Given an integer ‘n’, the task is to find the sum of first ‘n’ prime numbers.
First few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, ……
Examples:
Input: N = 4 Output: 17 2, 3, 5, 7 are first 4 prime numbers so their sum is equal to 17 Input: N = 40 Output: 3087
Approach:
- Create a sieve which will help us to identify if the number is prime or not in O(1) time.
- Run a loop starting from 1 until and unless we find n prime numbers.
- Add all the prime numbers and neglect those which are not prime.
- Then, display the sum of 1st N prime numbers.
Below is the implementation of the above solution
C++
// C++ implementation of above solution #include <bits/stdc++.h> 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 ;
}
}
} // find the sum of 1st N prime numbers int solve( int n)
{ // count of prime numbers
int count = 0, num = 1;
// sum of prime numbers
long long int sum = 0;
while (count < n) {
// if the number is prime add it
if (prime[num]) {
sum += num;
// increase the count
count++;
}
// get to next number
num++;
}
return sum;
} // Driver code int main()
{ // create the sieve
SieveOfEratosthenes();
int n = 4;
// find the value of 1st n prime numbers
cout << "Sum of 1st N prime numbers are :" << solve(n);
return 0;
} |
Java
// Java implementation of above solution public class Improve {
final static double 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 [( int ) (MAX + 1.0 )] ;
static void SieveOfEratosthenes()
{
for ( int i = 0 ; i <= MAX; 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 ( int i = p * 2 ; i <= MAX; i += p)
prime[i] = false ;
}
}
}
// find the sum of 1st N prime numbers
static int solve( int n)
{
// count of prime numbers
int count = 0 , num = 1 ;
// sum of prime numbers
long sum = 0 ;
while (count < n) {
// if the number is prime add it
if (prime[num]) {
sum += num;
// increase the count
count++;
}
// get to next number
num++;
}
return ( int ) sum;
}
// Driver code
public static void main(String args[])
{
// create the sieve
SieveOfEratosthenes();
int n = 4 ;
// find the value of 1st n prime numbers
System.out.println( "Sum of 1st N prime numbers are :" + solve(n));
}
// This Code is contributed by ANKITRAI1
} |
Python 3
# Python3 implementation of # above solution 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
i = p * 2
while (i < = MAX ):
prime[i] = False
i = i + p
# find the sum of 1st # N prime numbers def solve( n):
# count of prime numbers
count = 0
num = 1
# sum of prime numbers
total = 0
while (count < n):
# if the number is prime add it
if ( prime[num] ):
total = total + num
# increase the count
count = count + 1
# get to next number
num = num + 1
return total
# Driver code # create the sieve SieveOfEratosthenes() n = 4
# find the value of 1st # n prime numbers print ( "Sum of 1st N prime " +
"numbers are :" , solve(n))
# This code is contributed by ash264 |
C#
//C# implementation of above solution using System;
public class GFG{
static double 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 [( int )(MAX + 1.0)] ;
static void SieveOfEratosthenes()
{
for ( int i = 0; i <= MAX; 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 ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
// find the sum of 1st N prime numbers
static int solve( int n)
{
// count of prime numbers
int count = 0, num = 1;
// sum of prime numbers
long sum = 0;
while (count < n) {
// if the number is prime add it
if (prime[num]) {
sum += num;
// increase the count
count++;
}
// get to next number
num++;
}
return ( int ) sum;
}
// Driver code
static public void Main (){
// create the sieve
SieveOfEratosthenes();
int n = 4;
// find the value of 1st n prime numbers
Console.WriteLine( "Sum of 1st N prime numbers are :" + solve(n));
}
// This Code is contributed by ajit. } |
Javascript
<script> // javascript implementation of above solution var 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. var prime = Array.from({length: parseInt( (MAX + 1.0))}, (_, i) => false );
function SieveOfEratosthenes()
{ for (i = 0; i <= MAX; i++)
prime[i] = true ;
prime[1] = false ;
for (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 ;
}
}
} // find the sum of 1st N prime numbers function solve(n)
{ // count of prime numbers
var count = 0, num = 1;
// sum of prime numbers
var sum = 0;
while (count < n) {
// if the number is prime add it
if (prime[num]) {
sum += num;
// increase the count
count++;
}
// get to next number
num++;
}
return parseInt( sum);
} // Driver code //create the sieve SieveOfEratosthenes(); var n = 4;
// find the value of 1st n prime numbers document.write( "Sum of 1st N prime numbers are :" + solve(n));
// This code is contributed by 29AjayKumar </script> |
Output:
Sum of 1st N prime numbers are :17
Note(For competitive programming): In a problem which contains a large number of queries, a vector can be used to store all the prime numbers in the range of 10^8, this will take extra O(N) space. We can also use prefix array to store the sum of first N prime numbers in the range of 10^8.