基于私钥加密公钥解密的RSA算法C#实现方法

/* 
 功能：用指定的私钥(n,d)加密指定字符串source 
*/ 
private string EncryptString(string source, BigInteger d, BigInteger n) 
{ 
 int len = source.Length; 
 int len1 = 0; 
 int blockLen = 0; 
 if ((len % 128) == 0) 
 len1 = len / 128; 
 else 
 len1 = len / 128 + 1; 
 string block = ""; 
 string temp = ""; 
 for (int i = 0; i < len1; i++) 
 { 
 if (len >= 128) 
  blockLen = 128; 
 else 
  blockLen = len; 
 block = source.Substring(i * 128, blockLen); 
 byte[] oText = System.Text.Encoding.Default.GetBytes(block); 
 BigInteger biText = new BigInteger(oText); 
 BigInteger biEnText = biText.modPow(d, n); 
 string temp1 = biEnText.ToHexString(); 
 temp += temp1; 
 len -= blockLen; 
 } 
 return temp; 
} 
/* 
 功能：用指定的公钥(n,e)解密指定字符串source 
*/ 
private string DecryptString(string source, BigInteger e, BigInteger n) 
{ 
 int len = source.Length; 
 int len1 = 0; 
 int blockLen = 0; 
 if ((len % 256) == 0) 
 len1 = len / 256; 
 else 
 len1 = len / 256 + 1; 
 string block = ""; 
 string temp = ""; 
 for (int i = 0; i < len1; i++) 
 { 
 if (len >= 256) 
  blockLen = 256; 
 else 
  blockLen = len; 
 block = source.Substring(i * 256, blockLen); 
 BigInteger biText = new BigInteger(block, 16); 
 BigInteger biEnText = biText.modPow(e, n); 
 string temp1 = System.Text.Encoding.Default.GetString(biEnText.getBytes()); 
 temp += temp1; 
 len -= blockLen; 
 } 
 return temp; 
} 

/* 
 加密过程,其中d、n是RSACryptoServiceProvider生成的D、Modulus 
*/ 
private string EncryptProcess(string source, string d, string n) 
{ 
 byte[] N = Convert.FromBase64String(n); 
 byte[] D = Convert.FromBase64String(d); 
 BigInteger biN = new BigInteger(N); 
 BigInteger biD = new BigInteger(D); 
 return EncryptString(source, biD, biN); 
} 

/* 
 解密过程,其中e、n是RSACryptoServiceProvider生成的Exponent、Modulus 
*/ 
private string DecryptProcess(string source, string e, string n) 
{ 
 byte[] N = Convert.FromBase64String(n); 
 byte[] E = Convert.FromBase64String(e); 
 BigInteger biN = new BigInteger(N); 
 BigInteger biE = new BigInteger(E); 
 return DecryptString(source, biE, biN); 
}