Question

解密短信时遇到问题。示例：

明文：“晕兄弟”

密文：“Å¾iÌ=ßOÌÅbO”

明文：“haフo`bメothナメ”

k1：33 - ＆gt;第一个键

k2：125 - ＆gt;第二个键

我使用ASCII可打印＆amp; ASCII扩展字符总共设置了224个字符。

这是我的代码：

public class Affine {

//encyption method
public static String enkripsi(String pesan, int k1, int k2){

    //change text message into array
    char[] chars = pesan.toCharArray();

    //getting ASCII code from each characters index
    int[] ascii = new int[chars.length];
    for (int i = 0; i < chars.length; i++) {
        ascii[i] = (int) chars[i];
    }

    //Affine encryption formula
    int[] c = new int[ascii.length];
    for (int j = 0; j < ascii.length; j++) {
        c[j] = ((k1*ascii[j])+k2) % 224 ;
    }

    //change the decimal (ASCII code) value back to characters
    char[] charen = new char[c.length];
    for (int i = 0; i < c.length; i++) {
        charen[i] = (char)c[i];
    }

    //change characters to String
    String pesan_en = String.valueOf(charen);
    return pesan_en;
}

 //decryption method
public static String dekripsi(String isipesanMasuk, int k1, int k2){
    int j,g;
    int[] c;
    int[] f = new int [224];

    //change text message into array    
    char[] chars = isipesanMasuk.toCharArray();

    //getting ASCII code from each characters index
    int[] ascii = new int[chars.length];
    for (int i = 0; i < chars.length; i++) {
        ascii[i] = (int) chars[i];
    }


    //getting inverse from encryption formula of Affine
    //example 33f = 1 (mod) 224 -> f = (1+(224 * j)) / 5
    //g = (33 * f) mod 224
    //if g = 1 then stop
    for (j = 1; j < 224; j++) {
            f[j] = (1 +(224*j)) / k1;
            g = (k1*f[j]) % 224 ;
        if (g==1) {
            break;
        }
    }

    //Affine decrypion formula      
    c = new int[ascii.length];
    for (int k = 0; k < ascii.length; k++) {
        c[k] = (f[j]*(ascii[k]-k2)) % 224 ;
    }

    //change the decimal (ASCII code) value back to characters          
    char[] charde = new char[c.length];
    for (int i = 0; i < c.length; i++) {
        charde[i] = (char)c[i];
    }   

    //change characters to String
    String pesan_de = String.valueOf(charde);
    return pesan_de;
}   
    }

Answer 1

如果ascii[k]-k2给出负值，则解密公式会失效。要解决这个问题，请使用：

c[k] = (f[j]*(ascii[k]-k2+224)) % 224;

其他一些评论：

你不需要一个数组来计算k1的倒数，一个简单的整数变量就可以了。

加密可能导致控制字符（\ u0000到\ u000f和\ u007f到\ u009f）可能无法在所有通道中不加改变地传输。

仿射密码解密过程错误的结果

1 个答案: