Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Rewrite this program in C while keeping its functionality equivalent to the Groovy version. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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 ;
}
|
Change the programming language of this snippet from Groovy to C# without modifying what it does. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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 code in C# as shown below in Groovy. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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);
}
}
}}
|
Produce a functionally identical C++ code for the snippet given in Groovy. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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 ;
}
|
Write the same code in C++ as shown below in Groovy. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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);
}
}
|
Change the following Groovy code into Java without altering its purpose. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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 the following code from Groovy to Python, ensuring the logic remains intact. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+n**2 ;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Write the same code in Python as shown below in Groovy. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+n**2 ;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a VB translation of this Groovy snippet without changing its computational steps. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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
|
Rewrite the snippet below in VB so it works the same as the original Groovy code. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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
|
Write the same algorithm in Go as shown in this Groovy implementation. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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
}
}
}
|
Change the following Groovy code into Go without altering its purpose. | int n=104;
while( (n**2)%1000000 != 269696 )
{ if (n%10==4) n=n+2;
if (n%10==6) n=n+8;
}
println n+"^2== "+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
}
}
}
|
Port the provided Haskell code into C while preserving the original functionality. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Rewrite this program in C while keeping its functionality equivalent to the Haskell version. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
|
#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 ;
}
|
Rewrite this program in C# while keeping its functionality equivalent to the Haskell version. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Transform the following Haskell implementation into C#, maintaining the same output and logic. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| 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);
}
}
}}
|
Generate an equivalent C++ version of this Haskell code. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this Haskell snippet to C++ and keep its semantics consistent. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| #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 ;
}
|
Translate this program into Java but keep the logic exactly as in Haskell. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Convert the following code from Haskell to Java, ensuring the logic remains intact. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Ensure the translated Python code behaves exactly like the original Haskell snippet. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Port the following code from Haskell to Python with equivalent syntax and logic. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Keep all operations the same but rewrite the snippet in VB. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Change the following Haskell code into VB without altering its purpose. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| 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 Haskell code in Go. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Rewrite the snippet below in Go so it works the same as the original Haskell code. |
findBabbageNumber :: Integer
findBabbageNumber =
head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..])
main :: IO ()
main =
(putStrLn . unwords)
(zipWith
(++)
(show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])
| 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 C while keeping its functionality equivalent to the J version. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
|
#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 ;
}
|
Port the following code from J to C with equivalent syntax and logic. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
|
#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 ;
}
|
Convert this J block to C#, preserving its control flow and logic. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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);
}
}
}}
|
Produce a functionally identical C# code for the snippet given in J. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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 J to C++ with equivalent syntax and logic. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| #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 C++ as shown in this J implementation. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| #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 ;
}
|
Produce a language-to-language conversion: from J to Java, same semantics. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Generate a Java translation of this J snippet without changing its computational steps. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Translate this program into Python but keep the logic exactly as in J. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert this J snippet to Python and keep its semantics consistent. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Translate this program into VB but keep the logic exactly as in J. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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
|
Convert this J block to VB, preserving its control flow and logic. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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
|
Can you help me rewrite this code in Go instead of J, keeping it the same logically? | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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 J version. | square=: ^&2
modulo1e6=: 1000000&|
trythese=: i. 1000000
which=: I.
which 269696=modulo1e6 square trythese
25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
| 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
}
}
}
|
Write the same code in C as shown below in Julia. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
|
#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 ;
}
|
Transform the following Julia implementation into C, maintaining the same output and logic. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
|
#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 ;
}
|
Port the following code from Julia to C# with equivalent syntax and logic. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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 Julia snippet to C# and keep its semantics consistent. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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);
}
}
}}
|
Keep all operations the same but rewrite the snippet in C++. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| #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 ;
}
|
Rewrite this program in C++ while keeping its functionality equivalent to the Julia version. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| #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 an equivalent Java version of this Julia code. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Preserve the algorithm and functionality while converting the code from Julia to Java. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Translate this program into Python but keep the logic exactly as in Julia. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Write the same algorithm in Python as shown in this Julia implementation. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a VB translation of this Julia snippet without changing its computational steps. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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
|
Rewrite this program in VB while keeping its functionality equivalent to the Julia version. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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
|
Port the following code from Julia to Go with equivalent syntax and logic. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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 an equivalent Go version of this Julia code. | function babbage(x::Integer)
i = big(0)
d = floor(log10(x)) + 1
while i ^ 2 % 10 ^ d != x
i += 1
end
return i
end
| 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 C while keeping its functionality equivalent to the Lua version. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
|
#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 ;
}
|
Please provide an equivalent version of this Lua code in C. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
|
#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 ;
}
|
Rewrite the snippet below in C# so it works the same as the original Lua code. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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 Lua snippet to C# and keep its semantics consistent. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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);
}
}
}}
|
Transform the following Lua implementation into C++, maintaining the same output and logic. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| #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 Lua snippet to C++ and keep its semantics consistent. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| #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 Lua snippet to Java and keep its semantics consistent. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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 functionally identical Java code for the snippet given in Lua. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Can you help me rewrite this code in Python instead of Lua, keeping it the same logically? |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Ensure the translated Python code behaves exactly like the original Lua snippet. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Preserve the algorithm and functionality while converting the code from Lua to VB. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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
|
Rewrite the snippet below in VB so it works the same as the original Lua code. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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 Lua code in Go. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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
}
}
}
|
Produce a language-to-language conversion: from Lua to Go, same semantics. |
k = math.floor(math.sqrt(269696))
if k % 2 == 1 then
k = k - 1
end
while not ((k * k) % 1000000 == 269696) do
k = k + 2
end
io.write(string.format("%d * %d = %d\n", k, k, k * k))
| 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 C but keep the logic exactly as in Mathematica. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
|
#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 Mathematica snippet. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
|
#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 ;
}
|
Convert this Mathematica snippet to C# and keep its semantics consistent. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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);
}
}
}}
|
Ensure the translated C# code behaves exactly like the original Mathematica snippet. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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 provided Mathematica code into C++ while preserving the original functionality. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| #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 a version of this Mathematica function in C++ with identical behavior. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| #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 ;
}
|
Preserve the algorithm and functionality while converting the code from Mathematica to Java. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Keep all operations the same but rewrite the snippet in Java. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Generate an equivalent Python version of this Mathematica code. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Change the following Mathematica code into Python without altering its purpose. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Write the same code in VB as shown below in Mathematica. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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
|
Convert this Mathematica block to VB, preserving its control flow and logic. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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 Mathematica to Go. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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 Mathematica to Go without modifying what it does. | Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]
| 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
}
}
}
|
Convert this Nim block to C, preserving its control flow and logic. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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 ;
}
|
Convert this Nim block to C, preserving its control flow and logic. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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 ;
}
|
Rewrite this program in C# while keeping its functionality equivalent to the Nim version. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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);
}
}
}}
|
Produce a language-to-language conversion: from Nim to C#, same semantics. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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);
}
}
}}
|
Change the following Nim code into C++ without altering its purpose. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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 provided Nim code into C++ while preserving the original functionality. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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 the following code from Nim to Java, ensuring the logic remains intact. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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 the same code in Java as shown below in Nim. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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);
}
}
|
Please provide an equivalent version of this Nim code in Python. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo n
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Write the same algorithm in Python as shown in this Nim implementation. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo n
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Write the same code in VB as shown below in Nim. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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
|
Ensure the translated VB code behaves exactly like the original Nim snippet. | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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
|
Can you help me rewrite this code in Go instead of Nim, keeping it the same logically? | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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
}
}
}
|
Can you help me rewrite this code in Go instead of Nim, keeping it the same logically? | var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo 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
}
}
}
|
Ensure the translated C code behaves exactly like the original OCaml snippet. | let rec f a=
if (a*a) mod 1000000 != 269696
then f(a+1)
else a
in
let a= f 1 in
Printf.printf "smallest positive integer whose square ends in the digits 269696 is %d\n" a
|
#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 OCaml snippet. | let rec f a=
if (a*a) mod 1000000 != 269696
then f(a+1)
else a
in
let a= f 1 in
Printf.printf "smallest positive integer whose square ends in the digits 269696 is %d\n" a
|
#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 ;
}
|
Keep all operations the same but rewrite the snippet in C#. | let rec f a=
if (a*a) mod 1000000 != 269696
then f(a+1)
else a
in
let a= f 1 in
Printf.printf "smallest positive integer whose square ends in the digits 269696 is %d\n" a
| 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);
}
}
}}
|
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