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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
#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
#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
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