Special pythagorean triplet: Difference between revisions
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=={{header|Raku}}== |
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<lang perl6>hyper for 1..998 -> $a { |
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my $a2 = $a²; |
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for $a + 1 .. 999 -> $b { |
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my $c = 1000 - $a - $b; |
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last if $c < $b; |
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say "$a² + $b² = $c²\n$a + $b + $c = 1000" and exit if $a2 + $b² == $c² |
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} |
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}</lang> |
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{{out}} |
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<pre>200² + 375² = 425² |
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200 + 375 + 425 = 1000</pre> |
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=={{header|Ring}}== |
=={{header|Ring}}== |
Revision as of 11:34, 30 August 2021
Special pythagorean triplet is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
- The following problem is taken from Project Euler
- Related task
Go
<lang go>package main
import (
"fmt" "time"
)
func main() {
start := time.Now() for a := 3; ; a++ { for b := a + 1; ; b++ { c := 1000 - a - b if c <= b { break } if a*a+b*b == c*c { fmt.Printf("a = %d, b = %d, c = %d\n", a, b, c) fmt.Println("a + b + c =", a+b+c) fmt.Println("a * b * c =", a*b*c) fmt.Println("\nTook", time.Since(start)) return } } }
}</lang>
- Output:
a = 200, b = 375, c = 425 a + b + c = 1000 a * b * c = 31875000 Took 77.664µs
Julia
julia> [(a, b, c) for a in 1:1000, b in 1:1000, c in 1:1000 if a + b + c == 1000 && a^2 + b^2 == c^2] 2-element Vector{Tuple{Int64, Int64, Int64}}: (375, 200, 425) (200, 375, 425)
Python
Python 3.8.8 (default, Apr 13 2021, 15:08:03) Type "help", "copyright", "credits" or "license" for more information. >>> [(a, b, c) for a in range(1, 1000) for b in range(a, 1000) for c in range(1000) if a + b + c == 1000 and a*a + b*b == c*c] [(200, 375, 425)]
Raku
<lang perl6>hyper for 1..998 -> $a {
my $a2 = $a²; for $a + 1 .. 999 -> $b { my $c = 1000 - $a - $b; last if $c < $b; say "$a² + $b² = $c²\n$a + $b + $c = 1000" and exit if $a2 + $b² == $c² }
}</lang>
- Output:
200² + 375² = 425² 200 + 375 + 425 = 1000
Ring
<lang ring> load "stdlib.ring" see "working..." + nl
time1 = clock() for a = 1 to 1000
for b = a+1 to 1000 for c = b+1 to 1000 if (pow(a,2)+pow(b,2)=pow(c,2)) if a+b+c = 1000 see "a = " + a + " b = " + b + " c = " + c + nl exit 3 ok ok next next
next
time2 = clock() time3 = time2/1000 - time1/1000 see "Elapsed time = " + time3 + " s" + nl see "done..." + nl </lang>
- Output:
working... a = 200 b = 375 c = 425 Elapsed time = 497.61 s done...
Wren
Very simple approach, only takes 0.013 seconds even in Wren. <lang ecmascript>var a = 3 while (true) {
var b = a + 1 while (true) { var c = 1000 - a - b if (c <= b) break if (a*a + b*b == c*c) { System.print("a = %(a), b = %(b), c = %(c)") System.print("a + b + c = %(a + b + c)") System.print("a * b * c = %(a * b * c)") return } b = b + 1 } a = a + 1
}</lang>
- Output:
a = 200, b = 375, c = 425 a + b + c = 1000 a * b * c = 31875000