Program to find sum of prime numbers between 1 to n in C++

In this problem, we are given a number n. Our task is to create a program to find sum of prime numbers between 1 to n in C++.

Prime Numbers are those numbers that have only two factors. They are the number and 1.

Let’s take an example to understand the problem

Input

n = 15

Output

41

Explanation

Prime numbers between 1 to 15 are 2, 3, 5, 7, 11, 13. The sum is 41.

Solution Approach

A simple way to solve the problem is by using a loop and checking if each number is a prime number or not and add all those which are prime.

To check if a number i is a prime number. We will loop from i to i/2. check if there lies a number that can divide i. If yes, then the number is not a prime number.

Program to illustrate the working of our solution

Example

 Live Demo

#include <iostream>
using namespace std;
bool isPrime(int n){
   for(int i = 2; i < n/2; i++){
      if(n%i == 0){
         return false;
      }
   }
   return true;
}
int findPrimeSum(int n){
   int sumVal = 0;
   for(float i = 2; i <= n; i++){
      if(isPrime(i))
         sumVal += i;
   }
   return sumVal;
}
int main(){
   int n = 15;
   cout<<"The sum of prime number between 1 to "<<n<<" is "<<findPrimeSum(n);
   return 0;
}

Output

The sum of prime number between 1 to 15 is 45

An Effective solution is using Sieve of Eratosthenes to find prime numbers and adding them to find the required sum.

Program to illustrate the working of our solution

Example

 Live Demo

#include <iostream>
using namespace std;
int findPrimeSum(int n){
   int arr[n+1] = {0};
   for (int i = 2; i < n; i++) {
      for (int j = i * i; j < n; j+=i) {
         arr[j - 1] = 1;
      }
   }
   int sumVal;
   for (int i = 2; i < n; i++) {
      if (arr[i - 1] == 0)
         sumVal += i;
   }
   return sumVal;
}
int main(){
   int n = 15;
   cout<<"The sum of prime number between 1 to "<<n<<" is "<<findPrimeSum(n);
   return 0;
}

Output

The sum of prime number between 1 to 15 is 41