Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Preserve the algorithm and functionality while converting the code from Nim to PHP. | import strutils
var str = "little"
echo "Mary had a $
echo "Mary had a $
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Generate an equivalent PHP version of this OCaml code. | let extra = "little" in
Printf.printf "Mary had a %s lamb." extra
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Keep all operations the same but rewrite the snippet in PHP. | $extra = "little";
print "Mary had a $extra lamb.\n";
printf "Mary had a %s lamb.\n", $extra;
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Ensure the translated PHP code behaves exactly like the original PowerShell snippet. | $extra = "little"
"Mary had a {0} lamb." -f $extra
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Change the programming language of this snippet from Racket to PHP without modifying what it does. | #lang racket
(format "Mary had a ~a lamb" "little")
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Translate the given COBOL code snippet into PHP without altering its behavior. | IDENTIFICATION DIVISION.
PROGRAM-ID. interpolation-included.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 extra PIC X(6) VALUE "little".
PROCEDURE DIVISION.
DISPLAY FUNCTION SUBSTITUTE("Mary had a X lamb.", "X", extra)
GOBACK
.
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Transform the following REXX implementation into PHP, maintaining the same output and logic. |
options replace format comments java crossref savelog symbols
import java.text.MessageFormat
import java.text.FieldPosition
useBif()
useMessageFormat()
return
method useBif public static
st = "Mary had a %1$ lamb."
si = 'little'
say st.changestr('%1$', si)
return
method useMessageFormat public static
result = StringBuffer('')
args = Object [ -
Object Integer(7), -
Object Date(), -
Object 'a disturbance in the Force' -
]
msgfmt = MessageFormat('At {1, time} on {1, date}, there was {2} on planet {0, number, integer}.')
result = msgfmt.format(args, result, FieldPosition(0))
say result
return
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Convert the following code from Ruby to PHP, ensuring the logic remains intact. | irb(main):001:0> extra = 'little'
=> "little"
irb(main):002:0> "Mary had a
=> "Mary had a little lamb."
irb(main):003:0> "Mary had a %s lamb." % extra
=> "Mary had a little lamb."
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Please provide an equivalent version of this Scala code in PHP. |
fun main(args: Array<String>) {
val s = "little"
println("Mary had a $s lamb")
println("Mary had a ${s.toUpperCase()} lamb")
println("Mary had a ${s}r lamb")
}
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Produce a functionally identical PHP code for the snippet given in Swift. | let extra = "little"
println("Mary had a \(extra) lamb.")
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Port the following code from Tcl to PHP with equivalent syntax and logic. | set size "little"
puts "Mary had a $size lamb."
proc RandomWord {args} {
lindex $args [expr {int(rand()*[llength $args])}]
}
puts "Mary had a [RandomWord little big] lamb."
| <?php
$extra = 'little';
echo "Mary had a $extra lamb.\n";
printf("Mary had a %s lamb.\n", $extra);
?>
|
Rewrite the snippet below in Rust so it works the same as the original C code. | #include <stdio.h>
int main() {
const char *extra = "little";
printf("Mary had a %s lamb.\n", extra);
return 0;
}
| fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
|
Rewrite the snippet below in Rust so it works the same as the original C++ code. | #include <string>
#include <iostream>
int main( ) {
std::string original( "Mary had a X lamb." ) , toBeReplaced( "X" ) ,
replacement ( "little" ) ;
std::string newString = original.replace( original.find( "X" ) ,
toBeReplaced.length( ) , replacement ) ;
std::cout << "String after replacement: " << newString << " \n" ;
return 0 ;
}
| fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
|
Transform the following Go implementation into Rust, maintaining the same output and logic. | package main
import (
"fmt"
)
func main() {
str := "Mary had a %s lamb"
txt := "little"
out := fmt.Sprintf(str, txt)
fmt.Println(out)
}
| fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
|
Change the programming language of this snippet from Rust to Python without modifying what it does. | fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
| >>> original = 'Mary had a %s lamb.'
>>> extra = 'little'
>>> original % extra
'Mary had a little lamb.'
|
Translate the given Rust code snippet into VB without altering its behavior. | fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
| Dim name as String = "J. Doe"
Dim balance as Double = 123.45
Dim prompt as String = String.Format("Hello {0}, your balance is {1}.", name, balance)
Console.WriteLine(prompt)
|
Generate a Rust translation of this Java snippet without changing its computational steps. | String original = "Mary had a X lamb";
String little = "little";
String replaced = original.replace("X", little);
System.out.println(replaced);
System.out.printf("Mary had a %s lamb.", little);
String formatted = String.format("Mary had a %s lamb.", little);
System.out.println(formatted);
| fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
|
Transform the following C# implementation into Rust, maintaining the same output and logic. | class Program
{
static void Main()
{
string extra = "little";
string formatted = $"Mary had a {extra} lamb.";
System.Console.WriteLine(formatted);
}
}
| fn main() {
println!("Mary had a {} lamb", "little");
println!("{1} had a {0} lamb", "little", "Mary");
println!("{name} had a {adj} lamb", adj="little", name="Mary");
}
|
Generate an equivalent C# version of this Ada code. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Convert this Ada snippet to C# and keep its semantics consistent. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Can you help me rewrite this code in C instead of Ada, keeping it the same logically? |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Ensure the translated C code behaves exactly like the original Ada snippet. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Change the programming language of this snippet from Ada to C++ without modifying what it does. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this Ada block to C++, preserving its control flow and logic. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Can you help me rewrite this code in Go instead of Ada, keeping it the same logically? |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Translate the given Ada code snippet into Go without altering its behavior. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Maintain the same structure and functionality when rewriting this code in Java. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Please provide an equivalent version of this Ada code in Java. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Convert this Ada snippet to Python and keep its semantics consistent. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert the following code from Ada to Python, ensuring the logic remains intact. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Transform the following Ada implementation into VB, maintaining the same output and logic. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Maintain the same structure and functionality when rewriting this code in VB. |
with Ada.Text_IO;
procedure Babbage_Problem is
type Number is range 1 .. 99_736*99_736;
X: Number := 1;
begin
while not (((X*X) mod 1_000_000) = 269_696) loop
X := X+1;
end loop;
Ada.Text_IO.Put_Line(Number'Image(X));
end Babbage_Problem;
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Generate an equivalent C version of this Arturo code. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write the same algorithm in C as shown in this Arturo implementation. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Change the programming language of this snippet from Arturo to C# without modifying what it does. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write a version of this Arturo function in C# with identical behavior. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write a version of this Arturo function in C++ with identical behavior. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Port the following code from Arturo to C++ with equivalent syntax and logic. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this Arturo block to Java, preserving its control flow and logic. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Convert the following code from Arturo to Java, ensuring the logic remains intact. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Write a version of this Arturo function in Python with identical behavior. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Please provide an equivalent version of this Arturo code in Python. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Rewrite the snippet below in VB so it works the same as the original Arturo code. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Translate this program into VB but keep the logic exactly as in Arturo. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Generate an equivalent Go version of this Arturo code. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Rewrite this program in Go while keeping its functionality equivalent to the Arturo version. | n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n
print n
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Rewrite the snippet below in C so it works the same as the original AWK code. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Translate the given AWK code snippet into C without altering its behavior. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write the same algorithm in C# as shown in this AWK implementation. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Maintain the same structure and functionality when rewriting this code in C#. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write the same code in C++ as shown below in AWK. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert the following code from AWK to C++, ensuring the logic remains intact. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Change the programming language of this snippet from AWK to Java without modifying what it does. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Port the provided AWK code into Java while preserving the original functionality. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Maintain the same structure and functionality when rewriting this code in Python. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Can you help me rewrite this code in Python instead of AWK, keeping it the same logically? |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Translate this program into VB but keep the logic exactly as in AWK. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Preserve the algorithm and functionality while converting the code from AWK to VB. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Write a version of this AWK function in Go with identical behavior. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Please provide an equivalent version of this AWK code in Go. |
BEGIN {
n = 0
do {
n = n + 1
} while (n*n !~ /269696$/)
print("The smallest number whose square ends in 269696 is " n)
print("Its square is " n*n)
exit(0)
}
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Ensure the translated C code behaves exactly like the original BBC_Basic snippet. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write a version of this BBC_Basic function in C with identical behavior. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write a version of this BBC_Basic function in C# with identical behavior. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Port the following code from BBC_Basic to C# with equivalent syntax and logic. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write the same algorithm in C++ as shown in this BBC_Basic implementation. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this BBC_Basic snippet to C++ and keep its semantics consistent. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Keep all operations the same but rewrite the snippet in Java. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Write the same algorithm in Java as shown in this BBC_Basic implementation. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Convert this BBC_Basic block to Python, preserving its control flow and logic. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Keep all operations the same but rewrite the snippet in Python. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Transform the following BBC_Basic implementation into VB, maintaining the same output and logic. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Translate the given BBC_Basic code snippet into VB without altering its behavior. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Change the programming language of this snippet from BBC_Basic to Go without modifying what it does. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Translate this program into Go but keep the logic exactly as in BBC_Basic. |
LET n = 0
REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696
PRINT "The smallest number whose square ends in 269696 is" n
PRINT "Its square is" n^2
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Generate a C translation of this Clojure snippet without changing its computational steps. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Generate an equivalent C# version of this Clojure code. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Translate the given Clojure code snippet into C# without altering its behavior. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write the same algorithm in C++ as shown in this Clojure implementation. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Please provide an equivalent version of this Clojure code in C++. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Write the same algorithm in Java as shown in this Clojure implementation. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Please provide an equivalent version of this Clojure code in Java. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Convert this Clojure block to Python, preserving its control flow and logic. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert this Clojure block to Python, preserving its control flow and logic. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Please provide an equivalent version of this Clojure code in VB. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Generate an equivalent VB version of this Clojure code. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Generate a Go translation of this Clojure snippet without changing its computational steps. |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Can you help me rewrite this code in Go instead of Clojure, keeping it the same logically? |
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))
(first (filter babbage? (range)))
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Change the programming language of this snippet from Common_Lisp to C without modifying what it does. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write a version of this Common_Lisp function in C with identical behavior. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write a version of this Common_Lisp function in C# with identical behavior. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Port the following code from Common_Lisp to C# with equivalent syntax and logic. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Convert this Common_Lisp block to C++, preserving its control flow and logic. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this Common_Lisp snippet to C++ and keep its semantics consistent. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Generate a Java translation of this Common_Lisp snippet without changing its computational steps. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Change the programming language of this snippet from Common_Lisp to Java without modifying what it does. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Produce a language-to-language conversion: from Common_Lisp to Python, same semantics. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Maintain the same structure and functionality when rewriting this code in Python. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Rewrite the snippet below in VB so it works the same as the original Common_Lisp code. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Please provide an equivalent version of this Common_Lisp code in VB. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Produce a functionally identical Go code for the snippet given in Common_Lisp. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
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