Question

This is my another solution where the output is shown as expected

 public class Logic2 {
        public static void main(String[] args) {
            long sum = 0;
            calculation key = new calculation();
            sum = key.sum(3, 1000);
            System.out.print(sum);
        }
   }

class calculation {
    long total = 0;
    public long sum(int num, int limit) { //multples of num less than limit
        int number = Integer.valueOf(limit / num);
        if (limit % num == 0) {
            number -= 1;
        }

        total = (number / 2) * (2 * num + (number - 1) * num);
        return total;
    }
}

我自己编写了这段代码。似乎一切正常，但我没有得到所需的输出。为什么会这样？

Answer 1

您的数学似乎有点错误。尝试将其分解成较小的部分，以确保您得到了期望的结果。一个返回166833的工作示例

public static void main(String[] args) {
    int a = 3, N = 1000;
    System.out.println("Sum of multiples of " + a +
            " up to " + N + " = " +
            calculate_sum(a, N));

}

private static int calculate_sum(int a, int N) {
    // Number of multiples
    int m = N / a;

    // sum of first m natural numbers
    int sum = m * (m + 1) / 2;

    // sum of multiples
    return a * sum;
}

如果以同样的方式拆分方法，您将看到略微遗漏标记的位置。

Answer 2

public class Logic2 {
        public static void main(String[] args) {
            long sum = 0;
            calculation key = new calculation();
            sum = key.sum(3, 1000);
            System.out.print(sum);
        }
   }

class calculation {
    long total = 0;
    public long sum(int num, int limit) { //multples of num less than limit
        int number = Integer.valueOf(limit / num);
        if (limit % num == 0) {
            number -= 1;
        }

       total=((number)*(2*num+(number-1)*num))/2;
//previouslly total = (number / 2) * (2 * num + (number - 1) * num);
        return total;
    }
}

我自己发现了这个错误。在那，总计，如果我们写（number / 2），那么它将给出整数值，因此我没有得到所需的输出。无论如何，感谢大家在列表中查看我的帖子，对此我表示感谢。：）

查找小于指定数字的任何数字的倍数之和。我的错误在哪里？

2 个答案: