// C++ program to find XOR of numbers // from 1 to n. #include    using namespace std; int computeXOR(int n) {  int res = 0;  for (int i = 1; i <= n; i++) {  res = res ^ i;   }  return res; } int main() {  int n = 7;  cout << computeXOR(n) << endl;  return 0; } 

// Given a number n find the XOR from 1 to n for given n number import java.io.*; public class GfG {  static int computeXor(int n){  int res = 0;  for (int i = 1; i <= n; i++) {  res = res^i;   }  return res;  }  public static void main(String[] args) {  int n = 7;  System.out.println(computeXor(n));  } } 

#defining a function computeXOR def computeXOR(n): res = 0 for i in range(1n+1): res = res ^ i return res n = 7 print(computeXOR(n)) 

// C# program that finds the XOR // from 1 to n for a given number n using System; public class GFG {  static int computeXor(int n)  {  int res = 0;  for (int i = 1; i <= n; i++) {  res = res ^ i; // calculate XOR  }  return res;  }  // Driver code  public static void Main(string[] args)  {  int n = 7;  // Function call  int ans = computeXor(n);  Console.WriteLine(ans);  } } // This code is contributed by phasing17 

// JavaScript that for a number n // finds the XOR from 1 to n for given n number function computeXor(n){  var res = 0;  for (let i = 1; i <= n; i++)  res = res^i; // calculate XOR  return res; } // Driver Code let n = 7; console.log(computeXor(n)); 

// C++ program to find XOR of numbers // from 1 to n. #include   using namespace std; // Method to calculate xor int computeXOR(int n) {    // If n is a multiple of 4  if (n % 4 == 0)  return n;  // If n%4 gives remainder 1  if (n % 4 == 1)  return 1;  // If n%4 gives remainder 2  if (n % 4 == 2)  return n + 1;  // If n%4 gives remainder 3  return 0; } // Driver method int main() {  int n = 5;  cout<<computeXOR(n); } // This code is contributed by rutvik_56. 

// Java program to find XOR of numbers // from 1 to n. class GFG  {  // Method to calculate xor  static int computeXOR(int n)  {  // If n is a multiple of 4  if (n % 4 == 0)  return n;    // If n%4 gives remainder 1  if (n % 4 == 1)  return 1;    // If n%4 gives remainder 2  if (n % 4 == 2)  return n + 1;    // If n%4 gives remainder 3  return 0;  }    // Driver method  public static void main (String[] args)  {  int n = 5;  System.out.println(computeXOR(n));  } } 

# Python 3 Program to find  # XOR of numbers from 1 to n.  # Function to calculate xor  def computeXOR(n) : # Modulus operator are expensive  # on most of the computers. n & 3  # will be equivalent to n % 4. # if n is multiple of 4  if n % 4 == 0 : return n # If n % 4 gives remainder 1 if n % 4 == 1 : return 1 # If n%4 gives remainder 2  if n % 4 == 2 : return n + 1 # If n%4 gives remainder 3 return 0 # Driver Code if __name__ == '__main__' : n = 5 # function calling print(computeXOR(n)) # This code is contributed by ANKITRAI1 

// C# program to find XOR  // of numbers from 1 to n. using System; class GFG {    // Method to calculate xor  static int computeXOR(int n)  {  // If n is a multiple of 4  if (n % 4 == 0)  return n;    // If n%4 gives remainder 1  if (n % 4 == 1)  return 1;    // If n%4 gives remainder 2  if (n % 4 == 2)  return n + 1;    // If n%4 gives remainder 3  return 0;  }    // Driver Code  static public void Main ()  {  int n = 5;  Console.WriteLine(computeXOR(n));  } } // This code is contributed by ajit 

<script> // JavaScript program to find XOR of numbers  // from 1 to n.  // Function to calculate xor  function computeXOR(n)  {   // Modulus operator are expensive on most of the   // computers. n & 3 will be equivalent to n % 4.   // if n is multiple of 4   if(n % 4 == 0)   return n;   // If n % 4 gives remainder 1   if(n % 4 == 1)   return 1;   // If n % 4 gives remainder 2   if(n % 4 == 2)   return n + 1;   // If n % 4 gives remainder 3   if(n % 4 == 3)   return 0;    }  // Driver code   // your code goes here   let n = 5;   document.write(computeXOR(n));  // This code is constributed by Surbhi Tyagi. </script> 

 // PHP program to find XOR  // of numbers from 1 to n. // Function to calculate xor function computeXOR($n) { // Modulus operator are expensive  // on most of the computers. n & 3 // will be equivalent to n % 4.  switch($n & 3) // n % 4  { // if n is multiple of 4 case 0: return $n; // If n % 4 gives remainder 1  case 1: return 1; // If n % 4 gives remainder 2 case 2: return $n + 1; // If n % 4 gives remainder 3  case 3: return 0; } } // Driver code $n = 5; echo computeXOR($n); // This code is contributed by aj_36 ?>