Question

因为我是初学者，请尽可能简单地解释。接收整数n并返回大于的最小整数 n，其中数字之和可被11整除。例如，nextCRC（100）= 119，因为119是第一个大于的整数 100及其数字1 + 1 + 9 = 11的总和可被11整除。

1）我无法理解的第一件事是，为什么在开始时for循环中存在“true”。

2）如何计算大于11且可被11整除的下一个数字。

3）我如何对负数，数字＆lt; 0。

public static int nextCRC(int n)
{
    try
    {
        **for (int i = n + 1; true; i++)**
        {
            String number = String.valueOf(Math.abs(i));

            int count = 0;
            for (int j = 0; j < number.length(); j++)
            {
                count += Integer.parseInt(String.valueOf(number.charAt(j)));
            }

            if (count > 0 && count % 11 == 0) return  i;
        }
    }
    catch (Exception e)
    {
        return 0;
    }
}

public static void main(String[] args)
{
    System.out.println(nextCRC(100));
    System.out.println(nextCRC(-100));
}

}

Answer 1

我在代码中添加了注释，解释了代码的每个部分的作用：

public static int nextCRC(int n)
{
try
{
    for (int i = n + 1; true; i++) // the for loop will loop as long as the boolean is true. i starts at the inputted value and increments by 1 each iteration
//Since the value is "true" it will loop forever or until a value is returned
    {
        String number = String.valueOf(Math.abs(i)); // convert the number to string

        int count = 0;
        for (int j = 0; j < number.length(); j++) //iterate through each digit in the number
        {
            count += Integer.parseInt(String.valueOf(number.charAt(j))); // add the value of each digit to cout variable
        }

        if (count > 0 && count % 11 == 0) return  i; //return the number that was checked if the sum of its digits is divisible by 11
    }
}
catch (Exception e) // return a value of 0 if there is an exception
{
    return 0;
}
}

public static void main(String[] args)
{
System.out.println(nextCRC(100));
System.out.println(nextCRC(-100));
}
}

需要帮助为循环

1 个答案: