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 je 23 a 23 je 11 a 11 je 6 a 6 je 3 a 3 je 5 a 5 je 4 a 4 je magické číslo

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" }  

Komentáře

• NSStringCompareOptions.RegularExpressionSearch A myslel jsem si JS ' s String.fromCharCode byl upovídaný. : P
• Python a Ruby: string.replace.Swift: String.stringByReplacingOccurrencesOfString

Haskell, 285 – 180 = 105 bajtů

Ve skutečnosti neexistuje vůbec žádný integrovaný modul pro zobrazení čísla. Stále jsem nebyl spokojen se skóre. Neváhejte se vyjádřit. Budu však experimentovat dále. Skóre je stále lepší než skóre 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)  

využití

 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"  

Vysvětlení.

m je dostatečně triviální, ale c není. c je funkce, která spočítá počet znaků anglického názvu čísla.

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. 

Komentáře

• No jo? No, Swift má … uhm. .. vyšší skóre … (' nejsem moc dobrý v návratech)

C, 268 – 180 = 88 bajtů

 #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");}  

Vyzkoušejte zde .

Neoprávněné

 /* 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 bajtů

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)));}} 

De-golfed

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))); } } 

Komentáře

• Vím, že ' je to už déle než rok, ale můžete odstranit závorky v tomto ternárním přiřazení a změnit ==0 až <1. Takže: 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 bajtů )

QC, 265 – 30 – 150 = 85 bajtů

(✵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"+ 

Testovací sada

Neoprávněné:

(✵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"+ 

Komentáře

• Pokud ' nepoužíváte vestavěnou funkci pro získání délky číslo, takže ve skutečnosti můžete ze skóre odečíst dalších 150

JavaScript, 382 – 150 – 30 = 202 bajtů

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");})() 

Vstup je uveden jako parametr výrazu funkce okamžitě vyvolaného.

Testovací vstup:

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 

Od golfu:

// 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 bajtů

Přinejmenším není delší než Java! Je založen na toto kromě použití řetězců.

EDIT : I zapomněl jsem na 0 a že musím říct " 5 je 4 ", nepřeskočit na " 4 je magické číslo " – které trochu přidalo skóre.

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) 

Vyzkoušejte to zde!

Komentáře

• Nepoužijete ' nemusíte zobrazovat jméno, že?
• Toto je nesprávné. i(5) vytiskne 4 is the magic number namísto 5 is 4 and 4 is the magic number.

PHP, 168 – 30 = 138 bajtů

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 bajtů – 30 = 146/158

Verze s 188 bajty:

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 bajtů závislá na implementaci verze:

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" 

Obě jsou funkce.