Sum of the first N Prime numbers

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

# 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.

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