In this post, I will give you some ways to count how many primes number from 1 to R.

A prime number is a number larger than 1 and has exactly two distinct natural number divisors: the number 1 and itself.

Note: I will be using C++ throughout this algorithm.

Brute force

#include <bits/stdc++.h>

using namespace std;

// Algorithm:
// Check every possible from 1 to r
// If it is a prime number then increase variable count by 1

bool isPrime(long long nums) {

    // Base case
    if(nums < 2) return false;
    if(nums <= 3) return true;

    // check from 2 to sqrt(nums)
    // if nums cannot divide for numbers
    // then it is in fact a prime number
    for(long long i = 2; i * i <= nums; i++)
        if(nums % i == 0)
            return false;

    return true;
}

int main() {
    long long r; // using long long in case if number up to 1e18
    cin >> r;

    long long counts = 0; // counts prime numbers

    // Increment counts if current is a prime number
    // every number less than r
    for(long long i = 2; i <= r; i++)
        if(isPrime(i))
            counts++;

    // Print the result out
    cout << counts;
    return 0;
}

Sieve of Eratosthenes

Sieve of Eratosthenes is an ancient algorithm that can find prime numbers with a limit of up to 1e7 numbers in 1 second.

We mark every composite number that is a multiply of a prime number. If there are no numbers left that we can start marking with, we exit the algorithm.

E.g: for prime number 2, we will mark every even number, starting from 4 to any limit.

#include <bits/stdc++.h>

using namespace std;

int main() {
    int r;
    cin >> r;

    int counts = 0;

    // Create a variable from to mark numbers
    vector<bool> isPrime(r + 1, true);

    isPrime[0] = isPrime[1] = false;

    // Start with numbers 2 to sqrt(n)
    // If current number doesn't mark then
    // mark all the numbers which are divisible by it.
    for(int i = 2; i * i <= r; i++)
        if(isPrime[i])
            for(int j = i * i; j <= r; j += i)
                isPrime[j] = false;

    // Increment counts if current is a prime number
    // every number less than r
    for(int i = 2; i <= r; i++)
        if(isPrime[i])
            counts++;

    // Print the result out
    cout << counts;
    return 0;
}

Conclusion

These are some common ways to count prime numbers that I use.

For more information about the sieve of Eratosthenes, you can check its wiki here