Find the XOR of first N Prime Numbers
Last Updated :
18 Mar, 2022
Given a positive integer N, the task is to find the XOR of the first N prime numbers.
Examples:
Input: N = 3
Output: 4
First 3 prime numbers are 2, 3 and 5.
And 2 ^ 3 ^ 5 = 4
Input: N = 5
Output: 8
Approach:
- Create Sieve of Eratosthenes to identify if a number is prime or not in O(1) time.
- Run a loop starting from 1 until and unless we find N prime numbers.
- XOR all the prime numbers and neglect those which are not prime.
- Finally, print the XOR of the 1st N prime numbers.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAX 10000
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] == true ) {
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
int xorFirstNPrime( int n)
{
int count = 0, num = 1;
int xorVal = 0;
while (count < n) {
if (prime[num]) {
xorVal ^= num;
count++;
}
num++;
}
return xorVal;
}
int main()
{
SieveOfEratosthenes();
int n = 4;
cout << xorFirstNPrime(n);
return 0;
}
|
Java
class GFG
{
static final int MAX = 10000 ;
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] == true )
{
for (i = p * 2 ; i <= MAX; i += p)
prime[i] = false ;
}
}
}
static int xorFirstNPrime( int n)
{
int count = 0 , num = 1 ;
int xorVal = 0 ;
while (count < n)
{
if (prime[num])
{
xorVal ^= num;
count++;
}
num++;
}
return xorVal;
}
public static void main (String[] args)
{
SieveOfEratosthenes();
int n = 4 ;
System.out.println(xorFirstNPrime(n));
}
}
|
Python3
MAX = 10000
prime = [ True for i in range ( MAX + 1 )]
def SieveOfEratosthenes():
prime[ 1 ] = False
for p in range ( 2 , MAX + 1 ):
if (prime[p] = = True ):
for i in range ( 2 * p, MAX + 1 , p):
prime[i] = False
def xorFirstNPrime(n):
count = 0
num = 1
xorVal = 0
while (count < n):
if (prime[num]):
xorVal ^ = num
count + = 1
num + = 1
return xorVal
SieveOfEratosthenes()
n = 4
print (xorFirstNPrime(n))
|
C#
using System;
class GFG
{
static int MAX = 10000;
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] == true )
{
for (i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
static int xorFirstNPrime( int n)
{
int count = 0, num = 1;
int xorVal = 0;
while (count < n)
{
if (prime[num])
{
xorVal ^= num;
count++;
}
num++;
}
return xorVal;
}
static public void Main ()
{
SieveOfEratosthenes();
int n = 4;
Console.Write(xorFirstNPrime(n));
}
}
|
Javascript
<script>
let MAX = 10000;
let prime = new Array(MAX + 1);
function SieveOfEratosthenes()
{
let i;
for (i = 0; i < MAX + 1; i++)
{
prime[i] = true ;
}
prime[1] = false ;
for (let p = 2; p * p <= MAX; p++)
{
if (prime[p] == true )
{
for (i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
function xorFirstNPrime(n)
{
let count = 0, num = 1;
let xorVal = 0;
while (count < n)
{
if (prime[num])
{
xorVal ^= num;
count++;
}
num++;
}
return xorVal;
}
SieveOfEratosthenes();
let n = 4;
document.write(xorFirstNPrime(n));
</script>
|
Time Complexity: O(n + MAX*log(log(MAX)))
Auxiliary Space: O(MAX)
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