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<!--
14// ==================================================================
15// JavaScript RSA算法简单实现
16// Know Bugs:
17// 1. Math.pow存在计算精度问题，因此不能选取较大的质数，否则会溢出
18// By midea0978
19// date:2005-08-10
20// 参考文献
21// 1.RSA算法研究 http://www.cnblogs.com/midea0978/articles/65244.html
22// 2.PKCS #1: RSA Cryptography Specifications Version 2.0
23// http://www.faqs.org/rfcs/rfc2437.html
24// ==================================================================
25RSAAlgorithm()
26
function RSAAlgorithm(){
27 //1.首先选择两个简单的质数p,q
28 var p=5
29 var q=13
30 var n=p*q
31 //随机选择余数r,保证r与(p-1)*(q-1)=48互质
32 var r=7
33 //得到private key是p,q,r
34 document.write("----------------------------------<br>");
35 document.write("<B>Private Key：</B><br>");
36 document.write("p= "+p+"<br>");
37 document.write("q= "+q+"<br>");
38 document.write("r= "+r+"<br>");
39 document.write("----------------------------------<br>");
40 //2.计算public key
41 //找到m,保证rm ==1 mod (p-1)*(q-1)
42 var num=0
43 for(a=0;a<500;a++){
44 if(r*a%((p-1)*(q-1))==1){
45 num=a;
46 break;
47
}
48 }
49 var m=num
50 if(m>0){
51 document.write("<br>找到m= "+num+" ,计算余数="+(num*r)%((p-1)*(q-1))+",成功！<br>")
52 }
53 else{
54 document.write("找不到满足条件的m,请调整参数");
55 }
56 document.write("----------------------------------<br>");
57 document.write("<B>Public Key：<br></B>");
58 document.write("m= "+m+"<br>");
59 document.write("n= "+n+"<br>");
60 document.write("----------------------------------<br>");
61 document.write("<br>");
62 document.write("<B>#演示加密过程#</B><br>");
63 var text=20 //明文
64 document.write("明文="+text+"<br>");
65 //接下来, 计算 etext == text^r mod n, (0 <= b < n)
66 var etext=Math.pow(text,r)%n;
67 document.write("加密后的内容:"+etext+"<br>");
68 document.write("<br>");
69 document.write("#演示解密过程#<br>");
70 //text=etext^m mod pq
71 var text=Math.pow(etext,m)%n;
72 document.write("解密后的内容:"+text+"<br>");
73}
74/*
75JAVA下面的简单实现
76import java.math.BigInteger;
77public class RSAAlgorithm {
78 public static void main(String[] args) {
79 BigInteger p = new BigInteger("5"); //P
80 BigInteger q = new BigInteger("13"); //Q
81 BigInteger n = new BigInteger("65"); //N
82 BigInteger r = new BigInteger("7");
83
84 BigInteger biginteger = new BigInteger("20");
85 BigInteger biginteger1 = biginteger.modPow(r, n);
86 System.out.println("加密数据:" + biginteger1);
87 BigInteger m = r.modInverse(new BigInteger("48"));
88 System.out.println("m:" + m);
89 BigInteger biginteger2 = biginteger1.modPow(m, n);
90 System.out.println("解密数据:" + biginteger2);
91 }
92}
93*/
94
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<!--14
// ==================================================================15
// JavaScript RSA算法简单实现16
// Know Bugs:17
// 1. Math.pow存在计算精度问题，因此不能选取较大的质数，否则会溢出18
// By midea097819
// date:2005-08-1020
// 参考文献21
// 1.RSA算法研究 http://www.cnblogs.com/midea0978/articles/65244.html22
// 2.PKCS #1: RSA Cryptography Specifications Version 2.023
// http://www.faqs.org/rfcs/rfc2437.html 24
// ==================================================================25
RSAAlgorithm()26
function RSAAlgorithm(){
27
//1.首先选择两个简单的质数p,q28
var p=529
var q=1330
var n=p*q 31
//随机选择余数r,保证r与(p-1)*(q-1)=48互质32
var r=7 33
//得到private key是p,q,r34
document.write("----------------------------------<br>");35
document.write("<B>Private Key：</B><br>");36
document.write("p= "+p+"<br>");37
document.write("q= "+q+"<br>");38
document.write("r= "+r+"<br>");39
document.write("----------------------------------<br>");40
//2.计算public key41
//找到m,保证rm ==1 mod (p-1)*(q-1)42
var num=043
for(a=0;a<500;a++){44
if(r*a%((p-1)*(q-1))==1){45
num=a;46
break;47
}48
}49
var m=num 50
if(m>0){51
document.write("<br>找到m= "+num+" ,计算余数="+(num*r)%((p-1)*(q-1))+",成功！<br>")52
}53
else{54
document.write("找不到满足条件的m,请调整参数");55
}56
document.write("----------------------------------<br>");57
document.write("<B>Public Key：<br></B>");58
document.write("m= "+m+"<br>");59
document.write("n= "+n+"<br>");60
document.write("----------------------------------<br>");61
document.write("<br>");62
document.write("<B>#演示加密过程#</B><br>");63
var text=20 //明文64
document.write("明文="+text+"<br>");65
//接下来, 计算 etext == text^r mod n, (0 <= b < n)66
var etext=Math.pow(text,r)%n;67
document.write("加密后的内容:"+etext+"<br>");68
document.write("<br>");69
document.write("#演示解密过程#<br>");70
//text=etext^m mod pq71
var text=Math.pow(etext,m)%n;72
document.write("解密后的内容:"+text+"<br>");73
}74
/*75
JAVA下面的简单实现76
import java.math.BigInteger;77
public class RSAAlgorithm {78
public static void main(String[] args) {79
BigInteger p = new BigInteger("5"); //P80
BigInteger q = new BigInteger("13"); //Q81
BigInteger n = new BigInteger("65"); //N82
BigInteger r = new BigInteger("7");83
84
BigInteger biginteger = new BigInteger("20");85
BigInteger biginteger1 = biginteger.modPow(r, n);86
System.out.println("加密数据:" + biginteger1);87
BigInteger m = r.modInverse(new BigInteger("48"));88
System.out.println("m:" + m);89
BigInteger biginteger2 = biginteger1.modPow(m, n);90
System.out.println("解密数据:" + biginteger2);91
}92
}93
*/94
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