Question

我正在尝试使用rsa加密当前日期和时间，并使用此密文输入另一个应用程序来验证用户。

这是代码。我是java的noobie所以我从互联网上复制了它并且对它做了一些改动。

计划 -

import java.math.BigInteger;
import java.security.SecureRandom;
import java.util.Scanner;
import java.util.Date;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Calendar;


public class RSA3
{
public BigInteger n, d, e;

private int bitlen = 16;



 /** Create an instance that can both encrypt and decrypt. */

 public RSA3(int bits) 
{
     bitlen = bits;
    SecureRandom r = new SecureRandom();
 Scanner in = new Scanner(System.in);
 String number1, number2;
 System.out.println("Enter first number");
 number1 = in.nextLine();

 System.out.println("Enter second number");
 number2 = in.nextLine();

 BigInteger p  = new BigInteger(number1);
 BigInteger q = new BigInteger(number2);
 //BigInteger p = new BigInteger(bitlen / 2,100,r);
    //BigInteger q = new BigInteger(bitlen / 2,100,r);
    n = p.multiply(q);

 System.out.println("\n n="+n);
    BigInteger m =  (p.subtract(BigInteger.ONE)).multiply(q.subtract(BigInteger.ONE));
  System.out.println("\n m="+m);
    e = new BigInteger("3");
    while (m.gcd(e).intValue() > 1) 
 {
         e = e.add(new BigInteger("2"));
    }
  System.out.println("\n e="+e);
   d = e.modInverse(m);
  System.out.println("\n d="+d);
 }




  /** Encrypt the given plaintext message. */

  public synchronized BigInteger encrypt(BigInteger message) 
  {System.out.println("\n 2nd enc");
     return message.modPow(e, n);

   }



  /** Decrypt the given ciphertext message. */
  public synchronized BigInteger decrypt(BigInteger message) 
  {System.out.println("\n2nd dec");
   return message.modPow(d, n);

  }




  /** Trivial test program. */
 public static void main(String[] args) 
 {
  RSA3 rsa = new RSA3(1024);
 DateFormat dateFormat = new SimpleDateFormat("yyyyMMddHH");
   //get current date time with Date()
  Date date = new Date();
   System.out.println(dateFormat.format(date));


  String text1 = dateFormat.format(date);
  System.out.println("Plaintext: " + text1);
  BigInteger plaintext = new BigInteger(text1.getBytes());

   BigInteger ciphertext = rsa.encrypt(plaintext);
  System.out.println("Ciphertext: " + ciphertext);
  plaintext = rsa.decrypt(ciphertext);

    String text2 = new String(plaintext.toByteArray());
   System.out.println("Plaintext: " + text2);
  }
  }

如果我为p＆amp; q来自用户，就像我在代码中所做的那样。密文有问题。但如果我选择BigInteger（bitlen / 2,100，r）;它工作正常。但每次密文生成都是新的随机数。我不想这样我想要p＆amp;我个人的选择

我想我已经搞砸了某个地方的biginteger。如果有任何其他方法我可以告诉我。

我在这里找不到逻辑错误请帮忙谢谢

rsa与biginteger接受p＆amp; q来自用户

0 个答案: