ProEuler P_010d

package problems;

public class Problem_010d {


    /*
     * Wow, this one is also quite fast and it doesn't use an array at all! Very clever code from a guy called blub123... lol...
     *
     */

    // main
    public static void main(String[] args) {

        sumPrimesUnder(20000000,1);              // Print the final sum.
    }


    public static long sumPrimesUnder (long input, int a){
        long testrange = input;
        long answer = primeAdder(testrange);

        if(a == 1){ // Switch to suppress message or not. Good if being used in another code.
            System.out.print("The sum of all primes under " + testrange +" is: " + answer +".");
        }

        return answer;
    }



    public static long primeAdder(long testrange){

        // This method sends numbers into the isPrime method to check if they are prime, add them to a counter variable if they are and then return the sum.

        long result = 0;  // Create a counter.


        // algorithm
        for (int number = 0; number < testrange; number++) {  // For all Number = 0, up to 1000000, incrementing by two.
            if (isPrime(number)) {                          // Call the primechecking method on the number.
                result += number;                           // If the primechecking method returns true, then add the number to the counter variable "result".
            }
        }
        return result;

    }


    // functions
    // checks if a number is a prime
    public static boolean isPrime(int number) {                 // This is a boolean method, returns true or false.

        if (number < 2) {                                       // If the number is less than two, return false.
            return false;
        } else if (number == 2) {                               // If the number equals to, return true. Because 2 is prime, duh.
            return true;
        } else if (number % 2 == 0 || number % 3 == 0 || number % 5 == 0 || number % 7 == 0 || number % 11 == 0 || number % 13 == 0 || number % 5 == 17) {
            // If the number is divisible by 2,3 or 5, return false. (This makes checking if even numbers are prime fast.
            return false;
        } else {                                                // End of IF statement, start of ELSE.

            for (int divisor = 3; divisor <= Math.sqrt(number); divisor += 2) { // For divisor 3 up to and including the square-root of the number, increment by two and check following:
                if (number % divisor == 0) {                                    // If the number is divisble by the divisor we are checking, return false.
                    return false;                               // I was always trying to implement the root limit myself, but I always did it wrong and go some weird results.
                }
            }
            return true;                                                        // If the flow of execution reaches here, all divisors up to the root of the number we are testing for
                                                                                // have been tried, and therefore the number must be a prime.
        }
    }
}