java - 反转模数运算符以进行解密

Question

在学校，我有一个作业要根据以下要求加密一个四位数的整数。

A company wants to transmit data over the telephone, but they are concerned that their phones may be tapped. All of their data is transmitted as four-digit integers. They have asked you to write a method that will encrypt their data so that it may be transmitted more securely. Your method should read a four-digit integer and encrypt it as follows: Replace each digit by (the sum of that digit plus 7) modulus 10. Then, swap the second digit with the fourth. Then print the encrypted integer. Write a separate method that inputs an en-crypted four-digit integer, and decrypts it to form the original number.

四位数的加密不是问题，我把它转换成一个字符串，然后转换成一个char数组，然后根据需要单独加密数字。

我做的方法是这样的:

public int encrypt4DigitNumber(int number) throws Exception {
    String numberAsString = String.valueOf(number);
    if (numberAsString.length() != 4) {
        throw new Exception("The digit has to be 4 digits long");
    }
    int[] numbers = new int[4];

    char[] numbersAsChars = numberAsString.toCharArray();
    for (int index = 0; index < numbersAsChars.length; index++) {
        int currentNumber = Character.getNumericValue(numbersAsChars[index]);
        int numberToReplace = (currentNumber + 7) % 10;

        numbers[index] = numberToReplace;
    }

    int second = numbers[1];
    int fourth = numbers[3];

    numbers[1] = fourth;
    numbers[3] = second;

    StringBuilder encryptedNumberStringBuilder = new StringBuilder();
    for (int encryptedNumber : numbers) {
        encryptedNumberStringBuilder.append(encryptedNumber);
    }

    String encryptedNumberString = encryptedNumberStringBuilder.toString();
    return Integer.parseInt(encryptedNumberString);
}

当我必须解密加密的四位数代码时，问题就来了。

我知道我必须交换 2 个数组元素，并将 7 添加到数组中的每个数字。

我不知道该怎么做的是反转模运算符，我唯一能想到的就是将当前数字乘以 10，但这行不通。

在做了一些研究之后，我必须在某个地方搜索模数的剩余部分，但我完全不知道该怎么做。我是否还需要返回加密函数中的那些并将它们传递给解密函数？

谁能解释一下取反模运算符的过程？

Answer 1

如果您使用偏移量 7 加密从 0 到 9(= mod 10)的数字，您当然可以在解密期间从每个数字中减去偏移量，如果数字为负则环绕，但这不是很好:

int numberToReplace = currentNumber - 7;
if (numberToReplace < 0) {
    numberToReplace += 10;
}

如果加密偏移量为 7，则解密偏移量将为 3 (10 - 7)。您可以将每个数字加 3 并应用模 10 来解密它们。

int numberToReplace = (currentNumber + 3) % 10;