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Generate an equivalent C version of this Julia code.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Port the provided Julia code into C# while preserving the original functionality.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Generate an equivalent C# version of this Julia code.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Ensure the translated C++ code behaves exactly like the original Julia snippet.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Generate a C++ translation of this Julia snippet without changing its computational steps.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Maintain the same structure and functionality when rewriting this code in Java.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Write a version of this Julia function in Java with identical behavior.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Ensure the translated Python code behaves exactly like the original Julia snippet.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Translate this program into VB but keep the logic exactly as in Julia.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Produce a language-to-language conversion: from Julia to VB, same semantics.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Please provide an equivalent version of this Julia code in Go.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Rewrite this program in Go while keeping its functionality equivalent to the Julia version.
println([n for n = 1:5000 if sum(d^d for d in digits(n)) == n])
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Produce a functionally identical C code for the snippet given in Lua.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Generate a C translation of this Lua snippet without changing its computational steps.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Please provide an equivalent version of this Lua code in C#.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Change the programming language of this snippet from Lua to C# without modifying what it does.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Change the following Lua code into C++ without altering its purpose.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Maintain the same structure and functionality when rewriting this code in C++.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Convert this Lua block to Java, preserving its control flow and logic.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Can you help me rewrite this code in Java instead of Lua, keeping it the same logically?
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Convert the following code from Lua to Python, ensuring the logic remains intact.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Rewrite the snippet below in Python so it works the same as the original Lua code.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Translate the given Lua code snippet into VB without altering its behavior.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Generate a VB translation of this Lua snippet without changing its computational steps.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Translate this program into Go but keep the logic exactly as in Lua.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Maintain the same structure and functionality when rewriting this code in Go.
function isMunchausen (n) local sum, nStr, digit = 0, tostring(n) for pos = 1, #nStr do digit = tonumber(nStr:sub(pos, pos)) sum = sum + digit ^ digit end return sum == n end local function isMunchausen (n) local sum, digit, acc = 0, 0, n while acc > 0 do digit = acc % 10.0 sum = sum + digit ^ digit acc = acc // 10 end return sum == n end for i = 1, 5000 do if isMunchausen(i) then print(i) end end
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Convert this Mathematica snippet to C and keep its semantics consistent.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Convert the following code from Mathematica to C, ensuring the logic remains intact.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Write a version of this Mathematica function in C# with identical behavior.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Preserve the algorithm and functionality while converting the code from Mathematica to C#.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Convert the following code from Mathematica to C++, ensuring the logic remains intact.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Port the provided Mathematica code into C++ while preserving the original functionality.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Generate a Java translation of this Mathematica snippet without changing its computational steps.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Preserve the algorithm and functionality while converting the code from Mathematica to Java.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Convert this Mathematica block to Python, preserving its control flow and logic.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Maintain the same structure and functionality when rewriting this code in Python.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Change the following Mathematica code into VB without altering its purpose.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Translate the given Mathematica code snippet into VB without altering its behavior.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Change the programming language of this snippet from Mathematica to Go without modifying what it does.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Produce a functionally identical Go code for the snippet given in Mathematica.
Off[Power::indet]; Select[Range[5000], Total[IntegerDigits[#]^IntegerDigits[#]] == # &]
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Convert this Nim block to C, preserving its control flow and logic.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Port the following code from Nim to C with equivalent syntax and logic.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Can you help me rewrite this code in C# instead of Nim, keeping it the same logically?
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Write the same code in C# as shown below in Nim.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Write a version of this Nim function in C++ with identical behavior.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Port the following code from Nim to C++ with equivalent syntax and logic.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Convert the following code from Nim to Java, ensuring the logic remains intact.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Produce a functionally identical Java code for the snippet given in Nim.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Translate this program into Python but keep the logic exactly as in Nim.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Write a version of this Nim function in Python with identical behavior.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Transform the following Nim implementation into VB, maintaining the same output and logic.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Port the following code from Nim to VB with equivalent syntax and logic.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Rewrite this program in Go while keeping its functionality equivalent to the Nim version.
import math for i in 1..<5000: var sum: int64 = 0 var number = i while number > 0: var digit = number mod 10 sum += digit ^ digit number = number div 10 if sum == i: echo i
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Port the following code from OCaml to C with equivalent syntax and logic.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Please provide an equivalent version of this OCaml code in C.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Please provide an equivalent version of this OCaml code in C#.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Rewrite this program in C# while keeping its functionality equivalent to the OCaml version.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Write the same code in C++ as shown below in OCaml.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Rewrite this program in C++ while keeping its functionality equivalent to the OCaml version.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Write the same code in Java as shown below in OCaml.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Write the same algorithm in Java as shown in this OCaml implementation.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Translate this program into Python but keep the logic exactly as in OCaml.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Keep all operations the same but rewrite the snippet in Python.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Produce a language-to-language conversion: from OCaml to VB, same semantics.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Generate an equivalent Go version of this OCaml code.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Please provide an equivalent version of this OCaml code in Go.
let is_munchausen n = let pwr = [|1; 1; 4; 27; 256; 3125; 46656; 823543; 16777216; 387420489|] in let rec aux x = if x < 10 then pwr.(x) else aux (x / 10) + pwr.(x mod 10) in n = aux n let () = Seq.(ints 1 |> take 5000 |> filter is_munchausen |> iter (Printf.printf " %u")) |> print_newline
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Convert this Pascal block to C, preserving its control flow and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Convert this Pascal block to C, preserving its control flow and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Port the following code from Pascal to C# with equivalent syntax and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Convert this Pascal block to C#, preserving its control flow and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Preserve the algorithm and functionality while converting the code from Pascal to C++.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Produce a functionally identical C++ code for the snippet given in Pascal.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Translate the given Pascal code snippet into Java without altering its behavior.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Maintain the same structure and functionality when rewriting this code in Java.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Generate an equivalent Python version of this Pascal code.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Port the following code from Pascal to Python with equivalent syntax and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Convert the following code from Pascal to VB, ensuring the logic remains intact.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Convert this Pascal snippet to VB and keep its semantics consistent.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Maintain the same structure and functionality when rewriting this code in Go.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Port the following code from Pascal to Go with equivalent syntax and logic.
uses sysutils; type tdigit = byte; const base = 10; maxDigits = base-1; var DgtPotDgt : array[0..base-1] of NativeUint; cnt: NativeUint; function CheckSameDigits(n1,n2:NativeUInt):boolean; var dgtCnt : array[0..Base-1] of NativeInt; i : NativeUInt; Begin fillchar(dgtCnt,SizeOf(dgtCnt),#0); repeat i := n1;n1 := n1 div base;i := i-n1*base;inc(dgtCnt[i]); i := n2;n2 := n2 div base;i := i-n2*base;dec(dgtCnt[i]); until (n1=0) AND (n2= 0 ); result := true; For i := 0 to Base-1 do result := result AND (dgtCnt[i]=0); end; procedure Munch(number,DgtPowSum,minDigit:NativeUInt;digits:NativeInt); var i: NativeUint; begin inc(cnt); number := number*base; IF digits > 1 then Begin For i := minDigit to base-1 do Munch(number+i,DgtPowSum+DgtPotDgt[i],i,digits-1); end else For i := minDigit to base-1 do IF (number+i)<= (DgtPowSum+DgtPotDgt[i]) then IF CheckSameDigits(number+i,DgtPowSum+DgtPotDgt[i]) then iF number+i>0 then writeln(Format('%*d  %.*d', [maxDigits,DgtPowSum+DgtPotDgt[i],maxDigits,number+i])); end; procedure InitDgtPotDgt; var i,k,dgtpow: NativeUint; Begin DgtPotDgt[0]:= 0; For i := 1 to Base-1 do Begin dgtpow := i; For k := 2 to i do dgtpow := dgtpow*i; DgtPotDgt[i] := dgtpow; end; end; begin cnt := 0; InitDgtPotDgt; Munch(0,0,0,maxDigits); writeln('Check Count ',cnt); end.
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Translate the given Perl code snippet into C without altering its behavior.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Generate an equivalent C version of this Perl code.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Ensure the translated C# code behaves exactly like the original Perl snippet.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Rewrite this program in C# while keeping its functionality equivalent to the Perl version.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Translate this program into C++ but keep the logic exactly as in Perl.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Produce a language-to-language conversion: from Perl to C++, same semantics.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Generate an equivalent Java version of this Perl code.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Keep all operations the same but rewrite the snippet in Java.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
public class Main { public static void main(String[] args) { for(int i = 0 ; i <= 5000 ; i++ ){ int val = String.valueOf(i).chars().map(x -> (int) Math.pow( x-48 ,x-48)).sum(); if( i == val){ System.out.println( i + " (munchausen)"); } } } }
Convert this Perl snippet to Python and keep its semantics consistent.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Produce a language-to-language conversion: from Perl to Python, same semantics.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
for i in range(5000): if i == sum(int(x) ** int(x) for x in str(i)): print(i)
Produce a functionally identical VB code for the snippet given in Perl.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Produce a functionally identical VB code for the snippet given in Perl.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
Option Explicit Sub Main_Munchausen_numbers() Dim i& For i = 1 To 5000 If IsMunchausen(i) Then Debug.Print i & " is a munchausen number." Next i End Sub Function IsMunchausen(Number As Long) As Boolean Dim Digits, i As Byte, Tot As Long Digits = Split(StrConv(Number, vbUnicode), Chr(0)) For i = 0 To UBound(Digits) - 1 Tot = (Digits(i) ^ Digits(i)) + Tot Next i IsMunchausen = (Tot = Number) End Function
Preserve the algorithm and functionality while converting the code from Perl to Go.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Convert the following code from Perl to Go, ensuring the logic remains intact.
use List::Util "sum"; for my $n (1..5000) { print "$n\n" if $n == sum( map { $_**$_ } split(//,$n) ); }
package main import( "fmt" "math" ) var powers [10]int func isMunchausen(n int) bool { if n < 0 { return false } n64 := int64(n) nn := n64 var sum int64 = 0 for nn > 0 { sum += int64(powers[nn % 10]) if sum > n64 { return false } nn /= 10 } return sum == n64 } func main() { for i := 1; i <= 9; i++ { d := float64(i) powers[i] = int(math.Pow(d, d)) } fmt.Println("The Munchausen numbers between 0 and 500 million are:") for i := 0; i <= 500000000; i++ { if isMunchausen(i) { fmt.Printf("%d ", i) } } fmt.Println() }
Write the same algorithm in C as shown in this COBOL implementation.
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Generate an equivalent C version of this COBOL code.
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
#include <stdio.h> #include <math.h> int main() { for (int i = 1; i < 5000; i++) { int sum = 0; for (int number = i; number > 0; number /= 10) { int digit = number % 10; sum += pow(digit, digit); } if (sum == i) { printf("%i\n", i); } } return 0; }
Generate a C# translation of this COBOL snippet without changing its computational steps.
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Can you help me rewrite this code in C# instead of COBOL, keeping it the same logically?
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
Func<char, int> toInt = c => c-'0'; foreach (var i in Enumerable.Range(1,5000) .Where(n => n == n.ToString() .Sum(x => Math.Pow(toInt(x), toInt(x))))) Console.WriteLine(i);
Maintain the same structure and functionality when rewriting this code in C++.
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }
Convert this COBOL block to C++, preserving its control flow and logic.
IDENTIFICATION DIVISION. PROGRAM-ID. MUNCHAUSEN. DATA DIVISION. WORKING-STORAGE SECTION. 01 VARIABLES. 03 CANDIDATE PIC 9(4). 03 DIGITS PIC 9 OCCURS 4 TIMES, REDEFINES CANDIDATE. 03 DIGIT PIC 9. 03 POWER-SUM PIC 9(5). 01 OUTPUT-LINE. 03 OUT-NUM PIC ZZZ9. PROCEDURE DIVISION. BEGIN. PERFORM MUNCHAUSEN-TEST VARYING CANDIDATE FROM 1 BY 1 UNTIL CANDIDATE IS GREATER THAN 6000. STOP RUN. MUNCHAUSEN-TEST. MOVE ZERO TO POWER-SUM. MOVE 1 TO DIGIT. INSPECT CANDIDATE TALLYING DIGIT FOR LEADING '0'. PERFORM ADD-DIGIT-POWER VARYING DIGIT FROM DIGIT BY 1 UNTIL DIGIT IS GREATER THAN 4. IF POWER-SUM IS EQUAL TO CANDIDATE, MOVE CANDIDATE TO OUT-NUM, DISPLAY OUTPUT-LINE. ADD-DIGIT-POWER. COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** DIGITS(DIGIT)
#include <math.h> #include <iostream> unsigned pwr[10]; unsigned munch( unsigned i ) { unsigned sum = 0; while( i ) { sum += pwr[(i % 10)]; i /= 10; } return sum; } int main( int argc, char* argv[] ) { for( int i = 0; i < 10; i++ ) pwr[i] = (unsigned)pow( (float)i, (float)i ); std::cout << "Munchausen Numbers\n==================\n"; for( unsigned i = 1; i < 5000; i++ ) if( i == munch( i ) ) std::cout << i << "\n"; return 0; }