Bonjour!

Arrr ... Ahoy là, moi les copains! Dérouler la grand-voile! Plein à tribord! Ah, sens le vent dans tes cheveux!

Bon, mon coeur ... J'ai besoin d'un peu de compétences en codage! Mon équipage est un peu plus modernisé que moi-même ... Je préfère toujours les "points de la boussole" (voir ici pour plus d'informations, Arrr ...) tandis que mon équipage est toujours en train d'utiliser une tête. Je veux un moyen facile de convertir ce twaddle qu'ils appellent un headin en ce que je peux comprendre, Arrr!

Ce dont j'ai besoin, c'est d'un code qui prenne l'entrée d'un nombre (les décimales sont correctes) tel que ' 0 <= the headin' < 360et ça crache la réponse comme le point de boussole le plus proche! Arrr!

Voici quelques exemples:

> heading 0.1

North

> heading 11.25

North by East

> heading 22.7

North Northeast

> heading 44.99

Northeast

> heading 91

East

Maintenant, si vous êtes en train de donner de telle sorte qu'il tombe exactement au milieu de deux points cardinaux, ne vous inquiétez pas, mon cœur ... j'attends que le code crache Between <point 1> and <point 2>, par exemple heading 5.625dira que Between North and North by East cela ne se produit que pour une tête qui doit satisfaire l'équation H = 5.625 + 11.25(N)où H est la tête et N un entier compris entre 0 et 31 inclus ...

Deux restrictions ...

1) Je ne veux pas que vous utilisiez des tableaux pour stocker vos données pour les points ou les `` headin's '' ... t 'être calculé, jus' comme au bon vieux temps! Arrr!

2) Le code le plus court l'emporte, ou je vous fais marcher la planche ... Arrr!

Perl 5,10 en utilisant une substitution, 231 228 226 224

@c=(north,east,south,west);
@q=qw(P PbR P-Q QbP Q QbR R-Q RbP);
sub p{$j=$_[0]>>3&3;$_=$q[7&pop];s/P/$c[$j]/;s/Q/$c[$j+1&2]$c[$j|1]/;s/R/$c[$j+1&3]/;s/b/ by /;ucfirst}
$a=<>/11.25+.5;
say$a==int$a&&'Between '.p($a-1).' and ',p$a

Ajout de quatre nouvelles lignes pour plus de lisibilité.

Edit: Golfed 2 octets de plus en utilisant pop. Merci @Dom Hastings

Modifier: 2 octets de moins en utilisantqw()

Je passe trop de temps sur cette chasse au trésor ici, mais voici une solution en Java :

public class Aaaaarrrr {

    public static void main(String[] aaarrrgs) {

        float heading = Float.parseFloat(aaarrrgs[0]);

        final List<String> points = Arrays.asList("North",
                "North by east", "North-northeast", "Northeast by north",
                "Northeast", "Northeast by east", "East-northeast",
                "East by north", "East", "East by south", "East-southeast",
                "Southeast by east", "Southeast", "Southeast by south",
                "South-southeast", "South by east", "South", "South by west",
                "South-southwest", "Southwest by south", "Southwest",
                "Southwest by west", "West-southwest", "West by south", "West",
                "West by north", "West-northwest", "Northwest by west",
                "Northwest", "Northwest by north", "North-northwest",
                "North by west");


        float cp = heading / 360.0f * 32.0f;
        if (cp % 1 == 0.5f)
             System.out.print("Between " + points.get((int)Math.floor(cp)) + " and ");

         System.out.println(points.get(Math.round(cp)));

    }
}

modifier Si je minimise le code ci-dessus et le rend vraiment laid, il deviendrait ceci:

Java, 770 caractères

import java.util.*;class A{public static void main(String[] r){List<String> l=Arrays.asList("North","North by east","North-northeast","Northeast by north","Northeast","Northeast by east","East-northeast","East by north","East","East by south","East-southeast","Southeast by east","Southeast","Southeast by south","South-southeast","South by east","South","South by west","South-southwest","Southwest by south","Southwest","Southwest by west","West-southwest","West by south", "West","West by north","West-northwest","Northwest by west","Northwest","Northwest by north","North-northwest","North by west");float c=Float.parseFloat(r[0])/360.0f*32.0f;if (c%1==0.5f) System.out.print("Between "+l.get((int)Math.floor(c))+" and ");System.out.println(l.get(Math.round(c)));}}

Python, 264

n='north'
e='east'
s='south'
w='west'
b=' by '
def f(H):x,y,z=(n,e,s,w,e,s,w,n,n+e,s+e,s+w,n+w)[int(H%360/90)::4];return(x,x+b+y,x+'-'+z,z+b+x,z,z+b+y,y+'-'+z,y+b+x)[int(H%90*4/45)].capitalize()
h=input()+5.625
print h%11.25and f(h)or'Between '+f(h-1)+' and '+f(h)

Cela utilise la capitalisation selon la page wikipedia et devrait fonctionner pour n'importe quel nombre.

Arrr Python, 336

A,R,r=int,input()/360.*32,' by #South#north#West#East#south#North#west#east#-#/#Between#and'.split('#')
a=''.join(r[A(c,16)]for c in'6A608A6928A6802A68A6808A4928A402A4A405A4958A1808A18A1805A1958A108A1A107A1957A1705A17A1707A3957A305A3A302A3927A6707A67A6702A6927A607').split('/')
if R%1==.5:print r[11],a[A(R)],r[12],
print a[A(round(R))]

Merci @Jeen

Perl 5.10, 262 257 254

Un peu similaire à l'une des solutions Python:

$n=north;$e=east;$s=south;$w=west;
@d=($n,$n.$e,$e,$s.$e,$s,$s.$w,$w,$n.$w,$n);
sub p{$j=pop;$i=$j>>2;ucfirst(($d[$i],"$d[$i] by $d[$i+2&~1]","$d[$i+1&~1]-$d[$i|1]","$d[$i+1] by $d[$i&~1]")[$j&3])}
$a=<>/11.25+.5;
say$a==int$a&&'Between '.p($a-1).' and ',p$a

Ajout de quatre nouvelles lignes pour plus de lisibilité.

Edit: trois octets de moins grâce à @Dom Hastings