Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Generate a Go translation of this Erlang snippet without changing its computational steps. | is_perfect(X) ->
X == lists:sum([N || N <- lists:seq(1,X-1), X rem N == 0]).
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Produce a language-to-language conversion: from F# to C, same semantics. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Port the provided F# code into C# while preserving the original functionality. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Produce a functionally identical C++ code for the snippet given in F#. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Translate this program into Java but keep the logic exactly as in F#. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Translate this program into Python but keep the logic exactly as in F#. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Port the following code from F# to VB with equivalent syntax and logic. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Convert the following code from F# to Go, ensuring the logic remains intact. | let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Port the following code from Factor to C with equivalent syntax and logic. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Transform the following Factor implementation into C#, maintaining the same output and logic. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Rewrite this program in C++ while keeping its functionality equivalent to the Factor version. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Write the same code in Java as shown below in Factor. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Convert this Factor block to Python, preserving its control flow and logic. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Write the same algorithm in VB as shown in this Factor implementation. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Write the same algorithm in Go as shown in this Factor implementation. | USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Port the following code from Forth to C with equivalent syntax and logic. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Change the programming language of this snippet from Forth to C# without modifying what it does. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Change the following Forth code into C++ without altering its purpose. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Transform the following Forth implementation into Java, maintaining the same output and logic. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Maintain the same structure and functionality when rewriting this code in Python. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Produce a functionally identical VB code for the snippet given in Forth. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Maintain the same structure and functionality when rewriting this code in Go. | : perfect?
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Write the same code in C# as shown below in Fortran. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Write the same algorithm in C++ as shown in this Fortran implementation. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Change the programming language of this snippet from Fortran to C without modifying what it does. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Change the following Fortran code into Java without altering its purpose. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Write a version of this Fortran function in Python with identical behavior. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Keep all operations the same but rewrite the snippet in VB. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Convert this Fortran snippet to PHP and keep its semantics consistent. | FUNCTION isPerfect(n)
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect
| function is_perfect($number)
{
$sum = 0;
for($i = 1; $i < $number; $i++)
{
if($number % $i == 0)
$sum += $i;
}
return $sum == $number;
}
echo "Perfect numbers from 1 to 33550337:" . PHP_EOL;
for($num = 1; $num < 33550337; $num++)
{
if(is_perfect($num))
echo $num . PHP_EOL;
}
|
Generate a C translation of this Groovy snippet without changing its computational steps. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Convert the following code from Groovy to C#, ensuring the logic remains intact. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Preserve the algorithm and functionality while converting the code from Groovy to C++. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Transform the following Groovy implementation into Java, maintaining the same output and logic. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Rewrite the snippet below in Python so it works the same as the original Groovy code. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Preserve the algorithm and functionality while converting the code from Groovy to VB. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Produce a language-to-language conversion: from Groovy to Go, same semantics. | def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Change the programming language of this snippet from Haskell to C without modifying what it does. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Change the following Haskell code into C# without altering its purpose. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Produce a language-to-language conversion: from Haskell to C++, same semantics. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Convert this Haskell snippet to Java and keep its semantics consistent. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Translate the given Haskell code snippet into Python without altering its behavior. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Rewrite the snippet below in VB so it works the same as the original Haskell code. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Generate a Go translation of this Haskell snippet without changing its computational steps. | perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Port the following code from Icon to C with equivalent syntax and logic. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Transform the following Icon implementation into C#, maintaining the same output and logic. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Change the programming language of this snippet from Icon to C++ without modifying what it does. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Keep all operations the same but rewrite the snippet in Java. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Convert this Icon block to Python, preserving its control flow and logic. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Change the programming language of this snippet from Icon to VB without modifying what it does. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Write the same algorithm in Go as shown in this Icon implementation. | procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n)
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Write the same code in C as shown below in J. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Generate an equivalent C# version of this J code. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Convert the following code from J to C++, ensuring the logic remains intact. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Generate an equivalent Java version of this J code. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Produce a language-to-language conversion: from J to Python, same semantics. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Maintain the same structure and functionality when rewriting this code in VB. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Translate the given J code snippet into Go without altering its behavior. | is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Change the programming language of this snippet from Julia to C without modifying what it does. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Maintain the same structure and functionality when rewriting this code in C#. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Convert the following code from Julia to C++, ensuring the logic remains intact. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Rewrite the snippet below in Java so it works the same as the original Julia code. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Write the same code in Python as shown below in Julia. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Port the following code from Julia to VB with equivalent syntax and logic. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Transform the following Julia implementation into Go, maintaining the same output and logic. | isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Produce a language-to-language conversion: from Lua to C, same semantics. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Write the same code in C# as shown below in Lua. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Produce a language-to-language conversion: from Lua to C++, same semantics. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Write the same algorithm in Java as shown in this Lua implementation. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Rewrite this program in Python while keeping its functionality equivalent to the Lua version. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Translate the given Lua code snippet into VB without altering its behavior. | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Can you help me rewrite this code in Go instead of Lua, keeping it the same logically? | function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Write the same algorithm in C as shown in this Mathematica implementation. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Write the same code in C# as shown below in Mathematica. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Generate a C++ translation of this Mathematica snippet without changing its computational steps. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Translate the given Mathematica code snippet into Java without altering its behavior. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Convert the following code from Mathematica to Python, ensuring the logic remains intact. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Write the same algorithm in VB as shown in this Mathematica implementation. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Transform the following Mathematica implementation into Go, maintaining the same output and logic. | PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Convert this MATLAB snippet to C and keep its semantics consistent. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Change the following MATLAB code into C# without altering its purpose. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Ensure the translated C++ code behaves exactly like the original MATLAB snippet. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Keep all operations the same but rewrite the snippet in Java. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Preserve the algorithm and functionality while converting the code from MATLAB to Python. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Rewrite the snippet below in VB so it works the same as the original MATLAB code. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Convert this MATLAB snippet to Go and keep its semantics consistent. | function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Write a version of this Nim function in C with identical behavior. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Port the following code from Nim to C# with equivalent syntax and logic. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Change the programming language of this snippet from Nim to C++ without modifying what it does. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| #include <iostream>
using namespace std ;
int divisor_sum( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
}
|
Keep all operations the same but rewrite the snippet in Java. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Change the following Nim code into Python without altering its purpose. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Transform the following Nim implementation into VB, maintaining the same output and logic. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Port the provided Nim code into Go while preserving the original functionality. | import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Translate this program into C but keep the logic exactly as in OCaml. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Generate an equivalent C# version of this OCaml code. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
Write the same code in Java as shown below in OCaml. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| public static boolean perf(int n){
int sum= 0;
for(int i= 1;i < n;i++){
if(n % i == 0){
sum+= i;
}
}
return sum == n;
}
|
Rewrite this program in Python while keeping its functionality equivalent to the OCaml version. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| def perf1(n):
sum = 0
for i in range(1, n):
if n % i == 0:
sum += i
return sum == n
|
Change the programming language of this snippet from OCaml to VB without modifying what it does. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub
|
Change the following OCaml code into Go without altering its purpose. | let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n
| package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
|
Maintain the same structure and functionality when rewriting this code in C. | program PerfectNumbers;
function isPerfect(number: longint): boolean;
var
i, sum: longint;
begin
sum := 1;
for i := 2 to round(sqrt(real(number))) do
if (number mod i = 0) then
sum := sum + i + (number div i);
isPerfect := (sum = number);
end;
var
candidate: longint;
begin
writeln('Perfect numbers from 1 to 33550337:');
for candidate := 2 to 33550337 do
if isPerfect(candidate) then
writeln (candidate, ' is a perfect number.');
end.
| #include "stdio.h"
#include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1;
int tot = 1;
int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}
|
Port the following code from Pascal to C# with equivalent syntax and logic. | program PerfectNumbers;
function isPerfect(number: longint): boolean;
var
i, sum: longint;
begin
sum := 1;
for i := 2 to round(sqrt(real(number))) do
if (number mod i = 0) then
sum := sum + i + (number div i);
isPerfect := (sum = number);
end;
var
candidate: longint;
begin
writeln('Perfect numbers from 1 to 33550337:');
for candidate := 2 to 33550337 do
if isPerfect(candidate) then
writeln (candidate, ' is a perfect number.');
end.
| static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}
|
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