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Generate a PHP translation of this J snippet without changing its computational steps.
1 3 _5 +/ . * 4 _2 _1 3 dotp=: +/ . * 1 3 _5 dotp 4 _2 _1 3
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Produce a language-to-language conversion: from Julia to PHP, same semantics.
x = [1, 3, -5] y = [4, -2, -1] z = dot(x, y) z = x'*y z = x ⋅ y
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Write the same code in PHP as shown below in Julia.
x = [1, 3, -5] y = [4, -2, -1] z = dot(x, y) z = x'*y z = x ⋅ y
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the following Lua code into PHP without altering its purpose.
function dotprod(a, b) local ret = 0 for i = 1, #a do ret = ret + a[i] * b[i] end return ret end print(dotprod({1, 3, -5}, {4, -2, 1}))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the following Lua code into PHP without altering its purpose.
function dotprod(a, b) local ret = 0 for i = 1, #a do ret = ret + a[i] * b[i] end return ret end print(dotprod({1, 3, -5}, {4, -2, 1}))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Generate an equivalent PHP version of this Mathematica code.
{1,3,-5}.{4,-2,-1}
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Preserve the algorithm and functionality while converting the code from Mathematica to PHP.
{1,3,-5}.{4,-2,-1}
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Generate a PHP translation of this MATLAB snippet without changing its computational steps.
A = [1 3 -5] B = [4 -2 -1] C = dot(A,B)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Generate a PHP translation of this MATLAB snippet without changing its computational steps.
A = [1 3 -5] B = [4 -2 -1] C = dot(A,B)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Transform the following Nim implementation into PHP, maintaining the same output and logic.
proc dotp[T](a,b: T): int = doAssert a.len == b.len for i in a.low..a.high: result += a[i] * b[i] echo dotp([1,3,-5], [4,-2,-1]) echo dotp(@[1,2,3],@[4,5,6])
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the following Nim code into PHP without altering its purpose.
proc dotp[T](a,b: T): int = doAssert a.len == b.len for i in a.low..a.high: result += a[i] * b[i] echo dotp([1,3,-5], [4,-2,-1]) echo dotp(@[1,2,3],@[4,5,6])
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Rewrite this program in PHP while keeping its functionality equivalent to the OCaml version.
let dot = List.fold_left2 (fun z x y -> z +. x *. y) 0.
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Can you help me rewrite this code in PHP instead of OCaml, keeping it the same logically?
let dot = List.fold_left2 (fun z x y -> z +. x *. y) 0.
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Produce a language-to-language conversion: from Perl to PHP, same semantics.
sub dotprod { my($vec_a, $vec_b) = @_; die "they must have the same size\n" unless @$vec_a == @$vec_b; my $sum = 0; $sum += $vec_a->[$_] * $vec_b->[$_] for 0..$ return $sum; } my @vec_a = (1,3,-5); my @vec_b = (4,-2,-1); print dotprod(\@vec_a,\@vec_b), "\n";
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the programming language of this snippet from Perl to PHP without modifying what it does.
sub dotprod { my($vec_a, $vec_b) = @_; die "they must have the same size\n" unless @$vec_a == @$vec_b; my $sum = 0; $sum += $vec_a->[$_] * $vec_b->[$_] for 0..$ return $sum; } my @vec_a = (1,3,-5); my @vec_b = (4,-2,-1); print dotprod(\@vec_a,\@vec_b), "\n";
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Please provide an equivalent version of this PowerShell code in PHP.
function dotproduct( $a, $b) { $a | foreach -Begin {$i = $res = 0} -Process { $res += $_*$b[$i++] } -End{$res} } dotproduct (1..2) (1..2) dotproduct (1..10) (11..20)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Can you help me rewrite this code in PHP instead of PowerShell, keeping it the same logically?
function dotproduct( $a, $b) { $a | foreach -Begin {$i = $res = 0} -Process { $res += $_*$b[$i++] } -End{$res} } dotproduct (1..2) (1..2) dotproduct (1..10) (11..20)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Generate a PHP translation of this R snippet without changing its computational steps.
x <- c(1, 3, -5) y <- c(4, -2, -1) sum(x*y) x %*% y dotp <- function(x, y) { n <- length(x) if(length(y) != n) stop("invalid argument") s <- 0 for(i in 1:n) s <- s + x[i]*y[i] s } dotp(x, y)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Maintain the same structure and functionality when rewriting this code in PHP.
x <- c(1, 3, -5) y <- c(4, -2, -1) sum(x*y) x %*% y dotp <- function(x, y) { n <- length(x) if(length(y) != n) stop("invalid argument") s <- 0 for(i in 1:n) s <- s + x[i]*y[i] s } dotp(x, y)
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Convert the following code from Racket to PHP, ensuring the logic remains intact.
#lang racket (define (dot-product l r) (for/sum ([x l] [y r]) (* x y))) (dot-product '(1 3 -5) '(4 -2 -1)) (dot-product #(1 2 3) #(4 5 6))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Rewrite this program in PHP while keeping its functionality equivalent to the Racket version.
#lang racket (define (dot-product l r) (for/sum ([x l] [y r]) (* x y))) (dot-product '(1 3 -5) '(4 -2 -1)) (dot-product #(1 2 3) #(4 5 6))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Transform the following REXX implementation into PHP, maintaining the same output and logic.
options replace format comments java crossref savelog symbols binary whatsTheVectorVictor = [[double 1.0, 3.0, -5.0], [double 4.0, -2.0, -1.0]] dotProduct = Rexx dotProduct(whatsTheVectorVictor) say dotProduct.format(null, 2) return method dotProduct(vec1 = double[], vec2 = double[]) public constant returns double signals IllegalArgumentException if vec1.length \= vec2.length then signal IllegalArgumentException('Vectors must be the same length') scalarProduct = double 0.0 loop e_ = 0 to vec1.length - 1 scalarProduct = vec1[e_] * vec2[e_] + scalarProduct end e_ return scalarProduct method dotProduct(vecs = double[,]) public constant returns double signals IllegalArgumentException return dotProduct(vecs[0], vecs[1])
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Rewrite this program in PHP while keeping its functionality equivalent to the REXX version.
options replace format comments java crossref savelog symbols binary whatsTheVectorVictor = [[double 1.0, 3.0, -5.0], [double 4.0, -2.0, -1.0]] dotProduct = Rexx dotProduct(whatsTheVectorVictor) say dotProduct.format(null, 2) return method dotProduct(vec1 = double[], vec2 = double[]) public constant returns double signals IllegalArgumentException if vec1.length \= vec2.length then signal IllegalArgumentException('Vectors must be the same length') scalarProduct = double 0.0 loop e_ = 0 to vec1.length - 1 scalarProduct = vec1[e_] * vec2[e_] + scalarProduct end e_ return scalarProduct method dotProduct(vecs = double[,]) public constant returns double signals IllegalArgumentException return dotProduct(vecs[0], vecs[1])
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Generate an equivalent PHP version of this Ruby code.
class Vector property x, y, z def initialize(@x : Int64, @y : Int64, @z : Int64) end def dot_product(other : Vector) (self.x * other.x) + (self.y * other.y) + (self.z * other.z) end end puts Vector.new(1, 3, -5).dot_product Vector.new(4, -2, -1) class Array def dot_product(other) raise "not the same size!" if self.size != other.size self.zip(other).sum { |(a, b)| a * b } end end p [8, 13, -5].dot_product [4, -7, -11]
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Transform the following Ruby implementation into PHP, maintaining the same output and logic.
class Vector property x, y, z def initialize(@x : Int64, @y : Int64, @z : Int64) end def dot_product(other : Vector) (self.x * other.x) + (self.y * other.y) + (self.z * other.z) end end puts Vector.new(1, 3, -5).dot_product Vector.new(4, -2, -1) class Array def dot_product(other) raise "not the same size!" if self.size != other.size self.zip(other).sum { |(a, b)| a * b } end end p [8, 13, -5].dot_product [4, -7, -11]
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Convert this Scala block to PHP, preserving its control flow and logic.
fun dot(v1: Array<Double>, v2: Array<Double>) = v1.zip(v2).map { it.first * it.second }.reduce { a, b -> a + b } fun main(args: Array<String>) { dot(arrayOf(1.0, 3.0, -5.0), arrayOf(4.0, -2.0, -1.0)).let { println(it) } }
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Translate the given Scala code snippet into PHP without altering its behavior.
fun dot(v1: Array<Double>, v2: Array<Double>) = v1.zip(v2).map { it.first * it.second }.reduce { a, b -> a + b } fun main(args: Array<String>) { dot(arrayOf(1.0, 3.0, -5.0), arrayOf(4.0, -2.0, -1.0)).let { println(it) } }
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Convert the following code from Swift to PHP, ensuring the logic remains intact.
func dot(v1: [Double], v2: [Double]) -> Double { return reduce(lazy(zip(v1, v2)).map(*), 0, +) } println(dot([1, 3, -5], [4, -2, -1]))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Transform the following Swift implementation into PHP, maintaining the same output and logic.
func dot(v1: [Double], v2: [Double]) -> Double { return reduce(lazy(zip(v1, v2)).map(*), 0, +) } println(dot([1, 3, -5], [4, -2, -1]))
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the following Tcl code into PHP without altering its purpose.
package require math::linearalgebra set a {1 3 -5} set b {4 -2 -1} set dotp [::math::linearalgebra::dotproduct $a $b] proc pp vec {return \[[join $vec ,]\]} puts "[pp $a] \u2219 [pp $b] = $dotp"
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Change the following Tcl code into PHP without altering its purpose.
package require math::linearalgebra set a {1 3 -5} set b {4 -2 -1} set dotp [::math::linearalgebra::dotproduct $a $b] proc pp vec {return \[[join $vec ,]\]} puts "[pp $a] \u2219 [pp $b] = $dotp"
<?php function dot_product($v1, $v2) { if (count($v1) != count($v2)) throw new Exception('Arrays have different lengths'); return array_sum(array_map('bcmul', $v1, $v2)); } echo dot_product(array(1, 3, -5), array(4, -2, -1)), "\n"; ?>
Please provide an equivalent version of this C code in Rust.
#include <stdio.h> #include <stdlib.h> int dot_product(int *, int *, size_t); int main(void) { int a[3] = {1, 3, -5}; int b[3] = {4, -2, -1}; printf("%d\n", dot_product(a, b, sizeof(a) / sizeof(a[0]))); return EXIT_SUCCESS; } int dot_product(int *a, int *b, size_t n) { int sum = 0; size_t i; for (i = 0; i < n; i++) { sum += a[i] * b[i]; } return sum; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Generate an equivalent Rust version of this C# code.
static void Main(string[] args) { Console.WriteLine(DotProduct(new decimal[] { 1, 3, -5 }, new decimal[] { 4, -2, -1 })); Console.Read(); } private static decimal DotProduct(decimal[] vec1, decimal[] vec2) { if (vec1 == null) return 0; if (vec2 == null) return 0; if (vec1.Length != vec2.Length) return 0; decimal tVal = 0; for (int x = 0; x < vec1.Length; x++) { tVal += vec1[x] * vec2[x]; } return tVal; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Preserve the algorithm and functionality while converting the code from Java to Rust.
public class DotProduct { public static void main(String[] args) { double[] a = {1, 3, -5}; double[] b = {4, -2, -1}; System.out.println(dotProd(a,b)); } public static double dotProd(double[] a, double[] b){ if(a.length != b.length){ throw new IllegalArgumentException("The dimensions have to be equal!"); } double sum = 0; for(int i = 0; i < a.length; i++){ sum += a[i] * b[i]; } return sum; } }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Transform the following Go implementation into Rust, maintaining the same output and logic.
package main import ( "errors" "fmt" "log" ) var ( v1 = []int{1, 3, -5} v2 = []int{4, -2, -1} ) func dot(x, y []int) (r int, err error) { if len(x) != len(y) { return 0, errors.New("incompatible lengths") } for i, xi := range x { r += xi * y[i] } return } func main() { d, err := dot([]int{1, 3, -5}, []int{4, -2, -1}) if err != nil { log.Fatal(err) } fmt.Println(d) }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Can you help me rewrite this code in Python instead of Rust, keeping it the same logically?
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
def dotp(a,b): assert len(a) == len(b), 'Vector sizes must match' return sum(aterm * bterm for aterm,bterm in zip(a, b)) if __name__ == '__main__': a, b = [1, 3, -5], [4, -2, -1] assert dotp(a,b) == 3
Can you help me rewrite this code in Python instead of Rust, keeping it the same logically?
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
def dotp(a,b): assert len(a) == len(b), 'Vector sizes must match' return sum(aterm * bterm for aterm,bterm in zip(a, b)) if __name__ == '__main__': a, b = [1, 3, -5], [4, -2, -1] assert dotp(a,b) == 3
Maintain the same structure and functionality when rewriting this code in VB.
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Private Function dot_product(x As Variant, y As Variant) As Double dot_product = WorksheetFunction.SumProduct(x, y) End Function Public Sub main() Debug.Print dot_product([{1,3,-5}], [{4,-2,-1}]) End Sub
Write the same code in Rust as shown below in C.
#include <stdio.h> #include <stdlib.h> int dot_product(int *, int *, size_t); int main(void) { int a[3] = {1, 3, -5}; int b[3] = {4, -2, -1}; printf("%d\n", dot_product(a, b, sizeof(a) / sizeof(a[0]))); return EXIT_SUCCESS; } int dot_product(int *a, int *b, size_t n) { int sum = 0; size_t i; for (i = 0; i < n; i++) { sum += a[i] * b[i]; } return sum; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Preserve the algorithm and functionality while converting the code from C++ to Rust.
#include <iostream> #include <numeric> int main() { int a[] = { 1, 3, -5 }; int b[] = { 4, -2, -1 }; std::cout << std::inner_product(a, a + sizeof(a) / sizeof(a[0]), b, 0) << std::endl; return 0; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Port the following code from Java to Rust with equivalent syntax and logic.
public class DotProduct { public static void main(String[] args) { double[] a = {1, 3, -5}; double[] b = {4, -2, -1}; System.out.println(dotProd(a,b)); } public static double dotProd(double[] a, double[] b){ if(a.length != b.length){ throw new IllegalArgumentException("The dimensions have to be equal!"); } double sum = 0; for(int i = 0; i < a.length; i++){ sum += a[i] * b[i]; } return sum; } }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Port the provided Go code into Rust while preserving the original functionality.
package main import ( "errors" "fmt" "log" ) var ( v1 = []int{1, 3, -5} v2 = []int{4, -2, -1} ) func dot(x, y []int) (r int, err error) { if len(x) != len(y) { return 0, errors.New("incompatible lengths") } for i, xi := range x { r += xi * y[i] } return } func main() { d, err := dot([]int{1, 3, -5}, []int{4, -2, -1}) if err != nil { log.Fatal(err) } fmt.Println(d) }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Transform the following Rust implementation into VB, maintaining the same output and logic.
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Private Function dot_product(x As Variant, y As Variant) As Double dot_product = WorksheetFunction.SumProduct(x, y) End Function Public Sub main() Debug.Print dot_product([{1,3,-5}], [{4,-2,-1}]) End Sub
Write the same code in Rust as shown below in C++.
#include <iostream> #include <numeric> int main() { int a[] = { 1, 3, -5 }; int b[] = { 4, -2, -1 }; std::cout << std::inner_product(a, a + sizeof(a) / sizeof(a[0]), b, 0) << std::endl; return 0; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Please provide an equivalent version of this C# code in Rust.
static void Main(string[] args) { Console.WriteLine(DotProduct(new decimal[] { 1, 3, -5 }, new decimal[] { 4, -2, -1 })); Console.Read(); } private static decimal DotProduct(decimal[] vec1, decimal[] vec2) { if (vec1 == null) return 0; if (vec2 == null) return 0; if (vec1.Length != vec2.Length) return 0; decimal tVal = 0; for (int x = 0; x < vec1.Length; x++) { tVal += vec1[x] * vec2[x]; } return tVal; }
fn dot_product(a: &[i32], b: &[i32]) -> Option<i32> { if a.len() != b.len() { return None } Some( a.iter() .zip( b.iter() ) .fold(0, |sum, (el_a, el_b)| sum + el_a*el_b) ) } fn main() { let v1 = vec![1, 3, -5]; let v2 = vec![4, -2, -1]; println!("{}", dot_product(&v1, &v2).unwrap()); }
Write the same code in C# as shown below in Ada.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Convert this Ada snippet to C and keep its semantics consistent.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Convert the following code from Ada to C++, ensuring the logic remains intact.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Ensure the translated Go code behaves exactly like the original Ada snippet.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Rewrite the snippet below in Java so it works the same as the original Ada code.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Write a version of this Ada function in Python with identical behavior.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Translate this program into VB but keep the logic exactly as in Ada.
with Ada.Command_Line, Ada.Text_IO, Prime_Numbers; procedure Count is package Prime_Nums is new Prime_Numbers (Number => Natural, Zero => 0, One => 1, Two => 2); use Prime_Nums; procedure Put (List : Number_List) is begin for Index in List'Range loop Ada.Text_IO.Put (Integer'Image (List (Index))); if Index /= List'Last then Ada.Text_IO.Put (" x"); end if; end loop; end Put; N : Natural := 1; Max_N : Natural := 15; begin if Ada.Command_Line.Argument_Count = 1 then Max_N := Integer'Value (Ada.Command_Line.Argument (1)); end if; loop Ada.Text_IO.Put (Integer'Image (N) & ": "); Put (Decompose (N)); Ada.Text_IO.New_Line; N := N + 1; exit when N > Max_N; end loop; end Count;
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Ensure the translated C code behaves exactly like the original Arturo snippet.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Port the following code from Arturo to C# with equivalent syntax and logic.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Please provide an equivalent version of this Arturo code in C++.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Change the programming language of this snippet from Arturo to Java without modifying what it does.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Can you help me rewrite this code in Python instead of Arturo, keeping it the same logically?
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Change the following Arturo code into VB without altering its purpose.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Preserve the algorithm and functionality while converting the code from Arturo to Go.
loop 1..30 'x [ fs: [1] if x<>1 -> fs: factors.prime x print [pad to :string x 3 "=" join.with:" x " to [:string] fs] ]
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Keep all operations the same but rewrite the snippet in C.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Change the programming language of this snippet from AutoHotKey to C# without modifying what it does.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Change the following AutoHotKey code into C++ without altering its purpose.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Transform the following AutoHotKey implementation into Java, maintaining the same output and logic.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Translate this program into Python but keep the logic exactly as in AutoHotKey.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Transform the following AutoHotKey implementation into VB, maintaining the same output and logic.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Convert the following code from AutoHotKey to Go, ensuring the logic remains intact.
factorize(n){ if n = 1 return 1 if n < 1 return false result := 0, m := n, k := 2 While n >= k{ while !Mod(m, k){ result .= " * " . k, m /= k } k++ } return SubStr(result, 5) } Loop 22 out .= A_Index ": " factorize(A_index) "`n" MsgBox % out
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Ensure the translated C code behaves exactly like the original AWK snippet.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Preserve the algorithm and functionality while converting the code from AWK to C#.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Rewrite the snippet below in C++ so it works the same as the original AWK code.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Port the provided AWK code into Java while preserving the original functionality.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Convert this AWK block to Python, preserving its control flow and logic.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Produce a language-to-language conversion: from AWK to VB, same semantics.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Write the same algorithm in Go as shown in this AWK implementation.
BEGIN { fmt = "%d=%s\n" for (i=1; i<=16; i++) { printf(fmt,i,factors(i)) } i = 2144; printf(fmt,i,factors(i)) i = 6358; printf(fmt,i,factors(i)) exit(0) } function factors(n, f,p) { if (n == 1) { return(1) } p = 2 while (p <= n) { if (n % p == 0) { f = sprintf("%s%s*",f,p) n /= p } else { p++ } } return(substr(f,1,length(f)-1)) }
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Can you help me rewrite this code in C instead of BBC_Basic, keeping it the same logically?
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Convert this BBC_Basic block to C#, preserving its control flow and logic.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Write a version of this BBC_Basic function in C++ with identical behavior.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Convert the following code from BBC_Basic to Java, ensuring the logic remains intact.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Change the programming language of this snippet from BBC_Basic to Python without modifying what it does.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Change the following BBC_Basic code into VB without altering its purpose.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Change the following BBC_Basic code into Go without altering its purpose.
FOR i% = 1 TO 20 PRINT i% " = " FNfactors(i%) NEXT END DEF FNfactors(N%) LOCAL P%, f$ IF N% = 1 THEN = "1" P% = 2 WHILE P% <= N% IF (N% MOD P%) = 0 THEN f$ += STR$(P%) + " x " N% DIV= P% ELSE P% += 1 ENDIF ENDWHILE = LEFT$(f$, LEN(f$) - 3)
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Produce a language-to-language conversion: from Common_Lisp to C, same semantics.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Write a version of this Common_Lisp function in C# with identical behavior.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Write a version of this Common_Lisp function in C++ with identical behavior.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Convert this Common_Lisp block to Java, preserving its control flow and logic.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Write a version of this Common_Lisp function in Python with identical behavior.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Produce a functionally identical VB code for the snippet given in Common_Lisp.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Convert the following code from Common_Lisp to Go, ensuring the logic remains intact.
(ns listfactors (:gen-class)) (defn factors "Return a list of factors of N." ([n] (factors n 2 ())) ([n k acc] (cond (= n 1) (if (empty? acc) [n] (sort acc)) (>= k n) (if (empty? acc) [n] (sort (cons n acc))) (= 0 (rem n k)) (recur (quot n k) k (cons k acc)) :else (recur n (inc k) acc)))) (doseq [q (range 1 26)] (println q " = " (clojure.string/join " x "(factors q))))
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Transform the following D implementation into C, maintaining the same output and logic.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Please provide an equivalent version of this D code in C#.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Translate the given D code snippet into C++ without altering its behavior.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Rewrite this program in Java while keeping its functionality equivalent to the D version.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Generate an equivalent Python version of this D code.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Translate the given D code snippet into VB without altering its behavior.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine
Keep all operations the same but rewrite the snippet in Go.
int[] factorize(in int n) pure nothrow in { assert(n > 0); } body { if (n == 1) return [1]; int[] result; int m = n, k = 2; while (n >= k) { while (m % k == 0) { result ~= k; m /= k; } k++; } return result; } void main() { import std.stdio; foreach (i; 1 .. 22) writefln("%d: %(%d × %)", i, i.factorize()); }
package main import "fmt" func main() { fmt.Println("1: 1") for i := 2; ; i++ { fmt.Printf("%d: ", i) var x string for n, f := i, 2; n != 1; f++ { for m := n % f; m == 0; m = n % f { fmt.Print(x, f) x = "×" n /= f } } fmt.Println() } }
Change the following Delphi code into C without altering its purpose.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
#include <stdio.h> #include <stdlib.h> typedef unsigned long long ULONG; ULONG get_prime(int idx) { static long n_primes = 0, alloc = 0; static ULONG *primes = 0; ULONG last, p; int i; if (idx >= n_primes) { if (n_primes >= alloc) { alloc += 16; primes = realloc(primes, sizeof(ULONG) * alloc); } if (!n_primes) { primes[0] = 2; primes[1] = 3; n_primes = 2; } last = primes[n_primes-1]; while (idx >= n_primes) { last += 2; for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > last) { primes[n_primes++] = last; break; } if (last % p == 0) break; } } } return primes[idx]; } int main() { ULONG n, x, p; int i, first; for (x = 1; ; x++) { printf("%lld = ", n = x); for (i = 0, first = 1; ; i++) { p = get_prime(i); while (n % p == 0) { n /= p; if (!first) printf(" x "); first = 0; printf("%lld", p); } if (n <= p * p) break; } if (first) printf("%lld\n", n); else if (n > 1) printf(" x %lld\n", n); else printf("\n"); } return 0; }
Produce a functionally identical C# code for the snippet given in Delphi.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
using System; using System.Collections.Generic; namespace prog { class MainClass { public static void Main (string[] args) { for( int i=1; i<=22; i++ ) { List<int> f = Factorize(i); Console.Write( i + ": " + f[0] ); for( int j=1; j<f.Count; j++ ) { Console.Write( " * " + f[j] ); } Console.WriteLine(); } } public static List<int> Factorize( int n ) { List<int> l = new List<int>(); if ( n == 1 ) { l.Add(1); } else { int k = 2; while( n > 1 ) { while( n % k == 0 ) { l.Add( k ); n /= k; } k++; } } return l; } } }
Preserve the algorithm and functionality while converting the code from Delphi to C++.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
#include <iostream> #include <iomanip> using namespace std; void getPrimeFactors( int li ) { int f = 2; string res; if ( li == 1 ) res = "1"; else { while ( true ) { if( !( li % f ) ) { res += to_string(f); li /= f; if( li == 1 ) break; res += " x "; } else f++; } } cout << res << "\n"; } int main( int argc, char* argv[] ) { for ( int x = 1; x < 101; x++ ) { cout << right << setw( 4 ) << x << ": "; getPrimeFactors( x ); } cout << 2144 << ": "; getPrimeFactors( 2144 ); cout << "\n\n"; return system( "pause" ); }
Please provide an equivalent version of this Delphi code in Java.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
public class CountingInFactors{ public static void main(String[] args){ for(int i = 1; i<= 10; i++){ System.out.println(i + " = "+ countInFactors(i)); } for(int i = 9991; i <= 10000; i++){ System.out.println(i + " = "+ countInFactors(i)); } } private static String countInFactors(int n){ if(n == 1) return "1"; StringBuilder sb = new StringBuilder(); n = checkFactor(2, n, sb); if(n == 1) return sb.toString(); n = checkFactor(3, n, sb); if(n == 1) return sb.toString(); for(int i = 5; i <= n; i+= 2){ if(i % 3 == 0)continue; n = checkFactor(i, n, sb); if(n == 1)break; } return sb.toString(); } private static int checkFactor(int mult, int n, StringBuilder sb){ while(n % mult == 0 ){ if(sb.length() > 0) sb.append(" x "); sb.append(mult); n /= mult; } return n; } }
Write the same code in Python as shown below in Delphi.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
from functools import lru_cache primes = [2, 3, 5, 7, 11, 13, 17] @lru_cache(maxsize=2000) def pfactor(n): if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n] if __name__ == '__main__': mx = 5000 for n in range(1, mx + 1): factors = pfactor(n) if n <= 10 or n >= mx - 20: print( '%4i %5s %s' % (n, '' if factors != [n] or n == 1 else 'prime', 'x'.join(str(i) for i in factors)) ) if n == 11: print('...') print('\nNumber of primes gathered up to', n, 'is', len(primes)) print(pfactor.cache_info())
Ensure the translated VB code behaves exactly like the original Delphi snippet.
function Factorize(n : Integer) : String; begin if n <= 1 then Exit('1'); var k := 2; while n >= k do begin while (n mod k) = 0 do begin Result += ' * '+IntToStr(k); n := n div k; end; Inc(k); end; Result:=SubStr(Result, 4); end; var i : Integer; for i := 1 to 22 do PrintLn(IntToStr(i) + ': ' + Factorize(i));
Function CountFactors(n) If n = 1 Then CountFactors = 1 Else arrP = Split(ListPrimes(n)," ") Set arrList = CreateObject("System.Collections.ArrayList") divnum = n Do Until divnum = 1 For i = 0 To UBound(arrP)-1 If divnum = 1 Then Exit For ElseIf divnum Mod arrP(i) = 0 Then divnum = divnum/arrP(i) arrList.Add arrP(i) End If Next Loop arrList.Sort For i = 0 To arrList.Count - 1 If i = arrList.Count - 1 Then CountFactors = CountFactors & arrList(i) Else CountFactors = CountFactors & arrList(i) & " * " End If Next End If End Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End If End Function Function ListPrimes(n) ListPrimes = "" For i = 1 To n If IsPrime(i) Then ListPrimes = ListPrimes & i & " " End If Next End Function WScript.StdOut.Write "2 = " & CountFactors(2) WScript.StdOut.WriteLine WScript.StdOut.Write "2144 = " & CountFactors(2144) WScript.StdOut.WriteLine