解密程序中的奇怪错误

时间:2011-02-02 04:47:42

标签: java encryption shift

我一直在空闲时间写这个加密算法几天,我以为我终于让它工作了,但是当我对某些字符进行处理时它就开始出现故障了。我有这个设置用循环键替换字符的转换。问题是在翻译后的一个字符后会被删除。 解密代码如下:

import java.util.Scanner;
import java.io.*;
/* File CycleDeCipher.java*/

public class CycleDeCipher
{
    public static void main(String[] args)
    {
            new CycleDeCipher();
    }
    public CycleDeCipher()
    {
            String plainTxt;
            Scanner in = new Scanner(System.in);
            System.out.println("This program decrypts My Cyclical Substitution Algorithm. v0.2");
            System.out.println("Enter a multi digit number : ");
            Long mainKey = new Long(in.nextLong());;
            System.out.print("Enter your Cipher Text message :");
            in.nextLine();
            plainTxt = new String(in.next());
            in.nextLine();
            int[] keys = longParser(mainKey);
            String cipherTxt="";
            int j = 0;
            while(j < plainTxt.length())
            {
                    cipherTxt+=decryptCharacter(plainTxt.charAt(j),keys[j%4]);
                    j++;
                    System.out.println("char number " + j + " successfully translated!");
            }
            System.out.println("Your text is translated to :"+cipherTxt.toUpperCase());
    }   
    private String decryptCharacter(Character ch, int key)
    {
        System.out.println("Decrypting character "+ch.toString() + " with key "+key);
        if(Character.isLetter(ch)){
             ch = (char) ((int) Character.toLowerCase(ch) - key%10);
        }
        else {
            ch = (char) ((int) ch-key%10);
        }
        return(ch.toString());
    }
    public int[] longParser(Long key)
    {
        System.out.println("Parsing long to crypto keys...");
        int i = 0;
        int[] result;
        String sInput = new String(key.toString());
        char[] keys = new char[sInput.length()];
        for(i = 0; i < sInput.length(); i++)
        {
            keys[i] = sInput.charAt(i);
        }
        i = 0;
        result = new int[sInput.length()];
        for(i=0; i<keys.length; i++)
        {
            result[i] = (int) keys[i];
        }
        return result;
    }
}

import java.util.Scanner; import java.io.*; /* File CycleDeCipher.java*/ public class CycleDeCipher { public static void main(String[] args) { new CycleDeCipher(); } public CycleDeCipher() { String plainTxt; Scanner in = new Scanner(System.in); System.out.println("This program decrypts My Cyclical Substitution Algorithm. v0.2"); System.out.println("Enter a multi digit number : "); Long mainKey = new Long(in.nextLong());; System.out.print("Enter your Cipher Text message :"); in.nextLine(); plainTxt = new String(in.next()); in.nextLine(); int[] keys = longParser(mainKey); String cipherTxt=""; int j = 0; while(j < plainTxt.length()) { cipherTxt+=decryptCharacter(plainTxt.charAt(j),keys[j%4]); j++; System.out.println("char number " + j + " successfully translated!"); } System.out.println("Your text is translated to :"+cipherTxt.toUpperCase()); } private String decryptCharacter(Character ch, int key) { System.out.println("Decrypting character "+ch.toString() + " with key "+key); if(Character.isLetter(ch)){ ch = (char) ((int) Character.toLowerCase(ch) - key%10); } else { ch = (char) ((int) ch-key%10); } return(ch.toString()); } public int[] longParser(Long key) { System.out.println("Parsing long to crypto keys..."); int i = 0; int[] result; String sInput = new String(key.toString()); char[] keys = new char[sInput.length()]; for(i = 0; i < sInput.length(); i++) { keys[i] = sInput.charAt(i); } i = 0; result = new int[sInput.length()]; for(i=0; i<keys.length; i++) { result[i] = (int) keys[i]; } return result; } }
我不想这样做!“

我只是想知道是否有人可以修复代码,因此它不会放弃这些答案。

1 个答案:

答案 0 :(得分:0)

问题在于输入处理,而不是算法。默认情况下,java.util.Scanner在空白字符(包括输入字符串的第二个字符的空格)上分隔标记。因此,对in.next()的调用返回一个带有单个字符('R')的字符串,然后处理该字符串并返回单个输出字符。

解决这个问题的一个简单方法是使用Scanner.nextLine()而不是next来获取输入文本,这将获得该行上的所有字符(包括空格):

System.out.print("Enter your Cipher Text message :");
in.nextLine();
plainTxt = new String(in.nextLine());