我在java中實現RSA我遇到了一個代碼,下面給出了它在解密明文之後以數字形式顯示明文,但我希望它以簡單的英文輸入。我不想使用java api。使用java的RSA實現
TestRsa.Java
import java.io.IOException;
import java.math.BigInteger;
import java.util.Random;
public class TestRsa {
private BigInteger p, q;
private BigInteger n;
private BigInteger PhiN;
private BigInteger e, d;
public TestRsa() {
initialize();
}
public void initialize() {
int SIZE = 512;
/* Step 1: Select two large prime numbers. Say p and q. */
p = new BigInteger(SIZE, 15, new Random());
q = new BigInteger(SIZE, 15, new Random());
/* Step 2: Calculate n = p.q */
n = p.multiply(q);
/* Step 3: Calculate ø(n) = (p - 1).(q - 1) */
PhiN = p.subtract(BigInteger.valueOf(1));
PhiN = PhiN.multiply(q.subtract(BigInteger.valueOf(1)));
/* Step 4: Find e such that gcd(e, ø(n)) = 1 ; 1 < e < ø(n) */
do {
e = new BigInteger(2 * SIZE, new Random());
} while ((e.compareTo(PhiN) != 1)
|| (e.gcd(PhiN).compareTo(BigInteger.valueOf(1)) != 0));
/* Step 5: Calculate d such that e.d = 1 (mod ø(n)) */
d = e.modInverse(PhiN);
}
public BigInteger encrypt(BigInteger plaintext) {
return plaintext.modPow(e, n);
}
public BigInteger decrypt(BigInteger ciphertext) {
return ciphertext.modPow(d, n);
}
public static void main(String[] args) throws IOException {
TestRsa app = new TestRsa();
int plaintext;
System.out.println("Enter any character : ");
plaintext = System.in.read();
BigInteger bplaintext, bciphertext;
bplaintext = BigInteger.valueOf((long) plaintext);
bciphertext = app.encrypt(bplaintext);
System.out.println("Plaintext : " + bplaintext.toString());
System.out.println("Ciphertext : " + bciphertext.toString());
bplaintext = app.decrypt(bciphertext);
System.out.println("After Decryption Plaintext : "
+ bplaintext.toString());
}
}
-1張貼代碼,超過50%的空行。如果您希望人們閱讀您的代碼,則由您來決定是否可讀。 – EJP 2014-07-31 06:18:06