Make an Emergency Corridor

Question

In some nations there are recommendations or laws on how to form emergency corridors on streets that have multiple lanes per direction. (In the following we only consider the lanes going in the direction we are travelling.) These are the rules that hold in Germany:

• If there is only one lane, everyone should drive to the right such that the rescue vehicles can pass on the left.
• If there are two or more lanes, the cars on the left most lane should drive to the left, and everyone else should move to the right.

Challenge

Given the number N>0 of regular lanes, output the layout of the lanes when an emergency corridor is formed using a string of N+1 ASCII characters. You can use any two characters from ASCII code 33 up to 126, one to denote the emergency corridor, and one for denoting the cars. Trailing or leading spaces, line breaks etc are allowed.

Examples

Here we are using E for the emergency corridor, and C for the cars.

N  Output
1  EC
2  CEC
3  CECC
4  CECCC
5  CECCCC
6  CECCCCC
   etc

Answer 1

Python 2, 29 26 bytes

lambda n:10**n*97/30-1/n*9

Example:

>>> f(1)
23
>>> f(2)
323
>>> f(3)
3233

Answer 2

Python 3, 35 33 bytes

lambda N:'C'*(N>1)+'EC'+'C'*(N-2)

Edit: dropping f= to save 2 bytes, thanks to @dylnan's reminder.

Try it online!

To visualize it:

lambda N:''*(N>1)+''+''*(N-2)

Output:

1 
2 
3 
4 
5 
6 

Try online!

Python 3, 40 bytes

A straightforward solution:

lambda N:str(10**N).replace('100','010')

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Answer 3

C (gcc), 32 bytes

f(n){printf(".%.*f"+1%n,n-1,0);}

Try it online!

Uses 0 and . characters:

.0
0.0
0.00
0.000
0.0000

Answer 4

Japt, 5 4 bytes

Uses q for cars and + for the corridor.

ç¬iÄ

Try it

Credit to Oliver who golfed 4 bytes off at the same time as I did.

---

Explanation

A short solution but a tricky explanation!

The straightforward stuff first: The ç method, when applied to an integer, repeats its string argument that number of times. The i method takes 2 arguments (s & n) and inserts s at index n of the string it's applied to.

Expanding the 2 unicode shortcuts used gives us çq i+1, which, when transpiled to JS becomes U.ç("q").i("+",1), where U is the input. So we're repeating q U times and then inserting a + at index 1.

The final trick is that, thanks to Japt's index wrapping, when U=1, i will insert the + at index 0, whatever value you feed it for n.

Answer 5

R, 50 bytes

-11 thanks to Giuseppe!

pryr::f(cat("if"(x<2,12,c(21,rep(2,x-1))),sep=""))

Outputs 1 for emergency corridor and 2 for normal lanes

My attempt, 61 bytes

Nothing fancy to see here, but let's get R on the scoreboard =)

q=pryr::f(`if`(x<2,cat("EC"),cat("CE",rep("C",x-1),sep="")))

Usage:

q(5)
CECCCC

Answer 6

Haskell, 38 34 32 bytes

f 1="EC"
f n="CE"++("C"<*[2..n])

Try it online!

Answer 7

Python 2, 30 29 28 bytes

lambda n:`10/3.`[1/n:n-~1/n]

Print 3 instead of C and . instead of E.

Explanation:

Try it online.

lambda n:    # Method with integer parameter and string return-type
  `10/3.`    #  Calculate 10/3 as decimal (3.333333333) and convert it to a string
  [1/n       #   Take the substring from index 1 if `n=1`, 0 otherwise
   ,n-~      #   to index `n+1` +
       1/n]  #    1 if `n=1`, 0 otherwise

---

Python 2, 33 32 31 29 28 bytes

lambda n:1%n-1or'1-'+'1'*~-n

Prints 1 instead of C and - instead of E.

-2 bytes thanks to @ovs.
-1 byte thanks to @xnor.

Explanation:

Try it online.

lambda n:    # Method with integer parameter and string return-type
  1%n-1      #  If `n` is 1: Return '-1'
  or         #  Else:
    '1-'+    #   Return '1-', appended with:
    '1'*~-n  #   `n-1` amount of '1's

Answer 8

Pyth, 10 9 8 bytes

Xn1Q*NQZ

Uses 0 to denote the emergency corridor and ".
Try it here

Explanation

Xn1Q*NQZ
    *NQ     Make a string of <input> "s.
 n1Q         At index 0 or 1...
X      Z    ... Insert 0.

Answer 9

Octave (MATLAB*), 31 30 28 27 22 bytes

@(n)'CE'(1+(n>1==0:n))

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The program works as follows:

@(n)                   %Anonymous function to take input
            n>1==0:n   %Creates [1 0] if n is 1, or [0 1 (0 ...)] otherwise
         1+(        )  %Converts array of 0's and 1's to 1-indexed
    'CE'(            ) %Converts to ASCII by addressing in string

The trick used here is XNORing the seed array of 0:n with a check if the input is greater than 1. The result is that for n>1 the seed gets converted to a logical array of [0 1 (0 ...)] while for n==1 the seed becomes inverted to [1 0], achieving the necessary inversion.

The rest is just converting the seed into a string with sufficient appended cars.

---

(*) The TIO link includes in the footer comments an alternate solution for the same number of bytes that works in MATLAB as well as Octave, but it results in a sequence of '0' and '1' rather than 'E' and 'C'. For completeness, the alternate is:

@(n)['' 48+(n>1==0:n)]

---

• Saved 1 byte by using n==1~=0:1 rather than 0:1~=(n<2). ~= has precedence over <, hence the original brackets, but is seems that ~= and == are handled in order of appearance so by comparing with 1 we can save a byte.

• Saved 2 bytes by changing where the negation of 2:n is performed. This saves a pair of brackets. We also have to change the ~= to == to account for the fact that it will be negated later.

• Saved 1 byte using < again. Turns out that < has same precedence as == after all. Placing the < calculation before the == ensures correct order of execution.

• Saved 5 bytes by not creating two separate arrays. Instead relying on the fact that the XNOR comparison will convert a single range into logicals anyway.

Answer 10

brainfuck, 42 bytes

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

Try it online!

Takes input as char code and outputs as V being normal lanes and W being the cleared lane. (To test easily, I recommend replacing the , with a number of +s)

How it Works:

,[[>]+[<]>-] Turn input into a unary sequence of 1s on the tape
>>[<]<[<]    Move two cells left of the tape if input is larger than 1
             Otherwise move only one space
>+>+<        Add one to the two cells right of the pointer
             This transforms:
               N=1:  0 0' 1 0  -> 0 2' 1 0
               N>1:  0' 0 1 1* -> 0 1' 2 1*
[<-[--->+<]>.,>]  Add 86 to each cell to transform to Ws and Vs and print

Answer 11

Haskell, 35 33 32 bytes

2 bytes saved thanks to Angs, 1 byte saved thanks to Lynn

(!!)$"":"EC":iterate(++"C")"CEC"

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Haskell, 32 30 29 bytes

This is zero indexed so it doesn't comply with the challenge

g=(!!)$"EC":iterate(++"C")"CEC"

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Haskell, 30 bytes

This doesn't work because output needs to be a string

f 1=21
f 2=121
f n=10*f(n-1)+1

Try it online!

Here we use numbers instead of strings, 2 for the emergency corridor, 1 for the cars. We can add a 1 to the end by multiplying by 10 and adding 1. This is cheaper because we don't have to pay for all the bytes for concatenation and string literals.

It would be cheaper to use 0 instead of 1 but we need leading zeros, which end up getting trimmed off.

Answer 12

APL (Dyalog Unicode), 21 17 16 bytes

(-≠∘1)⌽'E',⍴∘'C'

Try it online!

Thanks to Erik for saving 4 bytes and Adám for one further byte.

How?

(-≠∘1)⌽'E',⍴∘'C' ⍝ Tacit function
           ⍴∘'C' ⍝ Repeat 'C', according to the input
       'E',      ⍝ Then append to 'E'
      ⌽          ⍝ And rotate
    1)           ⍝ 1
  ≠∘             ⍝ Different from the input? Returns 1 or 0
(-               ⍝ And negate. This rotates 0 times if the input is 1, and once if not.

Answer 13

Jelly, 11 9 bytes

’0ẋ⁾0E;ṙỊ

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Full program.

Uses 0 instead of C.

Answer 14

C (gcc), 39 bytes

f(n){printf("70%o"+!n,7|(1<<3*--n)-1);}

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Borrowed and adapted the printf trick from ErikF's answer.

Answer 15

Python 3, 32 bytes

lambda n:f"{'CE'[n<2:]:C<{n+1}}"

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Uses an f-string expression to format either'E'or'CE' padded on the right with'C'so it has width ofn+1.

f"{          :       }    a Python 3 f-string expression.
   'CE'[n<2:]             string slice based on value of n.
             :            what to format is before the ':' the format is after.
              C           padding character
               <          left align
                {n+1}     minimum field width based on n

Answer 16

Brain-Flak, 100 66 bytes

{({}[()]<((((()()()()){}){}){}())>)}{}(({}<>)())<>{<>{({}<>)<>}}<>

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Uses " as the emergency lane and ! as the normal lanes.

Answer 17

C#, 34 bytes

n=>n++<2?"EC":"CE".PadRight(n,'C')

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Answer 18

05AB1E, 7 bytes

Î>∍1I≠ǝ

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0 is C and 1 is E.

Explanation

Î>          # Push 0 and input incremented            -- [0, 4]
  ∍         # Extend a to length b                    -- [0000]
   1I≠      # Push 1 and input falsified (input != 1) -- [0000, 1, 1] 
      ǝ     # Insert b in a at location C             -- [0100]
            # Implicit display

Answer 19

Python 3, 30 29 bytes

lambda n:"CEC"[~n:]+"C"*(n-2)

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OK, there is a lot of Python answers already, but I think this is the first sub-30 byter among those still using "E" and "C" chars rather than numbers.

Answer 20

JavaScript (Node.js), 28 bytes

n=>n<2?21:"12".padEnd(n+1,1)

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Answer 21

Stax, 7 bytes

ü♣àj#F 

Run and debug it

This uses the characters "0" and "1". This works because when you rotate an array of size 1, it doesn't change.

Unpacked, ungolfed, and commented, it looks like this.

1]( left justify [1] with zeroes. e.g. [1, 0, 0, 0]
|)  rotate array right one place
0+  append a zero
$   convert to string

Run this one

Answer 22

APL+WIN, 20 16 bytes

4 bytes saved thanks to Adám

Prompts for integer n:

(-2≠⍴n)⌽n←1⎕/⍕10

1 for emergency corridor o for cars.

Answer 23

Perl 5 -p, 27 20 19 bytes

$_=1x$_;s/1?\K1/E1/

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Saved a byte by using 1 for the cars and E for the emergency corridor.

Answer 24

APL (Dyalog Unicode), 16 bytes

≡∘1⌽'CE',1↓⍴∘'C'

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Answer 25

JavaScript (Node.js), 19 bytes

n=>--n?"0"+10**n:10

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Answer 26

Whitespace, 141 104 103 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_number][T    T   T   _Retrieve][S S S T  S N
_Push_2][T  S S T   _Subtract][S N
S _Duplicate_input-2][N
T   T   N
_If_negative_Jump_to_Label_-1][S S S T  N
_Push_1][S N
S _Duplicate_1][T   N
S T _Print_as_integer][S S T    T   N
_Push_-1][T N
S T _Print_as_integer][T    S S T   _Subtract][N
S S T   N
_Create_Label_LOOP][S N
S _Duplicate][N
T   T   S N
_If_negative_Jump_to_EXIT][S S S T  N
_Push_1][S N
S _Duplicate_1][T   N
S T _Print_as_integer][T    S S T   _Subtract][N
S N
T   N
_Jump_to_LOOP][N
S S N
_Create_Label_-1][T N
S T _Print_as_integer][N
S S S N
_Create_Label_EXIT]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Prints 1 instead of C and - instead of E.

-1 byte thanks to @JoKing by suggesting the use of 1 and -1 instead of 0 and 1.

Explanation in pseudo-code:

Integer i = STDIN-input as integer - 2
If i is negative (-1):
  Print i (so print "-1")
Else:
  Print "1-1"
  Start LOOP:
    If i is negative:
      EXIT program
    Print "1"
    i = i-1
    Go to the next iteration of the LOOP

Example runs:

Input: 1

Command   Explanation                 Stack      Heap    STDIN   STDOUT   STDERR

SSSN      Push 0                      [0]
SNS       Duplicate top (0)           [0,0]
TNTT      Read STDIN as integer       [0]        {0:1}   1
TTT       Retrieve heap at 0          [1]        {0:1}
SSSTSN    Push 2                      [1,2]      {0:1}
TSST      Subtract top two            [-1]       {0:1}
SNS       Duplicate input-2           [-1,-1]    {0:1}
NTSN      If neg.: Jump to Label_-1   [-1]       {0:1}
NSSN      Create Label_-1             [-1]       {0:1}
TNST      Print top as integer        []         {0:1}           -1
NSSSN     Create Label_EXIT           []         {0:1}
                                                                         error

Try it online (with raw spaces, tabs and new-lines only).
Stops with error: Exit not defined.

Input: 4

Command   Explanation                   Stack      Heap    STDIN   STDOUT   STDERR

SSSN      Push 0                        [0]
SNS       Duplicate top (0)             [0,0]
TNTT      Read STDIN as integer         [0]        {0:4}   4
TTT       Retrieve heap at 0            [4]        {0:4}
SSSTSN    Push 2                        [4,2]      {0:4}
TSST      Subtract top two              [2]        {0:4}
SNS       Duplicate input-2             [2,2]      {0:4}
NTSN      If neg.: Jump to Label_-1     [2]        {0:4}
SSSTN     Push 1                        [2,1]      {0:4}
SNS       Duplicate top (1)             [2,1,1]    {0:4}
TNST      Print as integer              [2,1]      {0:4}           1
SSTTN     Push -1                       [2,1,-1]   {0:4}
TNST      Print as integer              [2,1]      {0:4}           -1
TSST      Subtract top two              [1]        {0:4}
NSSTN     Create Label_LOOP             [1]        {0:4}
 SNS      Duplicate top (1)             [1,1]      {0:4}
 NTTSN    If neg.: Jump to Label_EXIT   [1]        {0:4}
 SSSTN    Push 1                        [1,1]      {0:4}
 SNS      Duplicate top (1)             [1,1,1]    {0:4}
 TNST     Print as integer              [1,1]      {0:4}           1
 TSST     Subtract top two              [0]        {0:4}
 NSNTN    Jump to Label_LOOP            [0]        {0:4}

 SNS      Duplicate top (0)             [0,0]      {0:4}
 NTTSN    If neg.: Jump to Label_EXIT   [0]        {0:4}
 SSSTN    Push 1                        [0,1]      {0:4}
 SNS      Duplicate top (1)             [0,1,1]    {0:4}
 TNST     Print as integer              [0,1]      {0:4}           1
 TSST     Subtract top two              [-1]       {0:4}
 NSNTN    Jump to Label_LOOP            [-1]       {0:4}

 SNS      Duplicate top (-1)            [-1,-1]    {0:4}
 NTTSN    If neg.: Jump to Label_EXIT   [-1]       {0:4}
NSSSN     Create Label_EXIT             [-1]       {0:4}
                                                                            error

Try it online (with raw spaces, tabs and new-lines only).
Stops with error: Exit not defined.

Answer 27

Jelly, 6 bytes

⁵*ṾṙỊṙ

Displays car lanes as 0, the emergency lane as 1.

Try it online!

How it works

⁵*ṾṙỊṙ  Main link. Argument: n

⁵*      Compute 10**n.
  Ṿ     Uneval; get a string representation.
   ṙỊ   Rotate the string (n≤1) characters to the left.
     ṙ  Rotate the result n characters to the left.

Answer 28

AutoHotkey 32 bytes

Replaces the letter "C" with "EC" unless amount of C > 1, then it sends "CEC" and exits the app.

::C::EC
:*:CC::CEC^c
^c::ExitApp

C => EC
CC => CEC then exits the program. Any further Cs will be entered after the program exits.

Answer 29

J, 11 bytes

1":1&<=i.,]

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Based on ngn’s comment.

and :

1&<,~/@A.'',~''$~,&4

Answer 30

MathGolf, 7 6 bytes

ú░\┴╜╪

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Output 1 for E and 0 for C.

Explanation:

ú         # 10 to the power of the (implicit) input
          #  i.e. 1 → 10
          #  i.e. 4 → 10000
 ░        # Convert it to a string
          #  i.e. 10 → "10"
          #  i.e. 10000 → "10000"
  \       # Swap so the (implicit) input is at the top of the stack again
   ┴╜     # If the input is NOT 1:
     ╪    #  Rotate the string once towards the right
          #   i.e. "10000" and 4 → "01000"
          # Output everything on the stack (which only contains the string) implicitly

Answer 31

Canvas, 11 bytes

┤C×EC×╴╷？C×

Try it here!

Answer 32

[GNU sed 4.2.2], 14 bytes

• Score includes 1 byte for the -r sed parameter

• 3 bytes saved thanks to @Leo

Input integer is given in unary. Output characters are E for emergency vehicles and 1 for regular cars.

s/(.?)1/\1E1/

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Answer 33

Jelly, 14 bytes

⁵*×109:90+Ị9×Ɗ

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Port of orlp's python answer which only uses arithmetic.

---

Jelly, 15 bytes

ḊṬ1;,2Ṭ¤LÐṀị⁾CE

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Answer 34

C (gcc), 48 bytes

f(n){for(printf("CEC"+!--n);--n>0;putchar(67));}

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Answer 35

Python 3, 39 bytes

lambda n:'C'*(n>1)+'E'+'C'*(n-1+(n==1))

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Answer 36

C# (.NET Core), 39 bytes

n=>(n--<2?"EC":"CE")+new string('C',n);

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Answer 37

PowerShell, 49 37 35 bytes

param($n)'C'*(1%$n)+1+'C'*($n-1%$n)

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Saved a pair of quotes by just making the E into a number.

Answer 38

Gol><>, 10 bytes

IR1~I{lRn;

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One of a few 10-byte solutions I found. I like this one the most because it uses the - as the clear lane, where the - is a result of printing -1.

The other 10-byte solutions:

• IR0P{lPRn;
• IR`CP{`C}H

Pity Gol><> doesn't have a halt and output the stack numbers command, like it does for chars, or this could be 3 bytes smaller.

Answer 39

Java 10, 53 51 45 42 38 37 36 35 34 bytes

n->n+=Math.pow(10,n)*97/30-1/n*9-n

Prints 3 instead of C and 2 instead of E.

Try it online.

Port of @orlp's Python 2 answer.

n+=...-n is used instead of (long)... to save a byte.

---

Old 35 bytes answer:

n->(10/3f+"").substring(1/n,1/n-~n)

Prints 3 instead of C and . instead of E.

Explanation:

Try it online.

n->{               // Method with integer parameter and String return-type
  (10/3f+"")       //  Divide 10 by 3 and convert it to a String ("3.333333333")
   .substring(1/n  //  And take the substring from index 1 if `n=1`, 0 otherwise
    ,1/n           //  to index 1 if `n=1`, 0 otherwise
        -~n)       //  + `n+1`

Answer 40

Python 2, 30 bytes

lambda n:1%n*'0'+`10**(n-1%n)`

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If n is larger than 1, 1%n gets 1: 1*'0'+`10**(n-1)`
If n is equal to 1, 1%n gets 0: 0*'0'+`10**(1-0)`

---

Python 2, 31 bytes

lambda n:1/n*'.'+'%%.%df'%~-n%0

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1/n*'.'        # if n is 1, insert a dot ...
+              # before
'%%.%df'%~-n%0 # the decimal representation of 0 with n-1 digits after the decimal point

'%%.%df'%~-n expands to '%.<n-1>f'.

Answer 41

MATL, 7 bytes

Q:G2Xl=

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Uses 1 for emergency, 0 for normal traffic.

Q:      % Range 1...n+1, implicit input n
  G2Xl  % Explicitly push input, limit to 2 (so, either 1 or 2)
      = % Element-wise equality of range and limited input. 
        % Implicit display.

This answer assumes (like some other answers) that separating spaces are OK. If not, append VXz for ten bytes.

Answer 42

Excel, 29 bytes

=IF(A1=1,0,10&REPT(1,A1-2))&1

Uses 0 for Emergency lane. 1 for Cars:

1  01
2  101
3  1011
4  10111
5  101111

Answer 43

J, 20 bytes

'CECC'#~1(<,[,=,-~)]

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Answer 44

Python, 53 bytes

lambda n:-1*((n+1)//n-2)*'C'+'E'+(n+((n+1)//n-2))*'C'

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Answer 45

Excel, 28 bytes

=REPT(2,A1+1)-10^MAX(A1-1,1)

Uses 2 for cars, and 1 for Corridor

Makes a list of 2s n+1 long, then subtracts the larger of 10^(n-1) and 10:

n=1 : 22 - MAX(1,10) = 12
n=2 : 222 - MAX(10,10) = 212
n=3 : 2222 - MAX(100,10) = 2122
n=10 : 22222222222 - MAX(1000000000,10) = 21222222222

Answer 46

Haskell (interactive), 30 bytes

a 1=12;a 2=212;a n=a(n-1)*10+2

1 is a corridor, 2 is a car.

Example:

GHCi, version 8.2.2: http://www.haskell.org/ghc/  :? for help
Prelude> a 1=12;a 2=212;a n=a(n-1)*10+2
Prelude> a 1
12
Prelude> a 2
212
Prelude> a 3
2122
Prelude> a 4
21222
Prelude> a 5
212222

Answer 47

Perl 5 -lp, 19 bytes

#!/usr/bin/perl -lp
$_=--$_?CE.C x$_:EC

Try it online!

Answer 48

APL (Dyalog Classic), 11 bytes

⊃∘⍕¨1∘<=0,⍳

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uses ⎕io←1

Answer 49

Prolog (SWI), 63 51 bytes

1+`EC`.
2+`CEC`.
N+X:-M is N-1,M+Y,append(Y,`C`,X).

Try it online!

Answer 50

Octave, 36 34 33 bytes

x((n=input(''))+1)=0;x(1+(n>1))=1

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Answer 51

Windows Batch, 75 bytes

To be improved...

@if %1==1 echo EC&exit/b
@cd|set/p=CE
@for /l %%G in (2,1,%1)do @cd|set/p=C

Answer 52

Retina 0.8.2, 17 bytes

.+
E$&$*C
ECC
CEC

Try it online! I could save a byte by using 1 for a car.

Answer 53

Python 3, 31 bytes

lambda n:'ECE'[n>1:n+1]+'C'*~-n

Try it online!

At first I tried reversing EC if n>1, but converting a boolean to (-1,+1) turned out longer than using a different string instead.

Alternative 31 byte solutions:

• lambda n:'ECCE'[n>1::2]+'C'*~-n Try it online!
• lambda n:bin(4**n-2-1%n)[:n:-1] Try it online!

All of these also work in Python 2.

Answer 54

JavaScript, 28 bytes

f=n=>n-->2?f(n)+'1':n?121:21

Answer 55

Swift, 74 bytes

func r(i:Int)->String{return i>1 ?"CE"+(1..<i).map{_ in"C"}.joined():"EC"}

Try it online!

Answer 56

PHP, 103 Bytes

Try it online!

Code, recursive function

function f($n,$i=1,$s=''){if($i<$n){$s.="CE".str_repeat("C",$i)."
";f($n,$i+1,$s);}else{echo"EC
".$s;}}

Explanation

function f($n,$i=1,$s=''){
if($i<$n){                         #condition to get out, it counts to $n
    $s.="CE".str_repeat("C",$i)."  # "C" is repeated and concatenated 
    ";f($n,$i+1,$s);               #note that "\n" = "(Enter key)"  
}else{
    echo"EC                        #the first line "EC"
    ".$s;
}
}

Note that "\n" = " ", you save one byte

Output

f(5);
EC
CEC
CECC
CECCC
CECCCC

f(10);
EC
CEC
CECC
CECCC
CECCCC
CECCCCC
CECCCCCC
CECCCCCCC
CECCCCCCCC
CECCCCCCCCC

Answer 57

Jelly, 8 bytes

«2=‘R$;⁷

A full program printing the output with car lanes as 0 and the emergency lanes as 1.

Try it online!

Answer 58

Yabasic, 54 bytes

Input""n
If n>1Then?"C";Fi
?"E";
For i=1To n:?"C";Next

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Answer 59

Excel VBA, 32 bytes

An anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window.

?[If(A1-1,10,"01")&Rept(1,A1-1)]
''  or
?[If(A1-1,10,0)&Rept(1,A1-A1<1)]

Answer 60

><>, 26 bytes

::?v~1)?o"!">o<
00.>1-"~"}

Watch it run !

Results :

1.     !~
2.     ~!~
3.     ~!~~
4.     ~!~~~
etc.

Answer 61

Ruby, 25 bytes

->n{n>1?'CE'+?C*~-n:'EC'}

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Answer 62

Perl 6, 20 bytes

{1%$_*10~1 x$_-1%$_}

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Returns

1 => 01
2 => 101
3 => 1011
4 => 10111
5 => 101111

Another 20 byte solution with -p is $_=1 x$_;s/1?<(1/E1/ which is identical to Xcali's Perl 5 solution.

Answer 63

><>, 16 bytes

1-:?!v1$
n<}1{<>

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Prints 0 as the corridor and 1 as the cars.

Answer 64

Befunge-98 (FBBI), 14 bytes

Uses 0 for cars, and 1 for the empty lane.

12&:b0p`j\' k.

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This uses a & to end the program, stalling it until TIO shuts it off, instead of ending when it's finished. In order to avoid waiting, you can hit the run button again soon after it starts to end it prematurely and see the result it would give if it waited. In case ending like this is too iffy, here is a 15 byte solution that doesn't use this method:

12&:b0p`j\' k.@

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Explanation

This program uses the implicit zeroes at the bottom of the stack to print the cars at the end. This way we can avoid pushing N - 1 things, and only use the input for printing N + 1 things and determining whether N is 1 or not.

(stack terminology is bottom [a, b, c] top)

12                Push 1 and 2. The stack is [1, 2]
  &:              Take input and duplicate. [1, 2, N, N]
    b0p           Put N at (0, 11), which is after the '
       `          Pops the top 2 and pushes 1 if a>b, 0 if not. [1, 2>N]
        j\        Jumps the \ if 2>N (i.e. N = 1), otherwise switches the top 2.
                      The stack is [1] if N=1, otherwise [1, 0]
          '       Takes in N again from the put instruction earlier
            k.    Prints out N+1 characters: (01 or 10 depending on N) + (N-1 0s).
12                Wraps and adds irrelevant things to the stack.
  &               Waits for TIO to end the program.

Answer 65

PHP, 35 bytes

<?=str_pad(1<$argn?CE:E,C,$argn+1);

Run as pipe with -n or try it online.

Answer 66

C# (.NET Core), 74 bytes

Additional bytes for using System; and int L

var S="0X";for(int I=1;I<L;I++)S+='0';System.Console.Write(L<2?"X0":S);

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Answer 67

Pyt, 18 bytes

Đ1≤?ĉ01:ŕ01↔⁻⑴;áƑǰ

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Uses 0 for the emergency lane and 1 for the normal lanes

Explanation

                          Implicit input (N)
Đ                         Duplicate N on the stack
 1≤                       Is N≤1
   ?   :       ;          If top of stack is non-zero, continue execution until reaching 
                               colon, then jump to semicolon; otherwise, jump to colon 
                               and continue executing
   ?                      - Then (i.e. i≤1)
    ĉ                     Clear the stack
     01                   Push 0 followed by 1

       :                  - Else branch (i.e. N≥2)
        ŕ                 Pop the top of the stack
         01               Push 0 followed by 1
           ↔              Flip the stack
            ⁻⑴           Decrement N and then push an array of N-1 1's
               ;          - End of if statement

                á         Push contents of stack to an array
                 Ƒ        Flatten the resulting array
                  ǰ       Join the contents of the array with no delimiters

Answer 68

Pushy, 10 bytes

1wEs{t?};Q

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1             \ Push 1
 w            \ 'Mirror' the stack around the central item, yielding [n, 1, n]
  E           \ Pop 1 and n, and push 1 * 10 ** n
   s          \ Split into its digits
    {         \ Get input back on top of stack
     t?       \ If input > 1:
       };     \    Cyclically shift the stack right once
         Q    \ Collect all items on stack, interpret as indicies into uppercase alphabet,
              \ and print the resulting string.

Prints A for lanes and B for corridors.

Answer 69

ink, 55 33 55 bytes

=l(n)
~temp i=2
{n-1:C}EC<>
-(k){i<n:
~i++
C<>->k
}->->

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Starts by printing C if input is not 1, then EC, then prints C another n-2 times.