JavaScript RSA演算法簡單實現)

最後更新：2018-12-05 來源：互聯網

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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<TITLE> New Document </TITLE>
<META NAME="Generator" CONTENT="EditPlus">
<META NAME="Author" CONTENT="">
<META NAME="Keywords" CONTENT="">
<META NAME="Description" CONTENT="">
</HEAD>
<BODY>
<SCRIPT LANGUAGE="JavaScript">
<!--
// ==================================================================
// JavaScript RSA演算法簡單實現
// Know Bugs:
// 1. Math.pow存在計算精度問題，因此不能選取較大的質數，否則會溢出
// By midea0978
// date:2005-08-10
// 參考文獻
// 1.RSA演算法研究 http://www.cnblogs.com/midea0978/articles/65244.html
// 2.PKCS #1: RSA Cryptography Specifications Version 2.0
// http://www.faqs.org/rfcs/rfc2437.html
// ==================================================================
RSAAlgorithm()
function RSAAlgorithm(){
//1.首先選擇兩個簡單的質數p,q
var p=5
var q=13
var n=p*q
//隨機播放餘數r,保證r與(p-1)*(q-1)=48互質
var r=7
//得到private key是p,q,r
document.write("---------------------------------- ");
document.write("Private Key： ");
document.write("p= "+p+" ");
document.write("q= "+q+" ");
document.write("r= "+r+" ");
document.write("---------------------------------- ");
//2.計算public key
//找到m,保證rm ==1 mod (p-1)*(q-1)
var num=0
for(a=0;a<500;a++){
if(r*a%((p-1)*(q-1))==1){
num=a;
break;
}
}
var m=num
if(m>0){
document.write(" 找到m= "+num+" ,計算餘數="+(num*r)%((p-1)*(q-1))+",成功！ ")
}
else{
document.write("找不到滿足條件的m,請調整參數");
}
document.write("---------------------------------- ");
document.write("Public Key： ");
document.write("m= "+m+" ");
document.write("n= "+n+" ");
document.write("---------------------------------- ");
document.write(" ");
document.write("#示範加密過程# ");
var text=20 //明文
document.write("明文="+text+" ");
//接下來, 計算 etext == text^r mod n, (0 <= b < n)
var etext=Math.pow(text,r)%n;
document.write("加密後的內容:"+etext+" ");
document.write(" ");
document.write("#示範解密過程# ");
//text=etext^m mod pq
var text=Math.pow(etext,m)%n;
document.write("解密後的內容:"+text+" ");
}
/**//*
JAVA下面的簡單實現
import java.math.BigInteger;
public class RSAAlgorithm {
public static void main(String[] args) {
BigInteger p = new BigInteger("5"); //P
BigInteger q = new BigInteger("13"); //Q
BigInteger n = new BigInteger("65"); //N
BigInteger r = new BigInteger("7");
BigInteger biginteger = new BigInteger("20");
BigInteger bigintegerbiginteger1 = biginteger.modPow(r, n);
System.out.println("加密資料:" + biginteger1);
BigInteger m = r.modInverse(new BigInteger("48"));
System.out.println("m:" + m);
BigInteger biginteger2 = biginteger1.modPow(m, n);
System.out.println("解密資料:" + biginteger2);
}
}
*/
//-->
</SCRIPT>
</BODY>
</HTML>

本文章原先以中文撰寫並發佈於 aliyun.com，亦設英文版本，僅作資訊用途。本網站不對文章的準確性，完整性或可靠性或其任何翻譯作出任何明示或暗示的陳述或保證。如對該文章有任何疑慮或投訴，請傳送電郵至 info-contact@alibabacloud.com 並提供相關疑慮或投訴的詳細說明。職員會於 5 個工作天內與您聯絡，一經驗證之後，即會刪除該侵權內容。

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