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++;}}}
}