Desafío

Escriba un programa que emita la letra de 99 Bottles of Beer, pero en lugar de "beer", emita "fizz" si el número de botellas en la pared es múltiplo de 3, "zumbido" si es múltiplo de 5, y "fizzbuzz" si es un múltiplo de 3 y un múltiplo de 5. Si el número de botellas en la pared no es un múltiplo de 3 o 5, simplemente emita "cerveza" como de costumbre.

Letra

99 bottles of fizz on the wall, 99 bottles of fizz.
Take one down and pass it around, 98 bottles of beer on the wall.

98 bottles of beer on the wall, 98 bottles of beer.
Take one down and pass it around, 97 bottles of beer on the wall.

97 bottles of beer on the wall, 97 bottles of beer.
Take one down and pass it around, 96 bottles of fizz on the wall.

96 bottles of fizz on the wall, 96 bottles of fizz.
Take one down and pass it around, 95 bottles of buzz on the wall.

95 bottles of buzz on the wall, 95 bottles of buzz.
Take one down and pass it around, 94 bottles of beer on the wall.

....

3 bottles of fizz on the wall, 3 bottles of fizz.
Take one down and pass it around, 2 bottles of beer on the wall.

2 bottles of beer on the wall, 2 bottles of beer.
Take one down and pass it around, 1 bottle of beer on the wall.

1 bottle of beer on the wall, 1 bottle of beer.
Go to the store and buy some more, 99 bottles of fizz on the wall.

Este es el código de golf , por lo que gana la presentación más corta en cada idioma.

Python 2 , 263 253 245 bytes

i=99
x=''
while i:x+=', %s on the wall.\n\n%s on the wall, %s.\n'%(('%d bottle%s of %s'%(i,'s'*(i>1),(i%3<1)*'fizz'+(i%5<1)*'buzz'or'beer'),)*3)+'GToa kteo  otnhee  dsotwonr ea nadn dp absusy  isto maer omuonrde'[i>1::2];i-=1
print x[35:]+x[:33]

Pruébalo en línea!

C (GCC), 276 274 bytes

¡Gracias a Neil por guardar dos bytes!

#define w" on the wall"
#define c(i)printf("%d bottle%s of %s",i,"s"+!~-i,i%3?i%5?"beer":"buzz":i%5?"fizz":"fizzbuzz"),printf(
i;f(){for(i=99;i;c((i?:99))w".\n\n"))c(i)w", "),c(i)".\n"),printf(--i?"Take one down and pass it around, ":"Go to the store and buy some more, ");}

¿Quién no ama los paréntesis sin igual en las macro expansiones?

Sin golf:

#define c(i)                               \
    printf(                                \
        "%d bottle%s of %s",               \
        i,                   /* Number  */ \
        i-1 ? "s" : "",      /* Plural  */ \
        i % 3                /* FizzBuzz*/ \
            ? i % 5                        \
                ? "beer"                   \
                : "buzz"                   \
            : i % 5                        \
                ? "fizz"                   \
                : "fizzbuzz"               \
    )

i;
f() {
    for(i = 99; i; ) {
        c(i); printf(" on the wall, ");
        c(i); printf(".\n");
        printf(
            --i
                ? "Take one down and pass it around, "
                : "Go to the store and buy some more, "
        );

        // This has been stuffed into the for increment
        c((i?:99)); printf(" on the wall.\n\n");
    }
}

¡Véalo en vivo en Coliru!

Versión alternativa (276 bytes)

#define c(i)printf("%d bottle%s of %s",i,i-1?"s":"",i%3?i%5?"beer":"buzz":i%5?"fizz":"fizzbuzz"),printf(
i,*w=" on the wall";f(){for(i=99;i;c((i?:99))"%s.\n\n",w))c(i)"%s, ",w),c(i)".\n"),printf(--i?"Take one down and pass it around, ":"Go to the store and buy some more, ");}

Röda , 273 bytes

f{a=`bottle`f=` on the wall`g=`99 ${a}s of fizz`;[`$g$f, $g.
`];seq 98,1|{|b|d=`s`d=``if[b=1];c=``c=`fizz`if[b%3=0];c.=`buzz`if[b%5=0];c=`beer`if[c=``];e=`$b $a$d of $c`;[`Take one down and pass it around, $e$f.

$e$f, $e.
`]}_;[`Go to the store and buy some more, $g$f.`]}

Pruébalo en línea!

Jugará más golf en la mañana.

PHP, 242 bytes

function f($k){return"$k bottle".(s[$k<2])." of ".([fizz][$k%3].[buzz][$k%5]?:beer);}$w=" on the wall";for($b=f($c=99);$c;)echo"$b$w, $b.
",--$c?"Take one down and pass it around":"Go to the store and buy some more",", ",$b=f($c?:99),"$w.

";

Pruébalo en línea!

PHP, 244 bytes

for($e=s,$b=fizz,$c=99;$c;)echo strtr("301245, 30124.
6, 708295.

",[" bottle",$e," of ",$c,$b," on the wall",--$c?"Take one down and pass it around":"Go to the store and buy some more",$k=$c?:99,$e=s[2>$k],$b=[fizz][$k%3].[buzz][$k%5]?:beer]);

Pruébalo en línea!

utilizar la función strtr

PHP, 245 bytes

$f=function($k)use(&$b){$b="$k bottle".(s[$k<2])." of ".([fizz][$k%3].[buzz][$k%5]?:beer);};for($w=" on the wall",$f($c=99);$c;)echo"$b$w, $b.
",--$c?"Take one down and pass it around":"Go to the store and buy some more",", {$f($c?:99)}$b$w.

";

Pruébalo en línea!

use una función anónima en la cadena para obtener una cadena dependiendo del número entero

Expandido

$f=function($k)use(&$b){$b="$k bottle".(s[$k<2])." of ".([fizz][$k%3].[buzz][$k%5]?:beer);};
for($w=" on the wall",$f($c=99);$c;)
echo"$b$w, $b.
",--$c?"Take one down and pass it around":"Go to the store and buy some more"
,", {$f($c?:99)}$b$w.

";

05AB1E , 151 146 143 bytes

99LRv'¬ž“fizzÒÖ“#y35SÖÏJ‚˜1(è©y“ƒ¶€µ„‹€ƒî倕…¡, ÿÏꀂ ÿ€‰€€íÒ.“ªõ®y“ÿÏꀂ ÿ€‰€€íÒ, “D#4£ðýs…ÿÿ.}‚‚˜'Ïê'±¥:`)¦¦¬#7£ðý¨“‚œ€„€€ƒï€ƒ‚¥€ä€£, ÿ.“ª)˜»

Pruébalo en línea!

SOGL , 136 135 134 133 131 bytes

Ƨ, o▓k
"πFT+╔¡‘oW
³³q"'bμ⁸‘oH? so}5\;3\«+"ΞQv↑χāσκN⌡κYT¡‘_,S─‘oθwoX▓
MH∫}¹±{▓WkƧ.¶oH¡"sΗ─χpēGķ¶¾3Ζ^9f.⅟▒E┌Fρ_╬a→‘KΘw⁽oXH‽M}HkW">⁸‘p

En primer lugar, la tercera función:

                                    ▓  name this "▓" (example input: 15)                          [15]
³³                                     Create 4 extra copies of the top thing (number of things)  [15, 15, 15, 15, 15]
  q                                    output without popping one of them                         [15, 15, 15, 15, 15]
   "...‘o                              output " bottle"                                           [15, 15, 15, 15, 15]
         H?   }                        if pop-1 [isn't 0]                                         [15, 15, 15, 15]
            so                           output "s"                                               [15, 15, 15, 15]
               5\                      push if POP divides by 5                                   [15, 15, 15, 1]
                 ;                     swap [if divides & another number copy]                    [15, 15, 1, 15]
                  3\«                  push if POP divides by 3, multiplied by 2                  [15, 15, 1, 2]
                     +                 add those together                                         [15, 15, 3]
                      "...‘            push "buzz fizz fizzbuzz beer"                             [15, 15, 3, "buzz fizz fizzbuzz beer"]
                           ...‘o       output " of " (done over here to save a byte for a quote)  [15, 15, 3, "buzz fizz fizzbuzz beer"]
                                θ      split ["buzz fizz fizzbuzz beer"] on spaces                [15, 15, 3, ["buzz","fizz","fizzbuzz","beer"]]
                                 w     get the index (1-indexed, wrapping)                        [15, 15, ["buzz","fizz","fizzbuzz","beer"], "fizzbuzz"]
                                  o    output that string                                         [15, 15, ["buzz","fizz","fizzbuzz","beer"]]
                                   X   pop the array off of the stack                             [15, 15]

La primera función:

Ƨ, o▓k
     k  name this "function" "k"
Ƨ, o    output ", "
    ▓   execute the "bottleify" function

La segunda función:

"πFT+╔¡‘oW
         W  call this "W"
"πFT+╔¡‘    push " on the wall"
        o   output it

Y la parte principal:

MH∫}                                     repeat 99 times, each time pushing index
    ¹                                    wrap in an array
     ±                                   reverse it
      {                                  iterate over it
       ▓                                 execute that function
        W                                execute that function
         k                               execute that function
          Ƨ.¶o                           output ".\n"
              H¡                         push if POP-1 isn't 0 (aka 1 if pop <> 1, 0 if pop == 1)
                "...‘                    push "Stake one down and pass it aroundSgo to the store and buy some more"
                     K                   push the first letter of that string
                      Θ                  split ["take one down and pass it aroundSgo to the store and buy some more" with "S"]
                       w                 gets the xth (1-indexed, wrapping) item of that array
                        ⁽o               uppercase the 1st letter and output
                          X              pop the array off
                           H‽            if pop-1 [isn't 0]
                             M           push 100
                              }          ENDIF
                               H         decrease POP
                                k        execute that function
                                 W       execute that function
                                  ">⁸‘p  output ".\n\n"

Perdí un par de bytes debido a un error que Opone una nueva línea antes y después (y de alguna manera esto vuelve a V0.9 (este es el código V0.11))

Java, 344 340 339 bytes

(-4 bytes después del golf fizzbuzz; -1 byte eliminando el espacio en blanco perdido)

interface X{static void main(String[]a){for(int i=99;i>0;System.out.printf("%s on the wall, %s.%n%s, %s on the wall.%n%n",b(i),b(i--),i<1?"Go to the store and buy some more":"Take one down and pass it around",b(i<1?99:i)));}static String b(int i){return i+" bottle"+(i>1?"s":"")+" of "+(i%3<1?"fizz":"")+(i%5<1?"buzz":i%3<1?"":"beer");}}

Ligeramente sin golf (usando sangría de 1 espacio para eliminar el desplazamiento horizontal):

interface X {
 static void main(String[]a){
  for(int i=99;i>0;System.out.printf("%s on the wall, %s.%n%s, %s on the wall.%n%n",
   b(i),b(i--),
   i<1?"Go to the store and buy some more":"Take one down and pass it around",
   b(i<1?99:i)));
 }
 static String b(int i){
  return i+" bottle"+(i>1?"s":"")+" of "+(i%3<1?"fizz":"")+(i%5<1?"buzz":i%3<1?"":"beer");
 }
}

JavaScript (ES6), 316 309 bytes

Este es un programa completo en lugar de una función. Nada muy creativo, es solo el enfoque ingenuo (¡de ahí el bytecount!). Estoy usando en console.log()lugar de alert()porque muchos navegadores tienen un límite en la cantidad de caracteres que se pueden mostrar usando alert(). Tenga en cuenta que todos los espacios en blanco y las nuevas líneas son necesarios.

a="";for(i=99;i>0;i--){b=j=>"bottle"+(j>1?"s":"");d=a=>(a%3?"":"fizz")+(a%5?"":"buzz")||"beer");w=" on the wall";o=" of ";a+=`${i+" "+b(i)+o+d(i)+w+", "+i+" "+b(i)+o+d(i)}.
${i>1?"Take one down and pass it around, ":"Go to the store and buy some more, "}${(y=i-1?i-1:99)+" "+b(y)+o+d(y)+w}.

`;}console.log(a)

Sin golf:

let accumulator = "";
for(let i = 99; i>0; i--){
    let bottleString = j => "bottle"+(j>1?"s":""),
    drink = a =>(a%3?"":"fizz")+(a%5?"":"buzz")||"beer",
    wallString = " on the wall",
    of=" of ";
    accumulator += `${i+" "+bottleString(i)+of+drink(i)+wallString+", "+i+" "+bottleString(i)+of+drink(i)}.
${i>1?"Take one down and pass it around, ":"Go to the store and buy some more, "}${(y=i-1?i-1:99)+" "+bottleString(y)+of+drink(y)+wallString}.

`;
}

console.log(accumulator);

Aquí está el fragmento:

a="";for(i=99;i>0;i--){b=j=>"bottle"+(j>1?"s":"");d=a=>(a%3?"":"fizz")+(a%5?"":"buzz")||"beer";w=" on the wall";o=" of ";a+=`${i+" "+b(i)+o+d(i)+w+", "+i+" "+b(i)+o+d(i)}.
${i>1?"Take one down and pass it around, ":"Go to the store and buy some more, "}${(y=i-1?i-1:99)+" "+b(y)+o+d(y)+w}.

`;}console.log(a)

Por cierto, con esta respuesta, ¡me he ganado la insignia de bronce en code-golf ! ¡Nunca pensé que lograría esto nunca (aunque no es que sea un gran logro)!

Retina , 230 bytes

99$*_
_\B
Take one down and pass it around, $.'#.¶¶$.'#, $.'.¶
^
99#, 99.¶
_
Go to the store and buy some more, 99#.
#
 on the wall
1\b|(\d+)
$& bottle$#1$*s of $&$*_
\b(_{15})+\b
fizzbuzz
\b(_{5})+\b
buzz
\b(___)+\b
fizz
_+
beer

Pruébalo en línea! Explicación:

99$*_

Insertos 99 _s.

_\B
Take one down and pass it around, $.'#.¶¶$.'#, $.'.¶

Cambia todos menos el último _a la cadena Take one down and pass it around, $.'#.¶¶$.'#, $.'.¶, donde ¶es una nueva línea y $.'es el recuento de guiones bajos restantes. Esto efectivamente cuenta de 98 a 1.

^
99#, 99.¶

Agrega la primera línea del primer verso en formato "compacto".

_
Go to the store and buy some more, 99#.

Agrega la segunda línea del último verso. Por qué necesito saltar a través de aros para usar el _No lo sé, pero $parece coincidir dos veces, por lo que no puedo usar eso. Imagínate.

#
 on the wall

Sustituye una cadena que aparece varias veces en el verso.

1\b|(\d+)
$& bottle$#1$*s of $&$*_

Esto coincide con los números enteros en los versos, y sufija la (s) botella (s) apropiada (s) y se expande nuevamente a unario, en preparación para elegir la bebida. (Ahorro 1 byte en el 99s de esta manera).

\b(_{15})+\b
fizzbuzz
\b(_{5})+\b
buzz
\b(___)+\b
fizz
_+
beer

Reemplace los múltiplos exactos con la bebida adecuada.

sed , 468 459 456 bytes

s:^:99 bottles of fizz on the wall, 99 bottles of fizz.:
p
s:99:I8:g
s:fizz:XYYZ:g
x
s:^:Take one down and pass it around, I8 bottles of XYYZ on the wall.\n:
G
x
:
g
s:XXX:fizz:g
s:Y{5}:buzz:g
s:\bX*Y*Z:beer:g
s:[XYZ]::g
y:ABCDEFGHI:123456789:
s:\b0::g
/ 1 /bq
p
x
s:^::
tc
:c
s:(\S)0:\1@:g
Td
y:ABCDEFGHI:0ABCDEFGH:
:d
y:123456789@:0123456789:
s:(XXX)*(Y{5})*(Y*Z):XY\3:g
x
b
:q
s:es:e:g
aGo to the store and buy some more, 99 bottles of fizz on the wall.

Pruébalo en línea!

Requiere -r bandera.

Explicación

Hold space mantiene el patrón de dos líneas repetidas, con números representados como [A-I][0-9](dígitos separados para decenas y unidades ) y el tipo de bebida representada como X*Y*Z, donde Xrealiza un seguimiento de -N mod 3y Yde-N mod 5 .

En cada iteración posterior, los números se hacen decremented y las Xs y Ys se actualizan. Luego, el espacio de espera se copia en el espacio del patrón, se convierte en líneas de la canción y se imprime.

C, 349 345 344 bytes

#define b(x)x^1?" bottles":" bottle"
#define g i-1?"Take one down and pass it around":"Go to the store and buy some more"
*w=" on the wall";*s(x){return x?x%15?x%5?x%3?"beer":"fizz":"buzz":"fizzbuzz":"fizz";}i=100;main(){while(--i){printf("%d%s of %s%s, %d%s of %s.\n%s, %d%s of %s%s.\n",i,b(i),s(i),w,i,b(i),s(i),g,i-1?i-1:99,b(i-1),s(i-1),w);}}

Bueno, allá vas. Eso tomó una hora.

Pruébalo en línea!

Javascript (ES6), 236 234 233 232 bytes

for(i=99;i;console.log(z()+`, ${z(_)}.
${--i?'Take one down and pass it around':'Go to the store and buy some more'}, ${z()}.

`))z=(o=' on the wall',j=i||99)=>j+` bottle${j>1?'s':_} of `+((j%3?_:'fizz')+(j%5?_='':'buzz')||'beer')+o

Manifestación

// replace console.log to avoid 50-log limit in snippets:
console.log=_=>document.write(`<pre>${_}</pre>`)

for(i=99;i;console.log(z()+`, ${z(_)}.
${--i?'Take one down and pass it around':'Go to the store and buy some more'}, ${z()}.

`))z=(o=' on the wall',j=i||99)=>j+` bottle${j>1?'s':_} of `+((j%3?_:'fizz')+(j%5?_='':'buzz')||'beer')+o

Sin golf

i = 99  // start counter at 99

z = (   // define function z which takes arguments with defaults:
   o = ' on the wall', // o = defaults to ' on the wall'
   j = i || 99         // j = defaults to value of counter i - or 99 when i == 0
) => 
    j +                 // our current j counter
    ' bottle' +
    (j > 1 ? 's' : _) + // choose 's' when we have more than 1 bottle, or blank _
    (
        (j % 3 ? _ : 'fizz') +      // if j % 3 is 0, add 'fizz', otherwise blank _
        (j % 5 ? _ = '' : 'buzz')   // if j % 5 is 0, add 'buzz', otherwise blank _
                                    // _ gets defined here since it's the first place it's used
            ||                      // if no fizz or buzz, result is a falsey empty string
        'beer'                      // replace falsey value with 'beer'
    ) +
    o                               // append o

while (i) {         // while counter is non-zero
    console.log(    // output string:
        z() +       // call z without o argument
        ', ' +
        z(_) +      // call z with blank _ for o to block ' on the wall' here
        '.\n' +
        ( --i       // decrement i, if still non-zero:
            ? 'Take one down and pass it around'
                    // otherwise:
            : 'Go to the store and buy some more'
        ) + 
        ', ' +
        z() +       // another call to z without o
        '.\n\n'
    )
}

Rubí , 261 bytes

99.downto(1){|i|w=' on the wall'
f=->x{a='';x%3<1&&a+='fizz';x%5<1&&a+='buzz';a<?a&&a='beer';"%d bottle%s of %s"%[x,x<2?'':?s,a]}
puts [f[i]+w,f[i]+?.+$/+(i<2?'Take one down and pass it around':'Go to the store and buy some more'),f[i<2?99:i-1]+w+?.+$/*2]*', '}

Pruébalo en línea!

shortC , 314 312 bytes

Db(x)x^1?" bottles":" bottle"
Dg i-1?"Take one down and pass it around":"Go to the store and buy some more"
*w=" on the wall";*s(x){Tx?x%15?x%5?x%3?"beer":"fizz":"buzz":"fizzbuzz":"fizz";}i=100;AW--i){R"%d%s of %s%s, %d%s of %s.\n%s, %d%s of %s%s.\n",i,b(i),s(i),w,i,b(i),s(i),g,i-1?i-1:99,b(i-1),s(i-1),w

Lo siento, no hay explicación, pero olvidé por completo cómo funciona esto.

Carbón , 307 297 bytes

Ａ”｜‽2?{:×G↗”¦αＡ“6«eMηOU¶¿”¦ζＡ“9“e▷·gqε-g}”¦βＡ“9Ｂ{⦃⁺Ｂφ=；λO”¦ωＡfizz¦φＡbuzz¦γＡbeer¦ηＡ”↶C▶▶d℅d¬r·ＵS\λＴθＮevＴ◧→GM⁸ω┦τＡ“M↧k↓⁺*f÷,ψＺ¢▶\¿｜Ｐ“№κ×υpξＸoＷ”¦σＡ.¶πＦ⮌…¹¦¹⁰⁰«Ａ⎇⁻ι¹αζθ¿∧¬﹪鳬﹪ι⁵Ａ⁺φγ﹪ι³Ａφ﹪ι⁵ＡγεＡηε⁺⁺⁺⁺⁺⁺⁺ＩιθεβＩιθεπ¿⁻ι¹Ａ⁻ι¹λＡ⁹⁹λＡ⎇⁻λ¹αζθ¿∧¬﹪볬﹪λ⁵Ａ⁺φγ﹪λ³Ａφ﹪λ⁵ＡγεＡηε¿⁻ι¹ＡτδＡσδ⁺⁺⁺⁺δλθεω

Pruébalo en línea!

¡SI PODEMOS! Enlace a la versión detallada, esto se puede jugar mucho, estoy seguro.

tcl, 298

proc B i {set x " bottle[expr $i>1?"s":""] of [expr $i%3?$i%5?"beer":"":"fizz"][expr $i%5?"":"buzz"]"}
set i 99
time {puts "$i[B $i][set w " on the wall"], $i[B $i].
Take one down and pass it around, [incr i -1][B $i]$w."} 98
puts "1[B $i]$w, 1[B $i].
Go to the store and buy some more, 99[B 9]$w."

manifestación