Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Convert this Common_Lisp block to Go, preserving its control flow and logic. | (defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))
| 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 following code from D to C with equivalent syntax and logic. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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 ;
}
|
Translate this program into C but keep the logic exactly as in D. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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 ;
}
|
Generate a C# translation of this D snippet without changing its computational steps. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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);
}
}
}}
|
Ensure the translated C# code behaves exactly like the original D snippet. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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);
}
}
}}
|
Generate an equivalent C++ version of this D code. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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 ;
}
|
Rewrite the snippet below in C++ so it works the same as the original D code. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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 ;
}
|
Rewrite this program in Java while keeping its functionality equivalent to the D version. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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);
}
}
|
Generate a Java translation of this D snippet without changing its computational steps. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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);
}
}
|
Write the same algorithm in Python as shown in this D implementation. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", k, k, k*k);
}
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a Python translation of this D snippet without changing its computational steps. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", k, k, k*k);
}
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert this D snippet to VB and keep its semantics consistent. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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
|
Convert the following code from D to VB, ensuring the logic remains intact. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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
|
Keep all operations the same but rewrite the snippet in Go. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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
}
}
}
|
Please provide an equivalent version of this D code in Go. |
import std.math;
import std.stdio;
void main( )
{
int k = cast(int)(sqrt(269696.0));
if (k % 2 == 1)
k = k - 1;
while ((k * k) % 1000000 != 269696)
k = k + 2;
writefln("%d * %d = %d", 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
}
}
}
|
Convert this Elixir snippet to C and keep its semantics consistent. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Generate a C translation of this Elixir snippet without changing its computational steps. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Port the following code from Elixir to C# with equivalent syntax and logic. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Maintain the same structure and functionality when rewriting this code in C#. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Can you help me rewrite this code in C++ instead of Elixir, keeping it the same logically? | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Change the programming language of this snippet from Elixir to C++ without modifying what it does. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Convert this Elixir block to Java, preserving its control flow and logic. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Translate this program into Java but keep the logic exactly as in Elixir. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| public class Test {
public static void main(String[] args) {
int n = 0;
do {
n++;
} while (n * n % 1000_000 != 269696);
System.out.println(n);
}
}
|
Maintain the same structure and functionality when rewriting this code in Python. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Port the following code from Elixir to Python with equivalent syntax and logic. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Preserve the algorithm and functionality while converting the code from Elixir to VB. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Write the same code in VB as shown below in Elixir. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Translate this program into Go but keep the logic exactly as in Elixir. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Ensure the translated Go code behaves exactly like the original Elixir snippet. | defmodule Babbage do
def problem(n) when rem(n*n,1000000)==269696, do: n
def problem(n), do: problem(n+2)
end
IO.puts Babbage.problem(0)
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Write the same algorithm in C as shown in this Erlang implementation. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
|
#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 ;
}
|
Preserve the algorithm and functionality while converting the code from Erlang to C. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write the same code in C# as shown below in Erlang. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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);
}
}
}}
|
Rewrite this program in C++ while keeping its functionality equivalent to the Erlang version. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| #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 C++ but keep the logic exactly as in Erlang. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| #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 Erlang to Java, ensuring the logic remains intact. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang snippet. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Maintain the same structure and functionality when rewriting this code in Python. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Rewrite this program in VB while keeping its functionality equivalent to the Erlang version. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang version. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 Erlang code in Go. | -module(solution1).
-export([main/0]).
babbage(N,E) when N*N rem 1000000 == 269696 ->
io:fwrite("~p",[N]);
babbage(N,E) ->
case E of
4 -> babbage(N+2,6);
6 -> babbage(N+8,4)
end.
main()->
babbage(4,4).
| 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 F# to C. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
|
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
int current = 0,
square;
while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) {
current++;
}
if (square>+INT_MAX)
printf("Condition not satisfied before INT_MAX reached.");
else
printf ("The smallest number whose square ends in 269696 is %d\n", current);
return 0 ;
}
|
Write the same algorithm in C as shown in this F# implementation. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
|
#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 ;
}
|
Preserve the algorithm and functionality while converting the code from F# to C#. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| 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 F# to C#, same semantics. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| 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 F# to C++ without modifying what it does. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| #include <iostream>
int main( ) {
int current = 0 ;
while ( ( current * current ) % 1000000 != 269696 )
current++ ;
std::cout << "The square of " << current << " is " << (current * current) << " !\n" ;
return 0 ;
}
|
Please provide an equivalent version of this F# code in C++. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| #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 the snippet below in Java so it works the same as the original F# code. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| 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 F# code snippet into Java without altering its behavior. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| 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 F#. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a Python translation of this F# snippet without changing its computational steps. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Translate this program into VB but keep the logic exactly as in F#. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Change the programming language of this snippet from F# to VB without modifying what it does. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| Sub Baggage_Problem()
Dim i As Long
i = 520
Do While ((i * i) Mod 1000000) <> 269696
i = i + 4
Loop
Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _
"Its square is : " & i * i
End Sub
|
Maintain the same structure and functionality when rewriting this code in Go. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Translate this program into Go but keep the logic exactly as in F#. | Seq.initInfinite id
|> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696)
|> Seq.head |> printfn "%d"
| package main
import "fmt"
func main() {
const (
target = 269696
modulus = 1000000
)
for n := 1; ; n++ {
square := n * n
ending := square % modulus
if ending == target {
fmt.Println("The smallest number whose square ends with",
target, "is", n,
)
return
}
}
}
|
Generate a C translation of this Factor snippet without changing its computational steps. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
|
#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 ;
}
|
Preserve the algorithm and functionality while converting the code from Factor to C. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
|
#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 Factor code in C#. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| namespace Babbage_Problem
{
class iterateNumbers
{
public iterateNumbers()
{
long baseNumberSquared = 0;
long baseNumber = 0;
do
{
baseNumber += 1;
baseNumberSquared = baseNumber * baseNumber;
}
while (Right6Digits(baseNumberSquared) != 269696);
Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber);
Console.WriteLine("The square is " + baseNumberSquared);
}
private long Right6Digits(long baseNumberSquared)
{
string numberAsString = baseNumberSquared.ToString();
if (numberAsString.Length < 6) { return baseNumberSquared; };
numberAsString = numberAsString.Substring(numberAsString.Length - 6);
return long.Parse(numberAsString);
}
}
}}
|
Write a version of this Factor function in C# with identical behavior. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Factor snippet to C++ and keep its semantics consistent. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| #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 Factor code snippet into C++ without altering its behavior. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| #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 Factor function in Java with identical behavior. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Factor to Java, ensuring the logic remains intact. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Factor to Python. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Port the provided Factor code into VB while preserving the original functionality. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Factor snippet to VB and keep its semantics consistent. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 code in Go as shown below in Factor. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Go as shown below in Factor. |
USING: kernel math math.ranges prettyprint sequences ;
518 99,736 2 <range>
[ sq 1,000,000 mod 269,696 = ]
find
.
drop
| 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 Forth implementation. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
|
#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. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
|
#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 Forth to C#, same semantics. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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);
}
}
}}
|
Rewrite this program in C# while keeping its functionality equivalent to the Forth version. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Forth, keeping it the same logically? |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| #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 Forth snippet. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| #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 Java while keeping its functionality equivalent to the Forth version. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Forth code into Java while preserving the original functionality. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Python as shown below in Forth. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Keep all operations the same but rewrite the snippet in Python. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Translate this program into VB but keep the logic exactly as in Forth. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Forth block to VB, preserving its control flow and logic. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Forth to Go with equivalent syntax and logic. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Forth version. |
: BABBAGE
1
BEGIN
1+
DUP DUP
*
1000000 MOD
269696 =
UNTIL
. ;
BABBAGE
| 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 Fortran snippet. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 Fortran implementation. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 Fortran code into C++ without altering its purpose. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| #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 functionally identical C++ code for the snippet given in Fortran. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| #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 C version of this Fortran code. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
|
#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 Fortran code snippet into C without altering its behavior. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
|
#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 Fortran implementation into Go, maintaining the same output and logic. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 Java so it works the same as the original Fortran code. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 Fortran function in Java with identical behavior. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 following code from Fortran to Python with equivalent syntax and logic. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Generate a Python translation of this Fortran snippet without changing its computational steps. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
|
n=0
while n**2 % 1000000 != 269696:
n += 1
print(n)
|
Convert the following code from Fortran to VB, ensuring the logic remains intact. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 Fortran snippet without changing its computational steps. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| 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 PHP but keep the logic exactly as in Fortran. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| <?php
for (
$i = 1 ; // Initial positive integer to check
($i * $i) % 1000000 !== 269696 ; // While i*i does not end with the digits 269,696
$i++ // ... go to next integer
);
echo $i, ' * ', $i, ' = ', ($i * $i), PHP_EOL; // Print the result
|
Convert the following code from Fortran to PHP, ensuring the logic remains intact. | DO 3 N=1,99736
IF(MODF(N*N,1000000)-269696)3,4,3
3 CONTINUE
4 PRINT 5,N
5 FORMAT(I6)
STOP
| <?php
for (
$i = 1 ; // Initial positive integer to check
($i * $i) % 1000000 !== 269696 ; // While i*i does not end with the digits 269,696
$i++ // ... go to next integer
);
echo $i, ' * ', $i, ' = ', ($i * $i), PHP_EOL; // Print the result
|
Write the same algorithm in C 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 ;
|
#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 ;
}
|
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