Kto nie lubi twierdzenia Pitagorasa a² + b² = c²? Napisz najkrótszą możliwą metodę w dowolnym języku, który przyjmuje wartość aib i wypisuje „Przeciwprostokąt tego prawego trójkąta to” + c. Zachowaj c do tylko trzech miejsc po przecinku.

APL (54)

'The hypotenuse of this right triangle is',3⍕.5*⍨+/⎕*2

Test:

      'The hypotenuse of this right triangle is',3⍕.5*⍨+/⎕*2
⎕:
      9 10
The hypotenuse of this right triangle is 13.454

Wyjaśnienie:

• ⎕*2: podnieś wartości na wejściu do drugiej mocy
• +/: weź sumę
• .5*⍨: podnieś wynik do 0,5 potęgi
• 3⍕: zaokrąglić do 3 miejsc po przecinku

TI-BASIC, 76 55 53 52 bajty

Input :Disp "THE HYPOTENUSE OF THIS RIGHT TRIANGLE IS
Fix 3:R▶Pr(X,Y

Nie, nawiasy zamykające nie są wymagane. Ponadto mniej bajtów niż ta odpowiedź APL :)

Python 2.7 - 76 znaków

print'The hypotenuse of this right triangle is %.3f'%abs(input()+1j*input())

Wyjaśnienie

| a + ib | = √ (a ² + b ² ) = c

==> a ² + b ² = c ²

PJ na przeciwprostokątnej

Nauczyciel: „Czy możesz mi powiedzieć, co jest przeciwprostokątne?”

LJ: „Hipotenuza, proste pytanie. Jeśli wczoraj wieczorem odbyła się głośna impreza i przeczytałeś ją w gazecie, zatytułowanej High Party News”

Sclipting , 46 znaków

글坼各갠方終加감半方갾밈乘增貶껠矽녆둥긆둹댆뭴뉖멵댶넠닶눠덆둩댲걲늖덨덂건댦땡닦덬뉒걩댲밀⓶

Oczekuje danych wejściowych w postaci dwóch liczb (może być ułamkowych!) Oddzielonych spacją.

Jest to krótszy niż APL, pomimo konieczności użycia kilku niewygodnych sztuczek.

Wyjaśnienie

글坼 | split at space
各 | for each...
  갠方 | to the power of two
終
加 | add
감半方 | to the power of one half
갾밈乘 | multiply by 1000
增貶 | increment, then decrement (kludge for rounding)
껠矽 | insert '.' at 4th-last character position
녆둥긆둹댆뭴뉖멵댶넠닶눠덆둩댲걲늖덨덂건댦땡닦덬뉒걩댲밀⓶ | "The hypotenuse..."

dc 54

Styczne do wyniku odpowiedzi APL!

2^r2^+3kv[The hypotenuse of this right triangle is ]Pp

Test:

$ dc
3 4
2^r2^+3kv[The hypotenuse of this right triangle is ]Pp
The hypotenuse of this right triangle is 5.000

C, 77 lub 99

77 znaków, jeśli dane wejściowe mogą być tylko argumentami funkcji:

f(a,b){printf("The hypotenuse of this right triangle is %.3f\n",hypot(a,b));}

99, jeśli dane wejściowe należy odczytać ze standardowego wejścia:

a,b;f(){scanf("%d %d",&a,&b);printf("The hypotenuse of this right triangle is %.3f\n",hypot(a,b));}

Ogromne podziękowania dla @Yimin Rong!

Powershell

Just to see if i could...

echo "The hypotenuse of this right triangle is " ([math]::round([math]::sqrt(([math]::pow(([double](Read-Host -p "A")),2) + [math]::pow(([double](Read-Host -p "B")),2))),3))

Rubin, 94 90 82 znaki

p "The hypotenuse of this right triangle is %.3f"%(Math.sqrt(gets.to_i**2+gets.to_i**2))

Aktualizacja (dzięki za komentarze):

p "The hypotenuse of this right triangle is %.3f"%(gets.to_i**2+gets.to_i**2)**0.5

MATLAB 79 74

@(a,b)sprintf('The hypotenuse of this right triangle is %.3f',norm([a b]))

Python 2.7 - 80 chars

print'The hypotenuse of this right triangle is %.3f'%(input()**2+input()**2)**.5

C++ - 90

void h(int a,int b){printf("The hypotenuse of this right triangle is %.3f\n",hypot(a,b));}

Perl 6 (68 74 bytes)

{printf "The hypotenuse of this right triangle is %.3f
",sqrt [+] @_ X**2}

{} declares a lambda function. [+] is sum operator, X** is cross power operator (for example, 1, 2 X+ 10, 20 gives 11, 21, 12, 22). In this case, cross power operator takes one argument, so the result has the same length as @_. @_ contains all function arguments.

If it's disallowed to have function that may take wrong number of arguments (unsafe), it's possible to replace [+] @_ X**2 with $^a**2+$^b**2, where $^a and $^b are placeholder arguments.

Javascript (97)

x=prompt;a=x(),b=x();x('The hypotenuse of this right triangle is '+Math.sqrt(a*a+b*b).toFixed(3))

C, 100 chars (beats the other C solution by 1!)

A ridiculously inefficient algorithm.

x;f(a,b){for(;x-a*a-b*b;x=rand());printf("The hypotenuse of this right triangle is %.3f",sqrt(x));}

DELPHI / PASCAL

With indent (157)

program p;
{$APPTYPE CONSOLE}
var a,b:integer;
begin
     readln(a,b);
     writeln('the hypotenuse of this right triangle is',sqrt(b*b+a*a):2:3);
end.

EcmaScript 6, 82 79

f=(a,b)=>"The hypotenuse of this right triangle is "+Math.hypot(a,b).toFixed(3)

Usage:

f(3, 5)
> "The hypotenuse of this right triangle is 5"

Update: Switch to Math.hypot()

Golfscript (69 67 66 65)

This would be much easier if floating point was actually supported without resorting to workarounds... :)

~'The hypotenuse of this right triangle is '@.*@.*+2-1??+.'.'?4+<

A link to test it.

Python 2 (79)

def p(a,b):print'The hypotenuse of this right triangle is %.3d'%((a*a+b*b)**.5)

AWK — 84 78 characters

awk '{printf"The hypotenuse of this right triangle is %.3f\n",($1^2+$2^2)^.5}'

Thanks to Wasi for suggesting ^ operator and removing ()!

e.g.

$ echo 3 4 | awk '{printf"The hypotenuse of this right triangle is %.3f\n",($1^2+$2^2)^.5}'
The hypotenuse of this right triangle is 5.000

PowerShell: 111

Golfed Code

1..2|%{sv $_ (read-host)};"The hypotenuse of this right triangle is $("{0:N3}"-f[math]::sqrt($1/1*$1+$2/1*$2))"

Walkthrough

1..2|%{sv $_ (read-host)}; Gets two inputs interactively from the user, and stores them in $1 and $2. Might be able to cut some length by using arguments or pipeline inputs instead.

"The hypotenuse of this right triangle is Required text in the output, per the challenge specifications.

$(...)" Encapsulated code block will be processed as script before being included in the output.

"{0:N3}"-f Formats output from the next bit of code as a number with exactly three digits after the decimal point.

[math]::sqrt(...) Gets the square root of the encapsulated value.

$1/1*$1+$2/1*$2 Serves as our "a^2+b^2". Multiplying a number by itself is the shortest way to square it in PowerShell, but the variables need to be divided by 1 first to force them to integers. Otherwise, they are treated as text and 3*3+4*4 would be 3334444 instead of 25.

JavaScript: 83

i=prompt,'The hypotenuse of this right triangle is '+Math.hypot(i(),i()).toFixed(3)

Currently the shortest JS implementation using stdin :D
Works only on Firefox 27.0+ (EcmaScript 6)

JavaScript: 78

If we can use just two variables (as lot of scripts do here):

a=2,b=3,'The hypotenuse of this right triangle is '+Math.hypot(a,b).toFixed(3)

dc, 55

3k?d*?d*+v[The hypotenuse of this right triangle is ]Pp

Java, 112

(Also prints out a No Such Method error, though I'm not sure if this is against the rules)

class A{static{int a=1,b=1;System.out.printf("The hypotenuse of this right triangle is %.3f",Math.hypot(a,b));}}

Java, 149

(No error)

class A{static{int a=1,b=1;System.out.printf("The hypotenuse of this right triangle is %.3f",Math.hypot(a,b));}public static void main(String[] a){}}

C#

Method Only (114)

void H(double a, double b)
{
    Console.Write("The hypotenuse of this right triangle is {0:N3}", Math.Sqrt(a * a + b * b)); 
}

Complete Program (171)

using System;
class P
{        
   static void H(double a, double b)
   {
     Console.Write("The hypotenuse of this right triangle is {0:N3}", Math.Sqrt(a * a + b * b));
   }                
   static void Main()
   {
    H(3, 4);
   }
}

Complete Program (without using method - 141)

using System;class P{static void Main(){double a=3,b=4;Console.Write("The hypotenuse of this right triangle is {0:N3}",Math.Sqrt(a*a+b*b));}}

JavaScript 118 106 93

Unlike @micha's solution, mine takes in two variables via function and sends the alert of the result.

function(a,b){m=Math;c=d=>d*d,e=1e3;alert("The hypotenuse of this right triangle is "+m.round(m.sqrt(c(a)+c(b))*e)/e)}

function(a,b){e=1e3;alert("The hypotenuse of this right triangle is "+Math.round(Math.sqrt(a*a+b*b)*e)/e)}

Fat arrow functions to the rescue!

h=(a,b,e=1e3)=>"The hypotenuse of this right triangle is "+Math.round(Math.sqrt(a*a+b*b)*e)/e

c64 basic v2, 60 66 bytes

0inputa,b:?"The hypotenuse of this right triangle is";sQ(a*a+b*b)

Screenshot:

How to try it.

R, 61 76 bytes

cat("The hypotenuse of this right triangle is",round(sqrt(sum(scan()^2)),3))

cat displays its content to STDOUT.

The scan() function takes user's input from keyboard. This input exists as a vector, on which the ^2 is applied (^function is vectorized), and the sum() sums the elements of the vector. sqrt outputs the square-root, which is rounded to 3 decimal places by round(,3)

Thanks to @caird coinheringaahing for noticing that the previous answer didn't round.

ARBLE, 73 bytes

"The hypotenuse of this right triangle is "..floor(sqrt(a^2+b^2)*1e3)/1e3

Try it online!

OML, 57 bytes

"The hypotenuse of this right triangle is "shnhn+A6`*N3eD

Try it online!

Part 1

This simply outputs the string

"The hypotenuse of this right triangle is "s

Part 2

hnhn+A6`*N3eD
hn              take input and square it
  hn            take another input and square it
    +           add them
     A6`        push 10^6
        *       multiply the sum with that number
         N      take integer square root
          3eD   output with three places of precision
                implicit output

Jelly, 32 characters

,²S½ær3µ,“¡ÆC⁷⁺ɱSoṿȤç½?⁶Ẏtḍỵŀ»ṚK

Try it online!

There is probably a better string compression that allows me to get around needing to join with spaces but I was having trouble finding it.

Explanation:

,²S½ær3µ,“...»ṚK    Example inputs: 3, 4
,                   Pair the inputs. Result: [3, 4]
 ²                  Square them. Result: [9, 16]
  S                 Sum them. Result: 25
   ½                Get the square root of the sum. Result: 5
    ær3             Round to 3 decimal places. Result: 5
       µ            Take the result of that... Result: 5
         “...»       ...and  the compressed string Result: "The hypotenuse of this right triangle is"
        ,           And put them into a pair. Result: [5, "The hypotenuse of this right triangle is"]
              Ṛ     Reverse that. Result: ["The hypotenuse of this right triangle is", 5]
               k    Join it with spaces. Result: "The hypotenuse of this right triangle is 5.0"
                    Implicit output.