Finding the GCD

// Function to implement the Sieve of Eratosthenes
void sieve(bool prime[]) {

    // Step 1: Initialize all numbers as prime
    // Initially, assume all numbers from 2 to 1,000,000 are prime.
    for(int i=2; i<=1000000; i++){
        prime[i] = 1;  // Set all indices to true (prime)
    }

    // Step 2: Mark non-prime numbers
    // Loop through all numbers from 2 to 1,000,000
    for(int i = 2; i<=1000000; i++) {

        // If `i` is still marked as prime, process it
        if(prime[i]){
            // Mark all multiples of `i` as non-prime (set them to 0)
            // Starting from 2*i, 3*i, ..., up to 1,000,000
            for(int j = 2*i; j<=1000000; j+=i){
                prime[j] = 0;  // Mark multiple `j` as non-prime
            }
        }
    }

    // Step 3: Special cases for 0 and 1
    // Mark 0 and 1 as non-prime, because they are not prime by definition
    prime[0] = prime[1] = 0;
}

Example:

int main() {
    bool prime[1000001];  // Array to hold prime information for numbers from 0 to 1,000,000
    sieve(prime);  // Call the sieve function to populate the prime array
    // Check some prime numbers:
    cout << "Is 11 prime? " << (prime[11] ? "Yes" : "No") << endl;  // Output: Yes
    cout << "Is 15 prime? " << (prime[15] ? "Yes" : "No") << endl;  // Output: No
    cout << "Is 29 prime? " << (prime[29] ? "Yes" : "No") << endl;  // Output: Yes
    return 0;
}

Explanation of Steps:

1. Initialize all numbers:
   • The algorithm first marks all numbers from 2 to 1,000,000 as potentially prime.
2. Eliminate multiples:
   • Starting from 2 (the smallest prime), the algorithm marks all multiples of 2 (like 4, 6, 8, …) as non-prime.
   • It then moves to the next number, 3, and marks its multiples (like 6, 9, 12, …) as non-prime.
   • This process repeats for all numbers up to 1,000,000.
3. Check for prime status:
   • After the sieve runs, you can check if any number x is prime by checking the prime[x] array. If prime[x] == 1, it means x is prime; otherwise, it is not.