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Rewrite the snippet below in Java so it works the same as the original REXX code.
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
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 REXX.
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
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 REXX to Python.
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
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Generate a Python translation of this REXX snippet without changing its computational steps.
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
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Produce a language-to-language conversion: from REXX to VB, 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
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 REXX snippet to VB 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
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Change the following REXX code into Go without altering its purpose.
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
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 REXX to Go with equivalent syntax and logic.
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
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 Ruby to C, same semantics.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
#include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, square; while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); return 0 ; }
Convert this Ruby snippet to C and keep its semantics consistent.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
#include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, square; while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); return 0 ; }
Rewrite this program in C# while keeping its functionality equivalent to the Ruby version.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
namespace Babbage_Problem { class iterateNumbers { public iterateNumbers() { long baseNumberSquared = 0; long baseNumber = 0; do { baseNumber += 1; baseNumberSquared = baseNumber * baseNumber; } while (Right6Digits(baseNumberSquared) != 269696); Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber); Console.WriteLine("The square is " + baseNumberSquared); } private long Right6Digits(long baseNumberSquared) { string numberAsString = baseNumberSquared.ToString(); if (numberAsString.Length < 6) { return baseNumberSquared; }; numberAsString = numberAsString.Substring(numberAsString.Length - 6); return long.Parse(numberAsString); } } }}
Rewrite this program in C# while keeping its functionality equivalent to the Ruby version.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
namespace Babbage_Problem { class iterateNumbers { public iterateNumbers() { long baseNumberSquared = 0; long baseNumber = 0; do { baseNumber += 1; baseNumberSquared = baseNumber * baseNumber; } while (Right6Digits(baseNumberSquared) != 269696); Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber); Console.WriteLine("The square is " + baseNumberSquared); } private long Right6Digits(long baseNumberSquared) { string numberAsString = baseNumberSquared.ToString(); if (numberAsString.Length < 6) { return baseNumberSquared; }; numberAsString = numberAsString.Substring(numberAsString.Length - 6); return long.Parse(numberAsString); } } }}
Rewrite the snippet below in C++ so it works the same as the original Ruby code.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
#include <iostream> int main( ) { int current = 0 ; while ( ( current * current ) % 1000000 != 269696 ) current++ ; std::cout << "The square of " << current << " is " << (current * current) << " !\n" ; return 0 ; }
Write the same algorithm in C++ as shown in this Ruby implementation.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
#include <iostream> int main( ) { int current = 0 ; while ( ( current * current ) % 1000000 != 269696 ) current++ ; std::cout << "The square of " << current << " is " << (current * current) << " !\n" ; return 0 ; }
Change the programming language of this snippet from Ruby to Java without modifying what it does.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Port the following code from Ruby to Java with equivalent syntax and logic.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Ensure the translated Python code behaves exactly like the original Ruby snippet.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Can you help me rewrite this code in Python instead of Ruby, keeping it the same logically?
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Transform the following Ruby implementation into VB, maintaining the same output and logic.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Generate an equivalent VB version of this Ruby code.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Convert the following code from Ruby to Go, ensuring the logic remains intact.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
package main import "fmt" func main() { const ( target = 269696 modulus = 1000000 ) for n := 1; ; n++ { square := n * n ending := square % modulus if ending == target { fmt.Println("The smallest number whose square ends with", target, "is", n, ) return } } }
Write the same code in Go as shown below in Ruby.
n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n
package main import "fmt" func main() { const ( target = 269696 modulus = 1000000 ) for n := 1; ; n++ { square := n * n ending := square % modulus if ending == target { fmt.Println("The smallest number whose square ends with", target, "is", n, ) return } } }
Write the same code in C as shown below in Scala.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
#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 Scala code snippet into C without altering its behavior.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
#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 ; }
Maintain the same structure and functionality when rewriting this code in C#.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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 the following code from Scala to C#, ensuring the logic remains intact.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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++.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
#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 Scala code into C++ without altering its purpose.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
#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 Java code behaves exactly like the original Scala snippet.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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 Scala to Java.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Produce a language-to-language conversion: from Scala to Python, same semantics.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Generate a Python translation of this Scala snippet without changing its computational steps.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Produce a functionally identical VB code for the snippet given in Scala.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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 Scala.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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 Go while keeping its functionality equivalent to the Scala version.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
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 Scala code.
fun main(args: Array<String>) { var number = 520L var square = 520 * 520L while (true) { val last6 = square.toString().takeLast(6) if (last6 == "269696") { println("The smallest number is $number whose square is $square") return } number += 2 square = number * number } }
package main import "fmt" func main() { const ( target = 269696 modulus = 1000000 ) for n := 1; ; n++ { square := n * n ending := square % modulus if ending == target { fmt.Println("The smallest number whose square ends with", target, "is", n, ) return } } }
Translate this program into C but keep the logic exactly as in Swift.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
#include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, square; while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); return 0 ; }
Change the programming language of this snippet from Swift to C without modifying what it does.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
#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 Swift code.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
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); } } }}
Please provide an equivalent version of this Swift code in C++.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
#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 C++.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
#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 Swift implementation into Java, maintaining the same output and logic.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Please provide an equivalent version of this Swift code in Java.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
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 Swift to Python with equivalent syntax and logic.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Write the same algorithm in Python as shown in this Swift implementation.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Maintain the same structure and functionality when rewriting this code in VB.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
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 Swift to VB, same semantics.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Generate a Go translation of this Swift snippet without changing its computational steps.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
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 Swift snippet to Go and keep its semantics consistent.
import Swift for i in 2...Int.max { if i * i % 1000000 == 269696 { print(i, "is the smallest number that ends with 269696") break } }
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 C version of this Tcl code.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
#include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, square; while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); return 0 ; }
Keep all operations the same but rewrite the snippet in C.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
#include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, square; while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); return 0 ; }
Convert this Tcl snippet to C# and keep its semantics consistent.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
namespace Babbage_Problem { class iterateNumbers { public iterateNumbers() { long baseNumberSquared = 0; long baseNumber = 0; do { baseNumber += 1; baseNumberSquared = baseNumber * baseNumber; } while (Right6Digits(baseNumberSquared) != 269696); Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber); Console.WriteLine("The square is " + baseNumberSquared); } private long Right6Digits(long baseNumberSquared) { string numberAsString = baseNumberSquared.ToString(); if (numberAsString.Length < 6) { return baseNumberSquared; }; numberAsString = numberAsString.Substring(numberAsString.Length - 6); return long.Parse(numberAsString); } } }}
Ensure the translated C# code behaves exactly like the original Tcl snippet.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
namespace Babbage_Problem { class iterateNumbers { public iterateNumbers() { long baseNumberSquared = 0; long baseNumber = 0; do { baseNumber += 1; baseNumberSquared = baseNumber * baseNumber; } while (Right6Digits(baseNumberSquared) != 269696); Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber); Console.WriteLine("The square is " + baseNumberSquared); } private long Right6Digits(long baseNumberSquared) { string numberAsString = baseNumberSquared.ToString(); if (numberAsString.Length < 6) { return baseNumberSquared; }; numberAsString = numberAsString.Substring(numberAsString.Length - 6); return long.Parse(numberAsString); } } }}
Translate this program into C++ but keep the logic exactly as in Tcl.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
#include <iostream> int main( ) { int current = 0 ; while ( ( current * current ) % 1000000 != 269696 ) current++ ; std::cout << "The square of " << current << " is " << (current * current) << " !\n" ; return 0 ; }
Change the following Tcl code into C++ without altering its purpose.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
#include <iostream> int main( ) { int current = 0 ; while ( ( current * current ) % 1000000 != 269696 ) current++ ; std::cout << "The square of " << current << " is " << (current * current) << " !\n" ; return 0 ; }
Ensure the translated Java code behaves exactly like the original Tcl snippet.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Convert this Tcl block to Java, preserving its control flow and logic.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
public class Test { public static void main(String[] args) { int n = 0; do { n++; } while (n * n % 1000_000 != 269696); System.out.println(n); } }
Rewrite this program in Python while keeping its functionality equivalent to the Tcl version.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Write the same algorithm in Python as shown in this Tcl implementation.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
n=0 while n**2 % 1000000 != 269696: n += 1 print(n)
Change the programming language of this snippet from Tcl to VB without modifying what it does.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Convert this Tcl snippet to VB and keep its semantics consistent.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$i]"
Sub Baggage_Problem() Dim i As Long i = 520 Do While ((i * i) Mod 1000000) <> 269696 i = i + 4 Loop Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _ "Its square is : " & i * i End Sub
Translate the given Tcl code snippet into Go without altering its behavior.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$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 } } }
Generate an equivalent Go version of this Tcl code.
for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {} puts "$i squared is [expr $i*$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 } } }
Translate this program into PHP but keep the logic exactly as in Rust.
fn main() { let mut current = 0; while (current * current) % 1_000_000 != 269_696 { current += 1; } println!( "The smallest number whose square ends in 269696 is {}", current ); }
<?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
Translate this program into PHP but keep the logic exactly as in Rust.
fn main() { let mut current = 0; while (current * current) % 1_000_000 != 269_696 { current += 1; } println!( "The smallest number whose square ends in 269696 is {}", current ); }
<?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
Port the provided Ada code into PHP while preserving the original functionality.
with Ada.Text_IO; procedure Babbage_Problem is type Number is range 1 .. 99_736*99_736; X: Number := 1; begin while not (((X*X) mod 1_000_000) = 269_696) loop X := X+1; end loop; Ada.Text_IO.Put_Line(Number'Image(X)); end Babbage_Problem;
<?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 PHP as shown in this Ada implementation.
with Ada.Text_IO; procedure Babbage_Problem is type Number is range 1 .. 99_736*99_736; X: Number := 1; begin while not (((X*X) mod 1_000_000) = 269_696) loop X := X+1; end loop; Ada.Text_IO.Put_Line(Number'Image(X)); end Babbage_Problem;
<?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 this Arturo snippet to PHP and keep its semantics consistent.
n: new 0 while [269696 <> (n^2) % 1000000] -> inc 'n print n
<?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
Port the following code from Arturo to PHP with equivalent syntax and logic.
n: new 0 while [269696 <> (n^2) % 1000000] -> inc 'n print n
<?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 * $i), PHP_EOL; // Print the result
Rewrite the snippet below in PHP so it works the same as the original AWK code.
BEGIN { n = 0 do { n = n + 1 } while (n*n !~ /269696$/) print("The smallest number whose square ends in 269696 is " n) print("Its square is " n*n) exit(0) }
<?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 * $i), PHP_EOL; // Print the result
Can you help me rewrite this code in PHP instead of AWK, keeping it the same logically?
BEGIN { n = 0 do { n = n + 1 } while (n*n !~ /269696$/) print("The smallest number whose square ends in 269696 is " n) print("Its square is " n*n) exit(0) }
<?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 this BBC_Basic block to PHP, preserving its control flow and logic.
LET n = 0 REPEAT LET n = n + 1 UNTIL n^2 MOD 1000000 = 269696 PRINT "The smallest number whose square ends in 269696 is" n PRINT "Its square is" n^2
<?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
Preserve the algorithm and functionality while converting the code from BBC_Basic to PHP.
LET n = 0 REPEAT LET n = n + 1 UNTIL n^2 MOD 1000000 = 269696 PRINT "The smallest number whose square ends in 269696 is" n PRINT "Its square is" n^2
<?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
Produce a functionally identical PHP code for the snippet given in Clojure.
(defn babbage? [n] (let [square (* n n)] (= 269696 (mod square 1000000)))) (first (filter babbage? (range)))
<?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
Translate the given Clojure code snippet into PHP without altering its behavior.
(defn babbage? [n] (let [square (* n n)] (= 269696 (mod square 1000000)))) (first (filter babbage? (range)))
<?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
Generate a PHP translation of this Common_Lisp snippet without changing its computational steps.
(defun babbage-test (n) "A generic function for any ending of a number" (when (> n 0) (do* ((i 0 (1+ i)) (d (expt 10 (1+ (truncate (log n) (log 10))))) ) ((= (mod (* i i) d) n) i) )))
<?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
Ensure the translated PHP code behaves exactly like the original Common_Lisp snippet.
(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) )))
<?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
Port the following code from D to PHP 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); }
<?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
Change the programming language of this snippet from D to PHP without modifying what it does.
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); }
<?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 a version of this Elixir function in PHP with identical behavior.
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)
<?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 this Elixir block to PHP, 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)
<?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
Generate a PHP translation of this Erlang snippet without changing its computational steps.
-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).
<?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 PHP 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).
<?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 this F# block to PHP, preserving its control flow and logic.
Seq.initInfinite id |> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696) |> Seq.head |> printfn "%d"
<?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
Rewrite the snippet below in PHP 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"
<?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
Port the following code from Factor to PHP with equivalent syntax and logic.
USING: kernel math math.ranges prettyprint sequences ; 518 99,736 2 <range> [ sq 1,000,000 mod 269,696 = ] find . drop
<?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
Transform the following Factor implementation into PHP, maintaining the same output and logic.
USING: kernel math math.ranges prettyprint sequences ; 518 99,736 2 <range> [ sq 1,000,000 mod 269,696 = ] find . drop
<?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
Can you help me rewrite this code in PHP instead of Forth, keeping it the same logically?
: BABBAGE 1 BEGIN 1+ DUP DUP * 1000000 MOD 269696 = UNTIL . ; BABBAGE
<?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
Port the provided Forth code into PHP while preserving the original functionality.
: BABBAGE 1 BEGIN 1+ DUP DUP * 1000000 MOD 269696 = UNTIL . ; BABBAGE
<?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
Rewrite the snippet below in PHP 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
<?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
Port the provided Fortran code into PHP while preserving the original functionality.
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
Keep all operations the same but rewrite the snippet in PHP.
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 ;
<?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
Please provide an equivalent version of this Groovy code in PHP.
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 ;
<?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 * $i), PHP_EOL; // Print the result
Port the following code from Haskell to PHP with equivalent syntax and logic.
findBabbageNumber :: Integer findBabbageNumber = head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..]) main :: IO () main = (putStrLn . unwords) (zipWith (++) (show <$> ([id, (^ 2)] <*> [findBabbageNumber])) [" ^ 2 equals", " !"])
<?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 * $i), PHP_EOL; // Print the result
Generate an equivalent PHP version of this Haskell code.
findBabbageNumber :: Integer findBabbageNumber = head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..]) main :: IO () main = (putStrLn . unwords) (zipWith (++) (show <$> ([id, (^ 2)] <*> [findBabbageNumber])) [" ^ 2 equals", " !"])
<?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 this J snippet to PHP and keep its semantics consistent.
square=: ^&2 modulo1e6=: 1000000&| trythese=: i. 1000000 which=: I. which 269696=modulo1e6 square trythese 25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
<?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
Preserve the algorithm and functionality while converting the code from J to PHP.
square=: ^&2 modulo1e6=: 1000000&| trythese=: i. 1000000 which=: I. which 269696=modulo1e6 square trythese 25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736
<?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
Rewrite this program in PHP while keeping its functionality equivalent to the Julia version.
function babbage(x::Integer) i = big(0) d = floor(log10(x)) + 1 while i ^ 2 % 10 ^ d != x i += 1 end return i end
<?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 this Julia snippet to PHP and keep its semantics consistent.
function babbage(x::Integer) i = big(0) d = floor(log10(x)) + 1 while i ^ 2 % 10 ^ d != x i += 1 end return i end
<?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 a version of this Lua function in PHP with identical behavior.
k = math.floor(math.sqrt(269696)) if k % 2 == 1 then k = k - 1 end while not ((k * k) % 1000000 == 269696) do k = k + 2 end io.write(string.format("%d * %d = %d\n", k, k, k * k))
<?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