Bolos em rolo são deliciosos doces feitos com uma camada de pão de ló, uma camada de creme (ou sorvete), enrolada em espiral, consulte a Wikipedia .

Sua tarefa é obter uma sequência de frases não vazia da entrada padrão, remover os espaços e enrolá-la como um bolo, começando do centro da direita para a esquerda no sentido anti-horário, enquanto os espaços começam na parte inferior da primeira caractere de frase, do centro da esquerda para a direita, como no exemplo a seguir.

Observe que a frase será a camada de pão de ló e a camada de creme será representada como espaços vazios.

Input: Lorem ipsum dolor sit amet, consectetu adipiscing elit, sed do eiusmod

Output with spaces ' ' represented as _

ilegnicsip
t________i
,_srolod_d
s_i____m_a
e_t_oL_u_u
d_a_r__s_t
d_m_emip_e
o_e______t
e_t,consec
i____
usmod

Casos de teste

Input: Lorem ipsum dolor sit amet, consectetu adipiscing elit, sed do eiusmod

Output:

ilegnicsip
t        i
, srolod d
s i    m a
e t oL u u
d a r  s t
d m emip e
o e      t
e t,consec
i    
usmod

Input: Sandbox for Proposed Challenges

Output:

soporp
e    r
d aS o
c n  f
h dbox
a      
llenges

Input: Cake

Output:

aC 
k  
e

Input: it

Output:

ti

Regras:

• Isso é código-golfe, então o código mais curto vence.
• As brechas padrão são proibidas.

Python 2, 202 200 bytes

^{-2 bytes graças a LevitatingLion}

c=[];s=n=0
a=input().replace(' ','')
#Calculate the ammount of rotations
l=len(a)
while s<l:s=n*(1+n)/2;n+=1
#Add trailing spaces to not break the rolling
a+=' '*(s-l)
#Add a cream layer on top and a cake layer on bottom of the cake and roll 90 degrees
for i in range(1,n):c=[' '*i]+zip(*c[::-1])+[a[:i]];a=a[i:]
#Finish the rolling to match the pattern
exec~i%4*'c=zip(*c[::-1]);'
#Serve the cake
for i in c:print''.join(i)

Befunge, 260 250 bytes

"'"::00p09p19p35*::10p29pv
 v`*84:~<0p041p031p020p93<
v_0`!#v_^*!p04+g04:p03%4+g03:p02%4+g02:!%g04:+1pp01+%2-1g03p\3+<
>00g:: :09g\`8+0\p::19g`8+1\p20g1-2%+00p10g:::29g\`8+2\p::39g`8^
9g\-\v>29g:39g\->\:50p09g:1
$$1+\:!#@_1-55+,^>
,gg05<:+1\-1_:$#\^#

Experimente online!

A idéia básica é que construamos a espiral na metade inferior do campo de jogo, mantendo o controle das coordenadas mín e máx à medida que avançamos. Depois que todas as entradas foram processadas, iteramos no intervalo de coordenadas e escrevemos o conteúdo que foi salvo.

Explicação detalhada

Observe que os valores dx e dy estão no intervalo de 0 a 3, correspondendo aos deslocamentos relativos -1, 0, 1, 0. A atualização da direção é feita com a adição de 1, módulo 4. A conversão para um deslocamento relativo é feita por subtraindo 1, módulo 2.

"'"::00p09p19p      Set x, minx, and maxx to 39.
35*::10p29p39p      Set y, miny, and maxy to 15.
020p130p            Set dx to 0 (relative offset -1) and dy to 1 (relative offset 0).
140p                Set the segment length to 1.
0                   Set the index to 0 (on the stack).

                    -- The main input loop start here --
~                   Read a character from stdin.
:48*`_              If it's greater than 32 (i.e. not whitespace or EOL)...
  00g:              Get the current value of x and make a duplicate.
  ::09g\`8+0\p      If it's less than minx, update minx.
  ::19g`8+1\p       If it's greater than maxx, update maxx.
  20g1-2%           Calculate (dx-1)%2, which converts dx into the range -1 to 1.
  +00p              Add that to x and save (leaving the original x on the stack).
  10g:              Get the currrent value of y and make a duplicate.  
  ::29g\`8+2\p      If it's less than miny, update miny.
  ::39g`8+3\p       If it's greater than maxy, update maxy.
  30g1-2%           Calculate (dy-1)%2, which converts dy into the range -1 to 1.
  +10p              Add that to y and save (leaving the original y on the stack).
  p                 Store the character we read at the original x,y coordinates.
  1+                Increment the index by 1.
  :40g%!            Mod that with the current segment len; if zero it's time to turn.
  :20g+4%20p        Add the turn condition to the dx value, thus turning if needed.
  :30g+4%30p        Similarly add the turn condition to the dy value.
  :40g+40p          Also add it to the segment length, so it gets longer every turn.
  !*                Multiply the index by !turn, so it's reset to zero on every turn.
  ^                 Repeat the loop, reading another character.
0`!#v_^             If the char <= 32, we test for EOF and repeat the loop if not.

>                   If it is EOF, we start the output routine.
29g                 Initialise the y coordinate with miny.
:39g\-              Calculate maxy - miny to use as the height countdown.

                    -- The outer output loop starts here --
\:50p               Swap y to the top and save in a temporary variable.
09g                 Initialise the x coordinate with minx.
:19g\-              Calculate maxx - miny to use as our width countdown.
\                   Swap x to the top of the stack.

                    -- The inner output loop starts here --
  :50g              Take a copy of x and the y save in the temporary variable.
  g,                Read the character at those coordinate and write to stdout.
  \:_               Swap the width countdown to the top of the stack and break if zero.
  1-                Otherwise decrement the width countdown.
  \1+               Swap the x value back to the top and increment it.
  <                 Repeat the inner loop.

$$                  Clear the width countdown and x value from the stack.
1+                  Increment the y value.
\                   Swap the height countdown to the top of the stack.
:¬#@_               If it's zero, exit.
1-                  Otherwise decrement it.
55+,                Output a line break.
^                   Repeat the outer loop.