Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Please provide an equivalent version of this OCaml code 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);
}
}
}}
|
Can you help me rewrite this code in C++ instead of OCaml, keeping it the same logically? | 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 <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 OCaml code. | 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
| 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 OCaml snippet to Java and keep its semantics consistent. | 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
| 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 the given OCaml code snippet into Python without altering its behavior. | 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
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Produce a language-to-language conversion: from OCaml to Python, same semantics. | 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
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a VB translation of this OCaml snippet without changing its computational steps. | 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
| 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 VB translation of this OCaml snippet without changing its computational steps. | 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
| 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 OCaml. | 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
| 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 OCaml code into Go while preserving the original functionality. | 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
| 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 algorithm in C as shown in this Pascal implementation. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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 ;
}
|
Convert this Pascal snippet to C and keep its semantics consistent. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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 Pascal implementation into C#, maintaining the same output and logic. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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);
}
}
}}
|
Preserve the algorithm and functionality while converting the code from Pascal to C#. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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);
}
}
}}
|
Ensure the translated C++ code behaves exactly like the original Pascal snippet. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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 ;
}
|
Translate the given Pascal code snippet into C++ without altering its behavior. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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 ;
}
|
Change the following Pascal code into Java without altering its purpose. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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);
}
}
|
Convert this Pascal snippet to Java and keep its semantics consistent. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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);
}
}
|
Port the provided Pascal code into Python while preserving the original functionality. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
end.
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Keep all operations the same but rewrite the snippet in Python. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
end.
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Port the following code from Pascal to VB with equivalent syntax and logic. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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
|
Produce a functionally identical VB code for the snippet given in Pascal. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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
|
Can you help me rewrite this code in Go instead of Pascal, keeping it the same logically? | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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
}
}
}
|
Preserve the algorithm and functionality while converting the code from Pascal to Go. | program BabbageProblem;
var n : longint;
begin
n := 2;
repeat
n := n + 2
until (n * n) mod 1000000 = 269696;
write(n)
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
}
}
}
|
Translate the given Perl code snippet into C without altering its behavior. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 Perl block to C, preserving its control flow and logic. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 the snippet below in C# so it works the same as the original Perl code. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 the same code in C# as shown below in Perl. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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);
}
}
}}
|
Transform the following Perl implementation into C++, maintaining the same output and logic. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 ;
}
|
Ensure the translated C++ code behaves exactly like the original Perl snippet. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 Perl to Java with equivalent syntax and logic. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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);
}
}
|
Preserve the algorithm and functionality while converting the code from Perl to Java. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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 Perl to Python, ensuring the logic remains intact. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\n" ;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert the following code from Perl to Python, ensuring the logic remains intact. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\n" ;
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Can you help me rewrite this code in VB instead of Perl, keeping it the same logically? |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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
|
Port the following code from Perl to VB with equivalent syntax and logic. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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
|
Write the same algorithm in Go as shown in this Perl implementation. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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
}
}
}
|
Preserve the algorithm and functionality while converting the code from Perl to Go. |
use strict ;
use warnings ;
my $current = 0 ;
while ( ($current ** 2 ) % 1000000 != 269696 ) {
$current++ ;
}
print "The square of $current is " . ($current * $current) . " !\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
}
}
}
|
Write the same code in C as shown below in PowerShell. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
|
#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 PowerShell code snippet into C without altering its behavior. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
|
#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 PowerShell version. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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#. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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++. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| #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 the given PowerShell code snippet into C++ without altering its behavior. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| #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 PowerShell to Java, ensuring the logic remains intact. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 PowerShell to Java, ensuring the logic remains intact. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 PowerShell function in Python with identical behavior. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Translate this program into Python but keep the logic exactly as in PowerShell. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Transform the following PowerShell implementation into VB, maintaining the same output and logic. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 PowerShell. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 language-to-language conversion: from PowerShell to Go, same semantics. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 PowerShell code. |
$integer = 0
while (($integer * $integer) % 1000000 -ne 269696)
{
$integer++
}
$integer
| 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 C. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
|
#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 R code. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
|
#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 R block to C#, preserving its control flow and logic. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 R, keeping it the same logically? | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 R code. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| #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 C++ instead of R, keeping it the same logically? | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| #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 ;
}
|
Maintain the same structure and functionality when rewriting this code in Java. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 R to Java, ensuring the logic remains intact. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 Python. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Please provide an equivalent version of this R code in Python. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Please provide an equivalent version of this R code in VB. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 R. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 Go but keep the logic exactly as in R. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 R code. | babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]
| 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 Racket snippet. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
|
#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 Racket to C with equivalent syntax and logic. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
|
#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 a C# translation of this Racket snippet without changing its computational steps. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket to C# with equivalent syntax and logic. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket code into C++ without altering its purpose. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| #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 the given Racket code snippet into C++ without altering its behavior. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| #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 ;
}
|
Maintain the same structure and functionality when rewriting this code in Java. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket snippet to Java and keep its semantics consistent. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket code. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert this Racket block to Python, preserving its control flow and logic. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Preserve the algorithm and functionality while converting the code from Racket to VB. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket to VB with equivalent syntax and logic. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 language-to-language conversion: from Racket to Go, same semantics. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 Racket code. |
#lang racket
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))
(define square-ends-in-269696? (compose ends-in-269696? sqr))
(define first-number-that-when-squared-ends-in-269696
(for/first ((i
(in-range 1 (add1 99736)))
#:when (square-ends-in-269696? i))
i))
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
| 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 C. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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 the snippet below in C so it works the same as the original COBOL code. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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 ;
}
|
Generate an equivalent C# version of this COBOL code. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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);
}
}
}}
|
Convert this COBOL snippet to C# and keep its semantics consistent. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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 the same algorithm in C++ as shown in this COBOL implementation. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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 ;
}
|
Keep all operations the same but rewrite the snippet in C++. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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 ;
}
|
Transform the following COBOL implementation into Java, maintaining the same output and logic. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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);
}
}
|
Translate the given COBOL code snippet into Java without altering its behavior. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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);
}
}
|
Produce a functionally identical Python code for the snippet given in COBOL. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY N.
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Produce a functionally identical Python code for the snippet given in COBOL. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY N.
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert this COBOL snippet to VB and keep its semantics consistent. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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
|
Write the same algorithm in VB as shown in this COBOL implementation. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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
|
Rewrite the snippet below in Go so it works the same as the original COBOL code. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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
}
}
}
|
Produce a language-to-language conversion: from COBOL to Go, same semantics. | IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77 N PICTURE 99999.
* We know that 99,736 is a valid answer.
77 N-SQUARED PICTURE 9999999999.
77 LAST-SIX PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY 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
}
}
}
|
Please provide an equivalent version of this REXX code in C. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
|
#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 ;
}
|
Produce a language-to-language conversion: from REXX to C, same semantics. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
|
#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 provided REXX code into C# while preserving the original functionality. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
| 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 REXX snippet to C# and keep its semantics consistent. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
| 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 programming language of this snippet from REXX to C++ without modifying what it does. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
| #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 REXX snippet to C++ and keep its semantics consistent. |
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision
babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2 -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
-- "What is the smallest positive integer whose square ends in the digits 269,696?"
-- — Babbage, letter to Lord Bowden, 1837;
-- see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
-- (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)
method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end
return n -- end the function and return the result
| #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 ;
}
|
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