Java中的仿射密码输出

Question

我在Java中用BlueJ编写的程序解密了由26个小写字母（a，b，c，...，x，y，z）组成的字符串密码文本已使用affine cipher加密。它使用强力搜索，尝试所有可能的密钥。

程序正常工作，但我输出有问题。

我应该输出＆＃34;加密密钥＆＃34;和＆＃34;解密方程＆＃34;在每次暴力搜索之后，我似乎无法弄清楚如何。

这是我目前所拥有的示例输出：

Decrypting the cipher text:
vwduwljudeehghyhubwklqjlfrxogilqgsohdvhuhwxuqdqbeoxhsulqwviruydxowdqgdodupghvljqedvhgrqzklfkedqnbrxghflghrqldpvhwwlqjxsvdihkrxvhfr

Plaintext:
uvctvkitcddgfgxgtavjkpikeqwnfhkpfrngcugtgvwtpcpadnwgrtkpvuhqtxcwnvcpfcnctofgukipdcugfqpyjkejdcpmaqwfgekfgqpkcougvvkpiwruchgjqwugeq

Hit 0 to continue search or 1 to stop:
0

Plaintext:
yhsphmupsbbctczcpahdmfumkoqntlmftxncsycpchqpfsfabnqcxpmfhylopzsqnhsftsnspwtcymufbsyctofidmkdbsfeaoqtckmtcofmswychhmfuqxyslcdoqycko

Hit 0 to continue search or 1 to stop:
0

...
... many lines later
...

Plaintext:
gfyhfqshyxxuvuduhafrqlsqwkenvtqlvjnuyguhufehlylaxneujhqlfgtkhdyenfylvynyhmvugqslxyguvklcrqwrxyloakevuwqvuklqymguffqlsejgyturkeguwk

Hit 0 to continue search or 1 to stop:
0

Plaintext:
startigrabbedeverythingicouldfindpleasereturnanyblueprintsforvaultandalarmdesignbasedonwhichbankyoudecideoniamsettingupsafehouseco

Hit 0 to continue search or 1 to stop:
1

这是应该显示的示例输出：

Decrypting the cipher text:
vwduwljudeehghyhubwklqjlfrxogilqgsohdvhuhwxuqdqbeoxhsulqwviruydxowdqgdodupghvljqedvhgrqzklfkedqnbrxghflghrqldpvhwwlqjxsvdihkrxvhfr

Encryption key: a = 1 b = 1
Decryption equation: x = 1 *(y - 1)

Plaintext:
uvctvkitcddgfgxgtavjkpikeqwnfhkpfrngcugtgvwtpcpadnwgrtkpvuhqtxcwnvcpfcnctofgukipdcugfqpyjkejdcpmaqwfgekfgqpkcougvvkpiwruchgjqwugeq

Hit enter to continue search or 'S' key to stop:
Encryption key: a = 3 b = 1
Decryption equation: x = 9 *(y - 1)

Plaintext:
yhsphmupsbbctczcpahdmfumkoqntlmftxncsycpchqpfsfabnqcxpmfhylopzsqnhsftsnspwtcymufbsyctofidmkdbsfeaoqtckmtcofmswychhmfuqxyslcdoqycko

Hit enter to continue search or 'S' key to stop:
Encryption key: a = 5 b = 1
Decryption equation: x = 21 *(y - 1)

...
... many lines later
...

Hit enter to continue search or 'S' key to stop:
Encryption key: a = 25 b = 2
Decryption equation: x = 25 *(y - 2)

Plaintext:
hgzigrtizyyvwvevibgsrmtrxlfowurmwkovzhvivgfimzmbyofvkirmghuliezfogzmwzozinwvhrtmyzhvwlmdsrxsyzmpblfwvxrwvlmrznhvggrmtfkhzuvslfhvxl

Hit enter to continue search or 'S' key to stop:
Encryption key: a = 1 b = 3
Decryption equation: x = 1 *(y - 3)

Plaintext:
startigrabbedeverythingicouldfindpleasereturnanyblueprintsforvaultandalarmdesignbasedonwhichbankyoudecideoniamsettingupsafehouseco

Hit enter to continue search or 'S' key to stop: S

The plain text message below was encrypted with a = 1 and b = 3
startigrabbedeverythingicouldfindpleasereturnanyblueprintsforvaultandalarmdesignbasedonwhichbankyoudecideoniamsettingupsafehouseco

这是我的代码：

import java.util.Scanner;
import java.math.BigInteger;

public class AffineCipher{
/**
* Prompts the user to enter a string of cipher text, along with the ablility
* to continue to search or to stop the search. The output or plaintext is 
* printed each time.
*/
public static void main(String[] args){
    int count = 1;
    int choice = 0;
    int change = 0;
    int stop = 1;
    int cont = 0;
    Scanner kbd = new Scanner (System.in);
    System.out.printf("Decrypting the cipher text:%n");
    String input = kbd.next();
    System.out.printf("\n");

    while (choice != stop){
        while (choice == cont){
            if (count == 13){
                count = count + 2;
            }
            String deciphered = decrypt(input,change, count);
            System.out.println("Plaintext:");
            System.out.println(deciphered);
            System.out.printf("\n");

            if (count == 25){
                change = 1;
                count = 1;
            }
            System.out.printf("Hit 0 to continue search or 1 to stop:%n");
            choice = kbd.nextInt();
            System.out.printf("\n");
            count = count + 2;
        }
    }
}

/**
* Deciphers the string cipher text.
*/
static String decrypt(String input,int change, int count){
    if (change == 0){
        int firstKey = 0 + count;
        int secondKey = 1;
        int mod = 26;
        StringBuilder builder = new StringBuilder();
        BigInteger inverse = BigInteger.valueOf(count).modInverse(BigInteger.valueOf(mod));

        for (int in = 0; in < input.length(); in++){
            char character = input.charAt(in);                
            if (Character.isLetter(character)){
                int decoded = inverse.intValue() * (character - 'a' - secondKey + mod);                    
                character = (char) (decoded % mod + 'a');
            }
            builder.append(character);
        }
        return builder.toString();
    } else{
        int firstKey = 1;
        int mod = 26;
        StringBuilder builder = new StringBuilder();
        BigInteger inverse = BigInteger.valueOf(firstKey).modInverse(BigInteger.valueOf(mod));

        for (int in = 0; in < input.length(); in++){
            char character = input.charAt(in);
            if (Character.isLetter(character)){
                int decoded = inverse.intValue() * (character - 'a' - count + mod);
                character = (char) (decoded % mod + 'a');
            }
            builder.append(character);
        }
        return builder.toString();
    }
}
}

任何帮助将不胜感激。谢谢。