java - 有效数字 - 适当除数的算术平均值不大于该数字的根

Question

这个任务看起来很简单 - 在输入时我得到测试数量 (numOfTests)，然后是两个数字 (downBorder, upBorder) 和我必须找出这些数字(downBorder、upBorder)之间有多少数字是有效数字，其中有效数字是适当除数(除了 1 和之外的所有除数)的算术平均值相同的数字)小于或等于该数字的平方根。我写了代码，可能它能工作，但它太慢了。我的代码:

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws java.lang.Exception
    {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); //faster than Scanner
        int numOfTests = Integer.parseInt(bf.readLine());
        for(int i = 0; i < numOfTests; i++)
        {
            String[] borders = bf.readLine().split(" ");
            long downBorder = Long.parseLong(borders[0]);
            long upBorder = Long.parseLong(borders[1]);
            //System.out.println(String.format("down:  %s, up: %s", downBorder, upBorder));
            System.out.println(countNumberOfSignificantNumbers(downBorder, upBorder));
        }
    }

    /**
     * print numbers of significant numbers - (arithmetic average of all divisors that is not bigger than root of that number)
     * e.g 4 is significant but 6 is not
     * @param downBorder
     * @param upBorder
     */
    private static int countNumberOfSignificantNumbers(Long downBorder, Long upBorder) {
        int numberOfSignificantNumbers = 0;
        for(Long i = downBorder; i <= upBorder; i++)
        {
            if(i%2 != 0)
                continue;
            else
            {
                double avgOfProperDivisors = getAvgArithOfSumOfNumberDividers(i);
                if(avgOfProperDivisors != 0 && avgOfProperDivisors <= Math.sqrt(i))
                    numberOfSignificantNumbers++;
            }
        }
        return numberOfSignificantNumbers;
    }

    /**
     * method returns the arithemtic average of all proper divisors (all divisors except one and number itself)
     * @param number
     * @return
     */
    public static double getAvgArithOfSumOfNumberDividers(Long number)
    {
        long maxD = number/2;
        long sum=0;
        long numOfDivs = 0;
        for(long i = 2; i <= maxD; i++)
        {
            if(number % i == 0)
            {
                numOfDivs++;
                sum += i;
            }
        }
        return (numOfDivs > 0) ? (double)sum/numOfDivs : 0;
    }
}

此任务的瓶颈在于计算除数的平均值。我怎样才能让它变得更好更快？

Example input:
2
4 6
1 3

Example output:
1
0

Answer 1

Significant Number 永远是质数的完美平方，例如 4、9、25、49、121 等。您需要检查的是 upBorder 之间有多少个质数的完美平方> 和 downBorder。