1. cipher text

bmjs dtz uqfd ymj lfrj tk ymwtsjx dtz bns tw dtz inj ymjwj nx st rniiqj lwtzsi

刚看到需要解密的文本，就觉得它很像移位密码，于是开始试了一下。 源代码： 求最大公因数：Arithmetic.java

public class Arithmetic { int x = 0; int y = 0; public int euclid(int a,int b){ int first,second; first = a; second = b; int temp; if(first<second){ temp = first; first = second; second = temp; } while(first%second!=0){ temp = first%second; first = second; second = temp; } return second; } }

移位密码的解密：Affine.java

class Affine { String deciphering(String s, int a, int b){// 解密的实现 char[] ch = s.toCharArray(); int length = ch.length;// 密文长度 int[] in = new int[length]; for (int i = 0; i < ch.length; i++) { if(ch[i] == ' '){ //如果是空格就不用解密，直接跳过 } else { in[i] = ch[i] - 97;// 利用ascii变成0-25数字 in[i] = ((in[i] - b) * a) % 26; // 解密 if (in[i] < 0) { in[i] += 26; } ch[i] = (char) (in[i] + 97);// 将数字变成字母 } } return String.valueOf(ch);// 将字符串数字变成String类型的字符串，返回 } }

主类:Test.java

public class Test { public static void main(String[] args) { Arithmetic arithmetic = new Arithmetic(); final int MOD = 26; int [] gcd = new int[12]; int m = 0; String out = null; for(int i=1;i<MOD;i++){ if((arithmetic.euclid(i,MOD)) == 1) { //求与26互素的数 gcd[m] = (arithmetic.euclid_2(i,MOD)+26)%26; //求这些数mod26的逆,并把它加入gcd数组 m++; } } Scanner input = new Scanner(System.in); System.out.println("请输入需要解密的密文："); String s = input.nextLine();// 输入密文 Affine affine = new Affine(); int k =1; for(int i=0;i<12;i++) { for (int j = 0; j < 26; j++) { out = affine.deciphering(s, gcd[i], j); System.out.println("第"+k+"条明文为："+out); k++; } } } }

实验结果截图：

最后成功破解了由仿射密码加密的密文，得到了明文when you play the game of thrones you win or you die there is no middle ground。