# Find the XOR of first N Prime Numbers

• Last Updated : 16 Mar, 2021

Given a positive integer N, the task is to find the XOR of the first N prime numbers.
Examples:

Input: N = 3
Output:
First 3 prime numbers are 2, 3 and 5.
And 2 ^ 3 ^ 5 = 4
Input: N = 5
Output:

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Approach:

1. Create Sieve of Eratosthenes to identify if a number is prime or not in O(1) time.
2. Run a loop starting from 1 until and unless we find N prime numbers.
3. XOR all the prime numbers and neglect those which are not prime.
4. Finally, print the XOR of the 1st N prime numbers.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;``#define MAX 10000` `// Create a boolean array "prime[0..n]" and initialize``// all entries it as true. A value in prime[i] will``// finally be false if i is Not a prime, else true.``bool` `prime[MAX + 1];``void` `SieveOfEratosthenes()``{``    ``memset``(prime, ``true``, ``sizeof``(prime));` `    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= MAX; p++) {` `        ``// If prime[p] is not changed, then it is a prime``        ``if` `(prime[p] == ``true``) {` `            ``// Set all multiples of p to non-prime``            ``for` `(``int` `i = p * 2; i <= MAX; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return the xor of 1st N prime numbers``int` `xorFirstNPrime(``int` `n)``{``    ``// Count of prime numbers``    ``int` `count = 0, num = 1;` `    ``// XOR of prime numbers``    ``int` `xorVal = 0;` `    ``while` `(count < n) {` `        ``// If the number is prime xor it``        ``if` `(prime[num]) {``            ``xorVal ^= num;` `            ``// Increment the count``            ``count++;``        ``}` `        ``// Get to the next number``        ``num++;``    ``}``    ``return` `xorVal;``}` `// Driver code``int` `main()``{``    ``// Create the sieve``    ``SieveOfEratosthenes();` `    ``int` `n = 4;` `    ``// Find the xor of 1st n prime numbers``    ``cout << xorFirstNPrime(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``static` `final` `int` `MAX = ``10000``;` `// Create a boolean array "prime[0..n]"``// and initialize all entries it as true.``// A value in prime[i] will finally be false``// if i is Not a prime, else true.``static` `boolean` `prime[] = ``new` `boolean` `[MAX + ``1``];` `static` `void` `SieveOfEratosthenes()``{``    ``int` `i ;``    ``for` `(i = ``0``; i < MAX + ``1``; i++)``    ``{``        ``prime[i] = ``true``;``    ``}` `    ``prime[``1``] = ``false``;` `    ``for` `(``int` `p = ``2``; p * p <= MAX; p++)``    ``{` `        ``// If prime[p] is not changed,``        ``// then it is a prime``        ``if` `(prime[p] == ``true``)``        ``{` `            ``// Set all multiples of p to non-prime``            ``for` `(i = p * ``2``; i <= MAX; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return the xor of``// 1st N prime numbers``static` `int` `xorFirstNPrime(``int` `n)``{``    ``// Count of prime numbers``    ``int` `count = ``0``, num = ``1``;` `    ``// XOR of prime numbers``    ``int` `xorVal = ``0``;` `    ``while` `(count < n)``    ``{` `        ``// If the number is prime xor it``        ``if` `(prime[num])``        ``{``            ``xorVal ^= num;` `            ``// Increment the count``            ``count++;``        ``}` `        ``// Get to the next number``        ``num++;``    ``}``    ``return` `xorVal;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``// Create the sieve``    ``SieveOfEratosthenes();` `    ``int` `n = ``4``;` `    ``// Find the xor of 1st n prime numbers``    ``System.out.println(xorFirstNPrime(n));` `}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach``MAX` `=` `10000` `# Create a boolean array "prime[0..n]" and``# initialize all entries it as true.``# A value in prime[i] will finally be false +``# if i is Not a prime, else true.``prime ``=` `[``True` `for` `i ``in` `range``(``MAX` `+` `1``)]` `def` `SieveOfEratosthenes():` `    ``prime[``1``] ``=` `False` `    ``for` `p ``in` `range``(``2``, ``MAX` `+` `1``):` `        ``# If prime[p] is not changed,``        ``# then it is a prime``        ``if` `(prime[p] ``=``=` `True``):` `            ``# Set all multiples of p to non-prime``            ``for` `i ``in` `range``(``2` `*` `p, ``MAX` `+` `1``, p):``                ``prime[i] ``=` `False` `# Function to return the xor of``# 1st N prime numbers``def` `xorFirstNPrime(n):``    ` `    ``# Count of prime numbers``    ``count ``=` `0``    ``num ``=` `1` `    ``# XOR of prime numbers``    ``xorVal ``=` `0` `    ``while` `(count < n):` `        ``# If the number is prime xor it``        ``if` `(prime[num]):``            ``xorVal ^``=` `num` `            ``# Increment the count``            ``count ``+``=` `1` `        ``# Get to the next number``        ``num ``+``=` `1` `    ``return` `xorVal` `# Driver code` `# Create the sieve``SieveOfEratosthenes()` `n ``=` `4` `# Find the xor of 1st n prime numbers``print``(xorFirstNPrime(n))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `static` `int` `MAX = 10000;` `// Create a boolean array "prime[0..n]"``// and initialize all entries it as true.``// A value in prime[i] will finally be false``// if i is Not a prime, else true.``static` `bool` `[]prime = ``new` `bool` `[MAX + 1];` `static` `void` `SieveOfEratosthenes()``{``    ``int` `i ;``    ``for` `(i = 0; i < MAX + 1; i++)``    ``{``        ``prime[i] = ``true``;``    ``}` `    ``prime[1] = ``false``;` `    ``for` `(``int` `p = 2; p * p <= MAX; p++)``    ``{` `        ``// If prime[p] is not changed,``        ``// then it is a prime``        ``if` `(prime[p] == ``true``)``        ``{` `            ``// Set all multiples of p to non-prime``            ``for` `(i = p * 2; i <= MAX; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}``}` `// Function to return the xor of``// 1st N prime numbers``static` `int` `xorFirstNPrime(``int` `n)``{``    ``// Count of prime numbers``    ``int` `count = 0, num = 1;` `    ``// XOR of prime numbers``    ``int` `xorVal = 0;` `    ``while` `(count < n)``    ``{` `        ``// If the number is prime xor it``        ``if` `(prime[num])``        ``{``            ``xorVal ^= num;` `            ``// Increment the count``            ``count++;``        ``}` `        ``// Get to the next number``        ``num++;``    ``}``    ``return` `xorVal;``}` `// Driver code``static` `public` `void` `Main ()``{``    ` `    ``// Create the sieve``    ``SieveOfEratosthenes();``    ``int` `n = 4;` `    ``// Find the xor of 1st n prime numbers``    ``Console.Write(xorFirstNPrime(n));``}``}` `// This code is contributed by Sachin`

## Javascript

 ``
Output:
`3`

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