[Question]
  [
The famous C64 basic one liner
```
10 PRINT CHR$(205.5+RND(1)); : GOTO 10
```
prints a maze of slashes and backslashes.
```
\\/\\\//\/\////\\/\/
\/\///\\///////\//\/
/\\\//\//\////\\//\\
\////\//\//\/\\\\\\/
/\/\\///\\\\/\\\\/\\
\/\//\\\\\\//\/\////
/\//\\///\/\///\////
\/\\\//\\/\\\//\\/\/
//////\\/\\/\/\/\///
\\/\/\\////\/\/\\/\/
```
Read in such maze made of diagonal walls from stdin and print out the same maze with *horizontal* and *vertical* walls consisting of the wall character "#"
For example the small maze
```
/\\
\\/
///
```
translates to
```
     #####
     #   #
     # # # #
     # # # #
 ##### # # #
       #   #
   #########
      
     #####    
  
```
To be precise, each isolated wall segment has the length of five characters, adjacent wall segments share a corner. Moving a character to the right/left/top/down in the matrix of slashes and backslashes corresponds to a diagonal translation by 2 characters in vertical and 2 characters in horizontal direction in the #-matrix.
More context can be found in the book [10 print](https://10print.org/10_PRINT_121114.pdf).
  
[Answer]
# Python 3, ~~226~~ 224 bytes
My first Python golf, so probably very sub-optimal.
It produces a whole lot of trailing whitespace, but there are no preceding newlines, and at most two preceding spaces.
The input needs to be given by hand from command line (maybe someone knows a shorter way to get multiline input in Python...).
```
e,z,s=enumerate,'0',list(iter(input,""))
p=''.join(s)*5
r=[len(p)*[' ']for _ in p]
for y,l in e(s):
 for x,c in e(l):
  for i in range(-2,3):r[2*(x+y+(s>[z]))+i*(c>z)][2*(x+len(s)-y)+i*(c<z)]='#'
for l in r:print(''.join(l))
```
The idea is to initialize a huge array of spaces `r`, then iterate through the input and replace the spaces with `#` as needed, and finally print the whole array.
A trick I used is to compare characters to `z = '0'` instead of testing equality to `'/'` or `'\'`, which saves a bunch of bytes.
[Answer]
# Julia, 258 bytes
A functional solution...
```
A=split(readall(STDIN))
q(i,j)=fld(i-1,j)
n,^ =A[].(3),q
f(i,j)=try A[1+i^5][1+j^5]<'0'?(i+j)%5==1:(i-j)%5==0catch 0end
h(i,j)=f(i+i^4,j)|f(i+(i-1)^4,j)
g(i,j)=h(i,j+j^4)|h(i,j+(j-1)^4)
for i=1:6length(A),j=-n-5:2n;print(" #"[1+g(i-j,i+j)],j==2n?"\n":"")end
```
In order of appearance:
`f` covers '/' and '\' by their 5\*5 bit patters,
`h` folds every fifth and following line into a single line (recall "adjacent wall segments share a corner") and `g` does the same for the columns. Finally, `i-j,i+j` rotates the picture.
[Answer]
# JavaScript (ES6), 258
A function with the maze as a parameter, returning the output.
Unsure if it's valid, due to the input/output rules (it was fun anyway)
```
f=m=>([...m].map(c=>{if(c<' ')x=sx-=2,y=sy+=2;else for(x+=2,y+=2,d=c>'0',i=y-3*d,j=x-3*!d,c=5;c--;)o[i+=d][j+=!d]='#';},w=m.search`
`,h=m.match(/\n/g).length,sy=y=0,sx=x=h*2,o=Array(z=(w+h+1)*2).fill(' ').map(x=>Array(z).fill(x))),o.map(r=>r.join``).join`
`)
// LESS GOLFED
U=m=>(
  w=m.search`\n`,
  h=m.match(/\n/g).length,
  o=Array(z=(w+h+1)*2).fill(' ').map(x=>Array(z).fill(x)),
  sy=y=0,
  sx=x=h*2,
  [...m].forEach(c=>{
    if(c<' ')x=sx-=2,y=sy+=2
    else for(x+=2,y+=2,d=c>'0',i=y-3*d,j=x-3*!d,c=5;c--;)o[i+=d][j+=!d]='#';
  }),
  o.map(r=>r.join``).join`\n`  
)
// TEST
out=x=>O.innerHTML+=x+'\n'
test=`\\\\/\\\\\\//\\/\\////\\\\/\\/
\\/\\///\\\\///////\\//\\/
/\\\\\\//\\//\\////\\\\//\\\\
\\////\\//\\//\\/\\\\\\\\\\\\/
/\\/\\\\///\\\\\\\\/\\\\\\\\/\\\\
\\/\\//\\\\\\\\\\\\//\\/\\////
/\\//\\\\///\\/\\///\\////
\\/\\\\\\//\\\\/\\\\\\//\\\\/\\/
//////\\\\/\\\\/\\/\\/\\///
\\\\/\\/\\\\////\\/\\/\\\\/\\/`
out(test),out(f(test))
```
```
<pre id=O></pre>
```
]