Question

这就是我所拥有的。它正在解决问题，但要永远。我可以将最后一个循环0到28123分成两半并以某种方式同时运行它们以使其更快，然后在最后添加两个总和以获得最终结果吗？将“线程”帮助吗？我该怎么做才能使代码更快地解决？

/*
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
 */

public class power {

    public static void main(String[] args){
        System.out.println(bigSum());
    }

    public static boolean isPerfect(int a){
        boolean perfect = false;

        int sum = 0;

        for(int i = 1; i<(a/2)+1; i++)
        {
            if (a%i == 0)
            {
                sum = sum + i;
            }
        }

        if(a == sum){
            perfect = true;
        }

        return perfect;
    }

    public static boolean isAbundant(int a){
        boolean Abundant = false;

        int sum = 0;

        for(int i = 1; i<(a/2)+1; i++)
        {
            if (a%i == 0)
            {
                sum = sum + i;
            }
        }

        if(a < sum){
            Abundant = true;
        }

        return Abundant;
    }

    public static boolean isDeficient(int a){
        boolean Deficient = false;

        int sum = 0;

        for(int i = 1; i<(a/2)+1; i++)
        {
            if (a%i == 0)
            {
                sum = sum + i;
            }
        }

        if(a > sum){
            Deficient = true;
        }

        return Deficient;
    }

    public static boolean isSumOfTwoAbundant(int a){
        boolean SumOfTwoAbundant = false;

        for(int i = 1; i<a; i++){
            if(isAbundant(i) && isAbundant(a-i)){
                SumOfTwoAbundant = true;
            }
        }
        return SumOfTwoAbundant;
    }

    public static long bigSum(){
        int sum = 0;

        for(int i = 0; i<28123; i++){
            if(!isSumOfTwoAbundant(i)){
                sum = sum + i;
                System.out.println("i: " + i + "; " + "Sum: " + sum);
            }
        }
        return sum;
    }
}

Answer 1

你是否重新计算是否每个数字＆lt; a每次调用isSumOfTwoAbundant都很丰富。尝试保留一个丰富的数字列表，并在找到它时添加。然后你可以遍历该列表而不是重新检查数字的丰度＆lt;一个。类似的东西：

public static boolean isSumOfTwoAbundant(int a){
boolean SumOfTwoAbundant = false;
    if(isAbundant(a))
    {
        abundants.add(a);
    }

    for(int i = 0; i<abundants.length; i++) {
        for(int j = 0; j < abundants.length; j++) {
            if(a - abundants[i] == abundants[j]){
                SumOfTwoAbundant = true;
            }
        }
    }
    return SumOfTwoAbundant ;
}

private ArrayList<int> abundants;

还有很多其他方法可以让它变得更好，但欧拉计划是通过经验来学习这些。

如何更快地制作ProjectEuler 23解决方案

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