": 0":": 0 ";}return s+n+" "+e;}

With nicer formatting and variable names:

String nAry(int n) {
    if (n == 1) {
        return "1";
    }
    String start = "", end = "";
    for (int i = 1; i < n; i++) {
        start += i + " ? ";
        end += (i == n - 1) ? ": 0" : ": 0 ";
    }
    return start + n + " " + end;
}

Improved - Thanks to Jack Ammo's comments below:

String f(int n){String s="",e=s;for(int i=1;i<n;){s+=i+++" ? ";e+=" : 0";}return s+n+e;}

JavaScript (ES6), 59 bytes

Same approach as my CoffeeScript answer, using template strings help. String.prototype.repeat costs too many characters.

f=n=>{for(s=``;n;)s=` ? ${n--+s} : 0`;return s.slice(3,-4)}

Demo

Firefox only for now, as it is ES6.

f=n=>{for(s=``;n;)s=` ? ${n--+s} : 0`;return s.slice(3,-4)}

// DEMO
console.log = x => document.body.innerHTML += '<p>' + x

console.log(f(1));
console.log(f(3));
console.log(f(10));

K, 36 bytes

{(3_,/(" ? ",)'$!x),(4*-1+x)#" : 0"}

Python 2, 63 60 58 56

Try it here

Easy solution: (63)

n=input()
for i in range(n-1):print-~i,'?',
print`n`+' : 0'*~-n

Edit: I really wanted to try a recursive function. Here it is: (56)

f=lambda n,c=1:`c`+(' ? '+f(n,c+1)if c<n else~-n*' : 0')

Edit: Anyone know why this isn't working? I tried a list with an index of c<n, but that didn't work because of a stack overflow error. Same with this:

f=lambda n,c=1:`c`+((c<n)*(' ? '+f(n,c+1))or~-n*' : 0')

rs, 77 bytes

(\d+)/(_)^^(\1)
+^_(_+)/\1 _\1
_(_+)$/_\1( : 0)^^((^^\1))
(__+)/? (^^\1)
^./1

Live demo and test cases.

Explanation:

(\d+)/(_)^^(\1)

Expand the number into a series of N underscores.

+^_(_+)/\1 _\1

Repeatedly create a range of underscores, separated by spaces. e.g. This would turn ___ into _ __ ___.

_(_+)$/_\1( : 0)^^((^^\1))

Append to the last set of underscores (of length N) N-1 instances of : 0.

(__+)/? (^^\1)

Replace each group of underscores by it's length, preceded by ?, EXCEPT for the first one.

^./1

Replace the first one with the number 1.

Because of the format, this also handles 0 well: it just prints the empty string.

Swift, 79 75 bytes

let f={{$0+$1}((1..<$0).reduce(("1","")){($0.0+" ? \($1+1)",$0.1+" : 0")})}

f is implicitly declared as a function with one Int parameter which returns a String

Works with n >= 1 and crashes at runtime when n == 0. There are no trailing whitespaces

Edit: Managed to remove 2*2 characters, because string interpolation isn't always the shortest

Note to edit: This code takes forever (it doesn't stop) to compile, but it definitely would if the compiler would be able to handle it. Take a look at the version before this edit to get one that compiles

><>, 32 + 3 = 35 bytes

:l(?vln" ? "ooo0$
"ooo>nl?!;" : 

Note that there is a trailing space on the second line. The +3 is for the -v flag, e.g. run like

$ py -3 fish.py ternary.py -v 2
1 ? 2 : 0

Taking input as a code point like

i:l(?vln" ? "ooo0$!
 "ooo>nl?!;" :

is 34 bytes, but I prefer the above version since it's easier to test and it won't win anyway.

Explanation

There's quite a bit of pseudo-recursion and abuse going on, so let's take a look.

The first line prints the "1 ? 2 ? ... n-1 ? " part. The stack starts off with just the input n, thanks to the -v flag, and we do the following:

:l(?v           If (length of stack + 1 > n), go to the second line
ln              Print the length of the stack
" ? "ooo        Print the reverse of " ? " (but hey, palindromes)
0$              Push 0 and swap, keeping n on top and increasing the 
                length of the stack by 1

><> is toroidal, so the above executes in a loop until the stack consists of n at the top with n-1 zeroes below, at which point it moves to the second line.

The first time the second line is executed, the n instruction is run, printing the n at the top of the stack. This leaves just the n-1 zeroes, and we do the following, also in a loop:

l?!;            If the stack is empty, terminate
" : "ooo        Print the reverse of " : " (but hey, palin...)
n               Print one of the 0s, decreasing the stack's length by 1
                This reuses the same n instruction from before

Scala, 78 71 52 50 bytes

def f(n:Int)=(1 to n).mkString(" ? ")+" : 0"*(n-1)

Objective-C, 346 bytes

-(void)printTernaryOfInt:(int)ternary{NSMutableString *outString=@"".mutableCopy; for (int i=1;i<=ternary;i++) {[outString appendString:[NSString stringWithFormat:@" ? %i",i]];}[outString deleteCharactersInRange:NSMakeRange(0, 2)];for (int i=1;i<ternary;i++) {[outString appendString:[NSString stringWithFormat:@" : 0"]];}NSLog(@"%@",outString);}

Putting in 0 for the int or anything negative raises an NSRangeException due to outString containing nil. This should run on iOS 2.0 and later and many of the latest versions of Mac OS X.

A breakdown of the code:

-(void)printTernaryOfInt:(int)ternary{ ... }

Standard function declaration in Objective-C.

NSMutableString *outString=@"".mutableCopy;

Makes a string for output to go to, outString, and makes it mutable. (In other words, it can be read and written to.

for (int i=1;i<=ternary;i++) {[outString appendString:[NSString stringWithFormat:@" ? %i",i]];}

Adds the first part of the string to output.

[outString deleteCharactersInRange:NSMakeRange(0, 2)];

Cleans up the beginning of the string to make sure ? 1 is replaced with 1. Note: if 0 was given, this is where the NSRangeException would occur, due to there not being an index 1.

for (int i=1;i<ternary;i++) {[outString appendString:[NSString stringWithFormat:@" : 0"]];}

Adds the second part of the string to the string.

NSLog(@"%@",outString);}

Spits the string back out using NSLog and closes off the function.

Output:

Inputting 0 gives this crash log:

    2015-07-11 05:15:28.036 Example App[41665:2134488] *** Terminating app due to uncaught exception 'NSRangeException', reason: '-[__NSCFString deleteCharactersInRange:]: Range or index out of bounds'
*** First throw call stack:
(
    0   CoreFoundation                      0x009b5746 __exceptionPreprocess + 182
    1   libobjc.A.dylib                     0x0063ea97 objc_exception_throw + 44
    2   CoreFoundation                      0x009b566d +[NSException raise:format:] + 141
    3   CoreFoundation                      0x00981813 mutateError + 259
    4   CoreFoundation                      0x009818c1 -[__NSCFString deleteCharactersInRange:] + 65
    5   Example App                         0x000e3785 -[ViewController printTernaryOfInt:] + 277
    6   Example App                         0x000e3645 -[ViewController placeOrder:] + 133
    7   libobjc.A.dylib                     0x006547cd -[NSObject performSelector:withObject:withObject:] + 84
    8   UIKit                               0x00d75a40 -[UIApplication sendAction:to:from:forEvent:] + 99
    9   UIKit                               0x00d759d2 -[UIApplication sendAction:toTarget:fromSender:forEvent:] + 64
    10  UIKit                               0x00eb613a -[UIControl sendAction:to:forEvent:] + 69
    11  UIKit                               0x00eb6557 -[UIControl _sendActionsForEvents:withEvent:] + 598
    12  UIKit                               0x00eb57c1 -[UIControl touchesEnded:withEvent:] + 660
    13  UIKit                               0x00dcdcaa -[UIWindow _sendTouchesForEvent:] + 874
    14  UIKit                               0x00dce786 -[UIWindow sendEvent:] + 792
    15  UIKit                               0x00d8c681 -[UIApplication sendEvent:] + 242
    16  UIKit                               0x00d9cab8 _UIApplicationHandleEventFromQueueEvent + 21484
    17  UIKit                               0x00d702e7 _UIApplicationHandleEventQueue + 2300
    18  CoreFoundation                      0x008d706f __CFRUNLOOP_IS_CALLING_OUT_TO_A_SOURCE0_PERFORM_FUNCTION__ + 15
    19  CoreFoundation                      0x008ccb7d __CFRunLoopDoSources0 + 253
    20  CoreFoundation                      0x008cc0d8 __CFRunLoopRun + 952
    21  CoreFoundation                      0x008cba5b CFRunLoopRunSpecific + 443
    22  CoreFoundation                      0x008cb88b CFRunLoopRunInMode + 123
    23  GraphicsServices                    0x029e42c9 GSEventRunModal + 192
    24  GraphicsServices                    0x029e4106 GSEventRun + 104
    25  UIKit                               0x00d740b6 UIApplicationMain + 1526
    26  Example App                         0x000e3cfa main + 138
    27  libdyld.dylib                       0x02d76ac9 start + 1
    28  ???                                 0x00000001 0x0 + 1
)
libc++abi.dylib: terminating with uncaught exception of type NSException

1 gives this:

2015-07-11 05:06:02.360 Example App[41665:2134488]  1

2 gives this:

2015-07-11 05:06:07.613 Example App[41665:2134488]  1 ? 2 : 0

7 gives this:

2015-07-11 05:06:12.147 Example App[41665:2134488]  1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 : 0 : 0 : 0 : 0 : 0 : 0

200 gives this:

2015-07-11 05:06:35.552 Example App[41665:2134488]  1 ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ? 9 ? 10 ? 11 ? 12 ? 13 ? 14 ? 15 ? 16 ? 17 ? 18 ? 19 ? 20 ? 21 ? 22 ? 23 ? 24 ? 25 ? 26 ? 27 ? 28 ? 29 ? 30 ? 31 ? 32 ? 33 ? 34 ? 35 ? 36 ? 37 ? 38 ? 39 ? 40 ? 41 ? 42 ? 43 ? 44 ? 45 ? 46 ? 47 ? 48 ? 49 ? 50 ? 51 ? 52 ? 53 ? 54 ? 55 ? 56 ? 57 ? 58 ? 59 ? 60 ? 61 ? 62 ? 63 ? 64 ? 65 ? 66 ? 67 ? 68 ? 69 ? 70 ? 71 ? 72 ? 73 ? 74 ? 75 ? 76 ? 77 ? 78 ? 79 ? 80 ? 81 ? 82 ? 83 ? 84 ? 85 ? 86 ? 87 ? 88 ? 89 ? 90 ? 91 ? 92 ? 93 ? 94 ? 95 ? 96 ? 97 ? 98 ? 99 ? 100 ? 101 ? 102 ? 103 ? 104 ? 105 ? 106 ? 107 ? 108 ? 109 ? 110 ? 111 ? 112 ? 113 ? 114 ? 115 ? 116 ? 117 ? 118 ? 119 ? 120 ? 121 ? 122 ? 123 ? 124 ? 125 ? 126 ? 127 ? 128 ? 129 ? 130 ? 131 ? 132 ? 133 ? 134 ? 135 ? 136 ? 137 ? 138 ? 139 ? 140 ? 141 ? 142 ? 143 ? 144 ? 145 ? 146 ? 147 ? 148 ? 149 ? 150 ? 151 ? 152 ? 153 ? 154 ? 155 ? 156 ? 157 ? 158 ? 159 ? 160 ? 161 ? 162 ? 163 ? 164 ? 165 ? 166 ? 167 ? 168 ? 169 ? 170 ? 171 ? 172 ? 173 ? 174 ? 175 ? 176 ? 177 ? 178 ? 179 ? 180 ? 181 ? 182 ? 183 ? 184 ? 185 ? 186 ? 187 ? 188 ? 189 ? 190 ? 191 ? 192 ? 193 ? 194 ? 195 ? 196 ? 197 ? 198 ? 199 ? 200 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0 : 0

C, 84 78 bytes

C, even though it is not the shortest, as a function:

i;f(n){while(++i<n)printf("%i ? ",i);printf("%i",n);while(--i)printf(" : 0");}

In the name of golfing, the int type specifier is left off of i, f, and n because it is the default. i can be uninitialized because it is a global variable and defaults to zero. f does not return a value, but that only causes a warning. printf is not #include'd. To run, here is a full program version:

#include <stdio.h>

i;f(n){while(++i<n)printf("%i ? ",i);printf("%i",n);while(--i)printf(" : 0");}

int main(int argc, char *argv[]){
    if(argc != 2){
        return 1;
    }
    f(atoi(argv[1]));
    puts("");
}

C, 63 bytes

Reusable function, takes n as an argument.

i;f(n){for(i=1;i<2*n;i++)printf(i-1?n/i?" ? %d":" : 0":"1",i);}

Ungolfed and commented (pretty straightforward):

int f(int n) {
    int i;

    // 1 ... n, n+1 ... 2n-1
    for(i = 1; i < 2*n; i++) {
        // If i == 1, prints "1"
        // If i <= n, prints " ? %d", i (i = 2 ... n)
        // Else, prints " : 0" (i = n+1 ... 2n-1)
        printf(
            i-1 ?
                n/i ?
                    " ? %d" :
                    " : 0" :
                "1",
        i);
    }
}

Common Lisp, 84

(format t "~{~A ? ~}~@*~{~[~;~:;0~^ ? ~]~}" (loop for i from 1 to (read) collect i))

First, (loop for i from 1 to (read) collect i) generates a list of integers from 1 to whatever is put in, which is used as the only argument to the function. But the real magic of it is in the control string that looks like line noise. "~{~A ? ~}" iterates over the entire list stored within the first argument, outputting each number with the ? for the first half. ~@* resets the argument list to the first argument. ~{~[~;~:;0~^ ? ~]~} reiterates over the list, outputting 0 ? for each argument consumed, but outputting nothing if the argument is 0 or 1.