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Change the programming language of this snippet from Nim to C++ without modifying what it does.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Produce a language-to-language conversion: from Nim to Java, same semantics.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Convert this Nim block to Java, preserving its control flow and logic.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Please provide an equivalent version of this Nim code in Python.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Write the same algorithm in Python as shown in this Nim implementation.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Convert the following code from Nim to VB, ensuring the logic remains intact.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Change the programming language of this snippet from Nim to VB without modifying what it does.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Produce a language-to-language conversion: from Nim to Go, same semantics.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Rewrite this program in Go while keeping its functionality equivalent to the Nim version.
proc multifact(n, deg: int): int = result = (if n <= deg: n else: n * multifact(n - deg, deg)) proc multifactI(n, deg: int): int = result = n var n = n while n >= deg + 1: result *= n - deg n -= deg for i in 1..5: stdout.write "Degree ", i, ": " for j in 1..10: stdout.write multifactI(j, i), " " stdout.write('\n')
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Convert this OCaml snippet to C and keep its semantics consistent.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Rewrite this program in C while keeping its functionality equivalent to the OCaml version.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Port the provided OCaml code into C# while preserving the original functionality.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Please provide an equivalent version of this OCaml code in C#.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Preserve the algorithm and functionality while converting the code from OCaml to C++.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Write the same code in C++ as shown below in OCaml.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Rewrite the snippet below in Java so it works the same as the original OCaml code.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Can you help me rewrite this code in Java instead of OCaml, keeping it the same logically?
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Generate an equivalent Python version of this OCaml code.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Produce a language-to-language conversion: from OCaml to Python, same semantics.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Port the following code from OCaml to VB with equivalent syntax and logic.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Please provide an equivalent version of this OCaml code in VB.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Translate this program into Go but keep the logic exactly as in OCaml.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Transform the following OCaml implementation into Go, maintaining the same output and logic.
let multi_fac d n = let rec loop a x = if x < 2 then a else loop (a * x) (x - d) in loop n (n - d) let () = for i = 1 to 5 do Seq.(ints 1 |> take 10 |> map (multi_fac i) |> map string_of_int) |> List.of_seq |> String.concat " " |> print_endline done
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Write a version of this Perl function in C with identical behavior.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Port the provided Perl code into C while preserving the original functionality.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Keep all operations the same but rewrite the snippet in C#.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Can you help me rewrite this code in C# instead of Perl, keeping it the same logically?
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Write the same algorithm in C++ as shown in this Perl implementation.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Maintain the same structure and functionality when rewriting this code in C++.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert the following code from Perl to Java, ensuring the logic remains intact.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Rewrite this program in Java while keeping its functionality equivalent to the Perl version.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Preserve the algorithm and functionality while converting the code from Perl to Python.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Convert this Perl block to Python, preserving its control flow and logic.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Rewrite this program in VB while keeping its functionality equivalent to the Perl version.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Can you help me rewrite this code in VB instead of Perl, keeping it the same logically?
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Keep all operations the same but rewrite the snippet in Go.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Produce a functionally identical Go code for the snippet given in Perl.
{ my @cache; use bigint; sub mfact { my ($s, $n) = @_; return 1 if $n <= 0; $cache[$s][$n] //= $n * mfact($s, $n - $s); } } for my $s (1 .. 10) { print "step=$s: "; print join(" ", map(mfact($s, $_), 1 .. 10)), "\n"; }
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Convert this Racket snippet to C and keep its semantics consistent.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Port the provided Racket code into C while preserving the original functionality.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Translate the given Racket code snippet into C# without altering its behavior.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Write the same code in C# as shown below in Racket.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Port the following code from Racket to C++ with equivalent syntax and logic.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Change the following Racket code into C++ without altering its purpose.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Change the programming language of this snippet from Racket to Java without modifying what it does.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Ensure the translated Java code behaves exactly like the original Racket snippet.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Translate the given Racket code snippet into Python without altering its behavior.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Transform the following Racket implementation into Python, maintaining the same output and logic.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Generate an equivalent VB version of this Racket code.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Write a version of this Racket function in VB with identical behavior.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Please provide an equivalent version of this Racket code in Go.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Write the same algorithm in Go as shown in this Racket implementation.
#lang racket (define (multi-factorial-fn m) (lambda (n) (let inner ((acc 1) (n n)) (if (<= n m) (* acc n) (inner (* acc n) (- n m)))))) (for*/list ([m (in-range 1 (add1 5))] [mf-m (in-value (multi-factorial-fn m))]) (for/list ([n (in-range 1 (add1 10))]) (mf-m n))) (define (multi-factorial m n) ((multi-factorial-fn m) n)) (for/list ([m (in-range 1 (add1 5))]) (for/list ([n (in-range 1 (add1 10))]) (multi-factorial m n)))
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Rewrite this program in C while keeping its functionality equivalent to the REXX version.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Write a version of this REXX function in C with identical behavior.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Translate this program into C# but keep the logic exactly as in REXX.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Write the same code in C# as shown below in REXX.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Maintain the same structure and functionality when rewriting this code in C++.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert this REXX snippet to C++ and keep its semantics consistent.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Port the provided REXX code into Java while preserving the original functionality.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Translate the given REXX code snippet into Java without altering its behavior.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Write the same code in Python as shown below in REXX.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Translate this program into Python but keep the logic exactly as in REXX.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Produce a language-to-language conversion: from REXX to VB, same semantics.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Translate this program into VB but keep the logic exactly as in REXX.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Write the same algorithm in Go as shown in this REXX implementation.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Rewrite this program in Go while keeping its functionality equivalent to the REXX version.
numeric digits 1000 parse arg num deg . if num=='' | num=="," then num=15 if deg=='' | deg=="," then deg=10 say '═══showing multiple factorials (1 ──►' deg") for numbers 1 ──►" num say do d=1 for deg _= do f=1 for num _=_ Kfact(f, d) end say right('n'copies("!", d), 1+deg) right('['d"]", 2+length(num) )':' _ end exit Kfact: procedure; !=1; do j=arg(1) to 2 by -word(arg(2) 1,1); !=!*j; end; return !
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Please provide an equivalent version of this Ruby code in C.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Ensure the translated C code behaves exactly like the original Ruby snippet.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Generate an equivalent C# version of this Ruby code.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Generate an equivalent C# version of this Ruby code.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Generate a C++ translation of this Ruby snippet without changing its computational steps.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert the following code from Ruby to C++, ensuring the logic remains intact.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Can you help me rewrite this code in Java instead of Ruby, keeping it the same logically?
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Preserve the algorithm and functionality while converting the code from Ruby to Java.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Transform the following Ruby implementation into Python, maintaining the same output and logic.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Keep all operations the same but rewrite the snippet in Python.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Write the same code in VB as shown below in Ruby.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Ensure the translated VB code behaves exactly like the original Ruby snippet.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Maintain the same structure and functionality when rewriting this code in Go.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Maintain the same structure and functionality when rewriting this code in Go.
def multifact(n, d) n.step(to: 1, by: -d).product end (1..5).each {|d| puts "Degree
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Write the same code in C as shown below in Scala.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Maintain the same structure and functionality when rewriting this code in C.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Write the same algorithm in C# as shown in this Scala implementation.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Write the same algorithm in C# as shown in this Scala implementation.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Translate the given Scala code snippet into C++ without altering its behavior.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Keep all operations the same but rewrite the snippet in C++.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Maintain the same structure and functionality when rewriting this code in Java.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Ensure the translated Java code behaves exactly like the original Scala snippet.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Convert the following code from Scala to Python, ensuring the logic remains intact.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Please provide an equivalent version of this Scala code in Python.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Please provide an equivalent version of this Scala code in VB.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Translate the given Scala code snippet into VB without altering its behavior.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Write the same algorithm in Go as shown in this Scala implementation.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Produce a functionally identical Go code for the snippet given in Scala.
fun multifactorial(n: Long, d: Int) : Long { val r = n % d return (1..n).filter { it % d == r } .reduce { i, p -> i * p } } fun main(args: Array<String>) { val m = 5 val r = 1..10L for (d in 1..m) { print("%${m}s:".format( "!".repeat(d))) r.forEach { print(" " + multifactorial(it, d)) } println() } }
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Change the programming language of this snippet from Swift to C without modifying what it does.
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Transform the following Swift implementation into C, maintaining the same output and logic.
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Write the same algorithm in C# as shown in this Swift implementation.
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Change the following Swift code into C# without altering its purpose.
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Can you help me rewrite this code in C++ instead of Swift, keeping it the same logically?
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert this Swift block to C++, preserving its control flow and logic.
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Can you help me rewrite this code in Java instead of Swift, keeping it the same logically?
func multiFactorial(_ n: Int, k: Int) -> Int { return stride(from: n, to: 0, by: -k).reduce(1, *) } let multis = (1...5).map({degree in (1...10).map({member in multiFactorial(member, k: degree) }) }) for (i, degree) in multis.enumerated() { print("Degree \(i + 1): \(degree)") }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }