func c(var s:Int)->String{var r="";while(s != 4){r+="\(s)";let f=NSNumberFormatter();f.numberStyle=NSNumberFormatterStyle.SpellOutStyle;let v=f.stringFromNumber(s)!;s=v.stringByReplacingOccurrencesOfString("[- ]",withString:"",options:NSStringCompareOptions.RegularExpressionSearch,range:Range<String.Index>(start:v.startIndex,end:v.endIndex)).utf8.count+(s<0 ?3:0);r+=" is \(s) and "};return r+"4 is the magic number"}  

100003 é 23 e 23 é 11 e 11 é 6 e 6 é 3 e 3 é 5 e 5 é 4 e 4 é o número mágico

Ungolfed

 func a(var s: Int) -> String{ var r = "" while(s != 4){ r+="\(s)" let f = NSNumberFormatter() f.numberStyle = NSNumberFormatterStyle.SpellOutStyle let v = f.stringFromNumber(s)! s = v.stringByReplacingOccurrencesOfString( "[- ]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range: Range<String.Index>(start: v.startIndex, end: v.endIndex) ).utf8.count + (s < 0 ? 3 : 0) r+=" is \(s) and " } return r+"4 is the magic number" }  

Comentários

• NSStringCompareOptions.RegularExpressionSearch E pensei JS ' s String.fromCharCode foi prolixo. : P
• Python e Ruby: string.replace.Swift: String.stringByReplacingOccurrencesOfString

Haskell, 285 – 180 = 105 bytes

Na verdade, não há nenhum embutido para exibir o número. Ainda estou insatisfeito com o placar. Sinta-se à vontade para comentar. Vou experimentar mais, no entanto. Ainda assim, a pontuação é melhor do que a pontuação de Swift

 c n|n<0=8+c(-n)|n>999=r 1000+8|n>99=7+r 100|n>19=r 10+2-g[30..59]+g[20..29]|n>15=r 10-1|2>1=[0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7]!!n where{g=fromEnum.elem n;r k=c(mod n k)+c(div n k)} m 4="4 is the magic number" m 0="0 is 4 and "++m 4 m n=show n++" is "++show(c n)++" and "++m(c n)  

uso

 m 7 "7 is 5 and 5 is 4 and 4 is the magic number" m 999999 "999999 is 50 and 50 is 5 and 5 is 4 and 4 is the magic number"  

Explicação.

m é trivial o suficiente, no entanto, o c não é. c é a função para contar o número de caracteres do nome em inglês de número.

c n |n<0=8+c(-n) -- Add word "negative" in front of it, the length is 8 |n>999=r 1000+8 -- the english name for number with form xxx,yyy is xxx thousand yyy |n>99=7+r 100 -- the english name for number with form xyy is x hundred yy |n>19=r 10+2-g[30..59]+g[20..29] -- the english name for number with form xy with x more -- than 1 is x-ty. However *twoty>twenty, -- *threety>thirty, *fourty>forty, *fivety>fifty. |n>10=r 10-1-g(15:18:[11..13]) -- the english name for number with form 1x is x-teen. -- However, *oneteen>eleven, *twoteen>twelve, -- *threeteen>thirteen, *fiveteen>fifteen, -- *eightteen>eighteen |2>1=[0,3,3,5,4,4,3,5,5,4,3]!!n -- for number 0-10, the length is memorized. 0 is 0 -- because it is omitted. Input zero is handled -- separately. If we defined 0 to be 4, then -- 20 => twenty zero. where g =fromEnum.elem n -- Check if n is element of argument array, if true, 1 else 0 r k=c(mod n k)+c(div n k) -- Obvious. 

Comentários

• Ah sim? Bem, Swift tem … uhm. .. uma pontuação mais alta … (eu ' não sou muito bom em respostas)

C, 268 – 180 = 88 bytes

 #define t(x,o)n<x?o:f(n/x)+(n%x?f(n%x):0) char*i="4335443554366887798866555766";f(n){return t(1000,t(100,n<20?n<0?8+f(-n):i[n]-48:i[n/10+18]-48+(n%10?f(n%10):0))+7)+8;}main(n){for(scanf("%d",&n);n^4;n=f(n))printf("%d is %d and ",n,f(n));puts("4 is the magic number");}  

Experimente aqui .

Ungolfed

 /* Encode number length in string (shorter representation than array) */ char*i="4335443554366887798866555766"; f(n) { return n < 1000 ? n < 100 ? n < 20 ? n < 0 ? 8 + f(-n) /* "Negative x" */ : i[n] - 48 /* "x" */ : i[n/10+18] + (n%10 ? f(n%10) : 0) /* 20-99 */ : f(n/100) + (n%100 ? f(n%100) : 0) + 7 /* x hundred y */ : f(n/1000) + (n%1000 ? f(n%1000) : 0) + 8; /* x thousand y */ } main(n) { /* Keep printing until you get to the magic number */ for(scanf("%d",&n);n^4;n=f(n)) printf("%d is %d and ",n,f(n)); puts("4 is the magic number"); }  

Java, 800 – 150 = 650 bytes

class G{static String e="",i="teen",j="ty",k="eigh",y="thir",d="zero",l="one",n="two",m="three",h="four",s="five",c="six",t="seven",b=k+"t",g="nine",D="ten",L="eleven",N="twelve",M=y+i,H=h+i,S="fif"+i,C=c+i,T=t+i,B=k+i,G=g+i,o="twen"+j,p=y+j,q="for"+j,r="fif"+j,u=c+j,v=t+j,w=k+j,x=g+j,A=" ",O=" hundred ",z,E;public static void main(String a[]){z=e;int l=new Integer(a[0]);do{E=a(l,1,e);l=E.replace(A,e).length();z=z+E+" is "+a(l,1,e)+" and ";}while(l!=4);System.out.print(z+h+" is the magic number");}static String a(int P,int _,String Q){String[]f={e,l,n,m,h,s,c,t,b,g,D,L,N,M,H,S,C,T,B,G,e,D,o,p,q,r,u,v,w,x};int R=20,X=10,Y=100,Z=1000;return P==0?(_>0?d:e):(P<R?f[P]+Q:(P<Y?(f[R+(P/X)]+" "+a(P%X,0,e)).trim()+Q:(P<Z?a(P/Y,0,O)+a(P%Y,0,e)+Q:a(P/Z,0," thousand ")+a((P/Y)%X,0,O)+a(P%Y,0,e)+Q)));}} 

Jogado golfe

class G { static String e="",i="teen",j="ty",k="eigh",y="thir",d="zero",l="one",n="two",m="three",h="four",s="five",c="six",t="seven",b=k+"t",g="nine",D="ten",L="eleven",N="twelve",M=y+i,H=h+i,S="fif"+i,C=c+i,T=t+i,B=k+i,G=g+i,o="twen"+j,p=y+j,q="for"+j,r="fif"+j,u=c+j,v=t+j,w=k+j,x=g+j,A=" ",O=" hundred ",z,E; public static void main(String a[]){ z = e; int l = new Integer(a[0]); do { E = a(l,1,e); l = E.replace(A,e).length(); z = z+E+" is "+a(l,1,e)+" and "; } while(l!=4); System.out.println(z+h+" is the magic number"); } static String a(int P,int _,String Q) { String[] f = {e,l,n,m,h,s,c,t,b,g,D,L,N,M,H,S,C,T,B,G,e,D,o,p,q,r,u,v,w,x}; int R=20,X=10,Y=100,Z=1000; return P==0?(_>0?d:e):(P<R?f[P]+Q:(P<Y?(f[R+(P/X)]+" "+a(P%X,0,e)).trim()+Q:(P<Z?a(P/Y,0,O)+a(P%Y,0,e)+Q:a(P/Z,0," thousand ")+ a((P/Y)%X,0,O)+a(P%Y,0,e)+Q))); } } 

Comentários

• Eu sei que ' já faz mais de um ano, mas você pode remover alguns parênteses nessa atribuição ternária, bem como alterar ==0 para <1. Portanto: return P<1?_>0?d:e:P<R?f[P]+Q:P<Y?(f[R+(P/X)]+" "+a(P%X,0,e)).trim()+Q:P<Z?a(P/Y,0,O)+a(P%Y,0,e)+Q:a(P/Z,0," thousand ")+a((P/Y)%X,0,O)+a(P%Y,0,e)+Q; ( – 10 bytes )

QC, 265 – 30 – 150 = 85 bytes

(✵1:oaT%=ta100%=ha100/⌋T%=X[0 3 3 5 4 4 3 5 5 4 3 6 6 8 8 7 7 9 8 8]=Y[6 6 5 5 5 7 6 6]=a0≟4a20<Xt☌YtT/⌋2-☌Xo☌+▲▲hXh☌7+0▲+)(❆1:na0<8*=ba‖1000/⌋=ca1000%=nbb✵8+0▲a✵++){I4≠:EEI" is "++=II❆=EEI" and "++=E!}E"4 is the magic number"+ 

Conjunto de testes

Sem golfe:

(✵1: oaT%= # ones ta100%= # tens ha100/⌋T%= # hundreds X[0 3 3 5 4 4 3 5 5 4 3 6 6 8 8 7 7 9 8 8]= # length of "zero", "one", "two", ..., "nineteen" Y[6 6 5 5 5 7 6 6]= # length of "twenty", ..., "ninety" a0≟ 4 a20< Xt☌ YtT/⌋2-☌ Xo☌ + ▲ ▲ hXh☌7+0▲+) (❆1: na0<8*= # if negative, add 8 ba‖1000/⌋= # split aaaaaa into bbbccc ca1000%= n bb✵8+0▲ a✵ ++) {I4≠:EEI" is "++=II❆=EEI" and "++=E!}E"4 is the magic number"+ 

Comentários

• Se você ' não estiver usando uma função incorporada para obter o comprimento do número, para que você possa realmente subtrair outro 150 de sua pontuação

JavaScript, 382 – 150 – 30 = 202 bytes

var o=[0,3,3,5,4,4,3,5,5,4],f=s=>(s[1]==1?[3,6,6,8,8,7,7,9,8,8][s[0]]:o[s[0]]+(s.length>1?[0,3,6,6,5,5,5,7,6,6][s[1]]:0))+(s.length==3?(7+o[s[2]]-(o[s[2]]==0?7:0)):0),l=n=>{var s=(""+n).split("").reverse();return f(s.slice(0,3))+(s.length>3?(f(s.slice(3,6))+8):0)};(n=>{var s="";while(n!=4){s+=n+" is ";n=n>=0?l(n):(l(-n)+8);s+=n+" and ";}console.log(s+"4 is the magic number");})() 

A entrada é fornecida como o parâmetro para a expressão de função invocada imediatamente.

Teste de entrada:

999999 -> 999999 is 50 and 50 is 5 and 5 is 4 and 4 is the magic number 17 -> 17 is 9 and 9 is 4 and 4 is the magic number -404 -> -404 is 23 and 23 is 11 and 11 is 6 and 6 is 3 and 3 is 5 and 5 is 4 and 4 is the magic number 

Sem golfe:

// array of the lengths of digits in ones place: // one is 3, two is 3, three is 5, etc... zero is a special case // and is assigned zero length because zero is never written out in a number name var o=[0,3,3,5,4,4,3,5,5,4], // function that computes the length of a substring of the input // because the input is 6 digits, it can be broken into two 3 digit subsections // each of which can have it"s length calculated separately f=s=> ( s[1]==1? // check for if the tens digit is a one // when the tens is a one, pull the string length from an array that represents // ten, eleven, twelve, thirteen, etc... [3,6,6,8,8,7,7,9,8,8][s[0]] : // when the tens digit is not a one, add the ones digit normally and... o[s[0]] + // add the tens digit length from the array that represents // zero, ten, twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety (s.length>1?[0,3,6,6,5,5,5,7,6,6][s[1]]:0) ) + ( s.length==3? // check if the length is 3 and weren"t not accidentally trying to do something wierd with a minus sign // if so, then we have to add a hundred (7 characters) to the length and the // length of the ones digit that is in the hundreds place like // "one" hundred or "two" hundred (7+o[s[2]]- ( // also, if the hundreds place was a zero, subtract out those 7 characters // that were added because "hundred" isn"t added if there"s a zero in its // place o[s[2]]==0? 7 : 0 ) ) : // if the length wasn"t 3, then don"t add anything for the hundred 0 ), // function that computes the length of the whole six digit number l=n=>{ // coerce the number into a string and then reverse the string so that the // ones digit is the zeroth element instead of last element var s=(""+n).split("").reverse(); return // calculate the character length of the first 3 characters // like in the number 999888, this does the "888" f(s.slice(0,3)) + // then if there actually are any characters after the first 3 (s.length>3? // parse the character length of the second 3 characters (f(s.slice(3,6))+8) : 0 ) }; // lastly is the Immediately-Invoked Function Expression (n=>{ var s=""; // as long as we haven"t reached four, just keep going through the loop while(n!=4){ s+=n+" is "; n=n>=0?l(n):(l(-n)+8) // this handles negatives by only passing positive values to l and then just adding 8 onto the length for negatives s+=n+" and "; } // finally just say that "4 is the magic number" console.log(s+"4 is the magic number"); })(999999) 

Python 641-150 = 501 bytes

Pelo menos não é maior que Java! É baseado em isto exceto usando strings.

EDITAR : I esqueci 0 e que preciso dizer " 5 é 4 ", não pule para " 4 é o número mágico " – que acrescentou um pouco à pontuação.

w={0:"zero",1:"one",2:"two",3:"three",4:"four",5:"five",6:"six",7:"seven",8:"eight",9:"nine",10:"ten",11:"eleven",12:"twelve",13:"thirteen",14:"fourteen",15:"fifteen",16:"sixteen",17:"seventeen",18:"eighteen",19:"nineteen",20:"twenty",30:"thirty",40:"forty",50:"fifty",60:"sixty",70:"seventy",80:"eighty",90:"ninety"} s="" def i(n): global s e="" o=n%10 t=n%100 h=n/100%10 th=n/1000 if th: e+=i(th) e+="thousand" if h: e+=w[h] e+="hundred" if t: if t<20 or o==0: e+=w[t] else: e+=w[t-o] e+=w[o] if len(e)==4:s+="4 is the magic number";print s else: s+="%d is %d and "%(n,len(e));i(len(e)) In=input() i(In) 

Experimente aqui!

Comentários

• Você não ' não é necessário exibir o nome, certo?
• Isso está incorreto. i(5) imprime 4 is the magic number, em vez de 5 is 4 and 4 is the magic number.

PHP, 168 – 30 = 138 bytes

function m($i){$e=strlen(preg_replace("/[^a-z-]/","",(new NumberFormatter("en",5))->format($i)));echo($i==$e?"":"$i is $e and "),($e==4?"4 is the magic number":m($e));} 

Moo, 182 176 / 192 188 bytes – 30 = 146/158

Versão de 188 bytes:

u=$string_utils;s="";i=args[1];while(i-4)j=u:english_number(i);s=s+j+(s?" and "+j|"")+" is ";i=length(u:strip_chars(j,"- "}));endwhile;return s+(s?"four and "|"")+"four is the magic number" 

176 bytes dependente da implementação versão:

s="";i=args[1];while(i-4)j=#20:english_number(i);s=s+j+(s?" and "+j|"")+" is ";i=length(#20:strip_chars(j," -"));endwhile;return s+(s?"four and "|"")+"four is the magic number" 

Ambas são funções.