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package euler_21;

/**
 *
 * @author stevengreen22
 */
public class Euler_21 {

    /**
     * Let d(n) be defined as the sum of proper divisors of n (numbers less than
     * n which divide evenly into n).
     *
     * If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair
     * and each of a and b are called amicable numbers.
     *
     * For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
     * 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1,
     * 2, 4, 71 and 142; so d(284) = 220.
     *
     * Evaluate the sum of all the amicable numbers under 10000.
     *
     *     /*
      * TODO code application logic here
        i want to loop from 1 to 10000 with this number being the param of the method.
        i then want it to return the sum of above number.
        then use then sum as the parameter.
        if the sum of 2nd param == first param then save to array
        if not, increment loop
        not efficient as will be checking same numbers twice.

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

       long start = System.currentTimeMillis();

        int inLoopSum = 0;
        for (int i = 1; i < 10000; i++) {
            //set a var to equal the sum of the proper divisors
            int tempSum = sumOfProperDivisors(whatAreProperDivisorsOf(i));
            //set a var to calc sum of proper divosor of prev number
            int divOfTempSum = sumOfProperDivisors(whatAreProperDivisorsOf(tempSum));
            //check if they are pairs
            if (areAmicablePair(i, tempSum, tempSum, divOfTempSum) == true) {
                int index = 0;
                //System.out.println("\ntest if true and output numbers");
                System.out.printf("num 1: %d num 2: %d sum1: %d sume2: %d", i, tempSum, tempSum, divOfTempSum);
                inLoopSum = inLoopSum + tempSum;
                System.out.println("\nsums : " + i);
            }
        }

        System.out.println("inLoopSum (Answer) ..." + inLoopSum +"\tTime: "+(System.currentTimeMillis()-start)+" ms");
    }
    /*
     * Method to take in an int and store all the proper divisors in an array
     * test purposes uses max int of 284
     */
    public static int[] whatAreProperDivisorsOf(int x) {
        //array to store the values
        int[] temp = new int[x / 2];
        int index = 0;

        for (int i = 1; i < x; i++) {
            if (x % i == 0) {
                temp[index] = i;  //-1 as starting at one but index 0
                index++;
            }
        }
        return temp;
    }

    //simply add the divisors
    public static int sumOfProperDivisors(int[] a) {
        int sum = 0;
        for (int i : a) {    //i gets each value of array
            sum += i;
        }
        return sum;
    }

    //compare the sums to see if amicable pair
    public static boolean areAmicablePair(int num1, int num2, int sum1, int sum2) {

        boolean test = false;
        //running total of all amicable pairs
        if (num1 != num2) {
            // int sum = 0;
            if ((num1 == sum2) && (num2 == sum1)) {
                //  sum = sum + num1;
                test = true;
            }
        }
        //   System.out.println("\nsum: "+sum);
        return test;
    }
}