java实现RSA算法

package com.wllfengshu.test;

import java.math.BigInteger;
import java.util.Random;
import java.util.Scanner;

//RSA算法
public class TestRSA {
@SuppressWarnings("resource")
public static void main(String[] args) {
TestRSA rsa=new TestRSA();
BigInteger[] primeNumber = rsa.getPrimeNumber();
//step1:产生两个大素数
BigInteger p=primeNumber[0];
BigInteger q=primeNumber[1];
System.out.println("1、B产生的两个大素数是(保密)：p="+p+" q="+q);
//step2.1:计算n
BigInteger n=p.multiply(q);//n=p*q
System.out.println("2、计算的n是："+n);
//step2.2:计算sn
BigInteger sn=(p.subtract(new BigInteger("1"))).multiply(q.subtract(new BigInteger("1")));//sn=(p-1)*(q-1)
System.out.println("   计算的sn是："+sn);
//step3:随机选取e
BigInteger e=rsa.getE(sn);//0<e<sn && e和 sn互素
System.out.println("3、选取的e是："+e);
//step4:计算d
BigInteger d=rsa.getD(sn, e).mod(sn);//d同时与n和sn互素
System.out.println("4、计算的d是："+d);
//step5:B将n和e作为公钥公开
System.out.println("5、公钥：n="+n+" e="+e);
//step6:用户A获取到公钥
System.out.println("6、用户A已经获取到公钥了，可以开始发数据了...");
//step7:进行加密
System.out.println("7、请输入要发送的明文（仅数字）:");
Scanner scanner=new Scanner(System.in);
BigInteger m=new BigInteger(scanner.next());
BigInteger c=rsa.getC(m,e,n);//计算密文
System.out.println("加密后的密文是："+c);
//step8:进行解密
BigInteger mm=rsa.getDecrypt(c,n,d);//进行解密
System.out.println("8、解密后的的结果是："+mm);
}
/**
* 随机产生两个大素数: res[0]、res[1]
* @return
*/
public BigInteger[] getPrimeNumber(){
BigInteger p=null;
BigInteger q=null;
BigInteger[] res=new BigInteger[2];
Random random = new Random();
p=BigInteger.probablePrime(64, random);//为了效率，此处设为64
q=BigInteger.probablePrime(64, random);
res[0]=p;
res[1]=q;
return res;
}
/**
* 随机选取e
* 0<e<sn && e和 sn互素
* @param sn
* @return
*/
public BigInteger getE(BigInteger sn){
BigInteger e = null;
//说明:此处把产生的e位数-2，是防止 nextProbablePrime()方法产生的素数大于sn
int length = sn.toString().length()-2;// length为随机数位数
e=new BigInteger(sn.toString().subSequence(0, length-2).toString()).nextProbablePrime();
return e;
}
/**
* 选取d
* d同时与n和sn互素
* @param n
* @param sn
* @return
*/
public BigInteger getD(BigInteger sn,BigInteger e){//Euclid算法
BigInteger[] ret = new BigInteger[3];
BigInteger u = BigInteger.valueOf(1), u1 = BigInteger.valueOf(0);
BigInteger v = BigInteger.valueOf(0), v1 = BigInteger.valueOf(1);
if (e.compareTo(sn) > 0) {
BigInteger tem = sn;
sn = e;
e = tem;
}
while (e.compareTo(BigInteger.valueOf(0)) != 0) {
BigInteger tq = sn.divide(e); // tq = sn / e
BigInteger tu = u;
u = u1;
u1 = tu.subtract(tq.multiply(u1)); // u1 =tu - tq * u1
BigInteger tv = v;
v = v1;
v1 = tv.subtract(tq.multiply(v1)); // v1 = tv - tq * v1
BigInteger tsn = sn;
sn = e;
e = tsn.subtract(tq.multiply(e)); // e = tsn - tq * e
ret[0] = u;
ret[1] = v;
ret[2] = sn;
}
return ret[1];
}
/**
* 计算密文
* c=(m^e)%n
* @param m 明文
* @param e
* @param n
* @return
*/
public BigInteger getC(BigInteger m,BigInteger e,BigInteger n){
BigInteger c=null;
c=m.modPow(e, n);//返回其值为 (m^e  mod n)
return c;
}
/**
* 计算解密
* c^d=m%n m为解密后的结果
* @param c 密文
* @param n 公钥
* @param d 私钥
* @return
*/
public BigInteger getDecrypt(BigInteger c,BigInteger n,BigInteger d){
BigInteger m=null;
m=c.modPow(d, n);//返回其值为 (c^d  mod n)
return m;
}
}