Private Sub Command3_Click() Dim ni%, m1%, m%, n%, r% n1 = Text1.Text m1 = Text2.Text If m1 n1 Then ' mn m = m1: n = n1 Else m = n1: n = m1 End If Do r = m Mod n If r = 0 Then Exit Do m = n n = r Loop Text3.Text = n Text4.Text = m1 * n1 / n End Sub
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我覺得你的并不是RSA加密解密算法。
在點(diǎn)虐 的有一個(gè)System.Security.Cryptography的命名空間，里面有一RSACryptoServiceProvider的類用來對(duì)byte進(jìn)行RSA加密解密。
具體例子如下：
using System;
using System.Security.Cryptography;
using System.Text;
class RSACSPSample
{
static void Main()
{
try
{
//Create a UnicodeEncoder to convert between byte array and string.
UnicodeEncoding ByteConverter = new UnicodeEncoding();
//Create byte arrays to hold original, encrypted, and decrypted data.
byte[] dataToEncrypt = ByteConverter.GetBytes("Data to Encrypt");
byte[] encryptedData;
byte[] decryptedData;
//Create a new instance of RSACryptoServiceProvider to generate
//public and private key data.
RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
//Pass the data to ENCRYPT, the public key information
//(using RSACryptoServiceProvider.ExportParameters(false),
//and a boolean flag specifying no OAEP padding.
encryptedData = RSAEncrypt(dataToEncrypt,RSA.ExportParameters(false), false);
//Pass the data to DECRYPT, the private key information
//(using RSACryptoServiceProvider.ExportParameters(true),
//and a boolean flag specifying no OAEP padding.
decryptedData = RSADecrypt(encryptedData,RSA.ExportParameters(true), false);
//Display the decrypted plaintext to the console.
Console.WriteLine("Decrypted plaintext: {0}", ByteConverter.GetString(decryptedData));
}
catch(ArgumentNullException)
{
//Catch this exception in case the encryption did
//not succeed.
Console.WriteLine("Encryption failed.");
}
}
static public byte[] RSAEncrypt(byte[] DataToEncrypt, RSAParameters RSAKeyInfo, bool DoOAEPPadding)
{
try
{
//Create a new instance of RSACryptoServiceProvider.
RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
//Import the RSA Key information. This only needs
//toinclude the public key information.
RSA.ImportParameters(RSAKeyInfo);
//Encrypt the passed byte array and specify OAEP padding.
//OAEP padding is only available on Microsoft Windows XP or
//later.
return RSA.Encrypt(DataToEncrypt, DoOAEPPadding);
}
//Catch and display a CryptographicException
//to the console.
catch(CryptographicException e)
{
Console.WriteLine(e.Message);
return null;
}
}
static public byte[] RSADecrypt(byte[] DataToDecrypt, RSAParameters RSAKeyInfo,bool DoOAEPPadding)
{
try
{
//Create a new instance of RSACryptoServiceProvider.
RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
//Import the RSA Key information. This needs
//to include the private key information.
RSA.ImportParameters(RSAKeyInfo);
//Decrypt the passed byte array and specify OAEP padding.
//OAEP padding is only available on Microsoft Windows XP or
//later.
return RSA.Decrypt(DataToDecrypt, DoOAEPPadding);
}
//Catch and display a CryptographicException
//to the console.
catch(CryptographicException e)
{
Console.WriteLine(e.ToString());
return null;
}
}
}
[Visual Basic]
Try
'Create a new RSACryptoServiceProvider object.
Dim RSA As New RSACryptoServiceProvider()
'Export the key information to an RSAParameters object.
'Pass false to export the public key information or pass
'true to export public and private key information.
Dim RSAParams As RSAParameters = RSA.ExportParameters(False)
Catch e As CryptographicException
'Catch this exception in case the encryption did
'not succeed.
Console.WriteLine(e.Message)
End Try
[C#]
try
{
//Create a new RSACryptoServiceProvider object.
RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
//Export the key information to an RSAParameters object.
//Pass false to export the public key information or pass
//true to export public and private key information.
RSAParameters RSAParams = RSA.ExportParameters(false);
}
catch(CryptographicException e)
{
//Catch this exception in case the encryption did
//not succeed.
Console.WriteLine(e.Message);
}
：將方程x^4-px^3+q=0移項(xiàng)，得
x^4+q=px^3
可見，x^4≥0，則x^4+q0，所以px^30，即x0，本題也就是要求出使方程x^4-px^3+q=0有正整數(shù)解的素?cái)?shù)p、q；
且素?cái)?shù)p必定是奇素?cái)?shù)，否則是偶素?cái)?shù)的話，那么p=2，則方程成為：x^4+q=2x^3，即q=2x^3-x^4=x^3×(2-x)0，得出2-x0，即x2，則只能是x=1，代入方程：1^4+q=2×1^3，即1+q=2，解得q=1，不是素?cái)?shù)，故p必定是奇素?cái)?shù)。
分兩種情形討論：
情形一：當(dāng)x為偶數(shù)時(shí)，設(shè)為x=2n，則有
(2n)^4+q=p×(2n)^3
16n^4+q=p×8n^3
上式右端是偶數(shù)，則左端的q必須為偶數(shù)，否則：左端奇偶相加得奇，不符。
而q作為素?cái)?shù)，唯一的偶素?cái)?shù)就是2，即q=2，則上式成為
16n^4+2=p×8n^3
兩邊同時(shí)除以2，得：8n^4+1=p×4n^3，顯然，左端奇偶相加得奇，但右端為偶，矛盾。所以方程無偶整數(shù)解；
情形二：當(dāng)x為奇數(shù)時(shí)，設(shè)為x=2n-1，則有
(2n-1)^4+q=p×(2n-1)^3
觀察上式，右端為奇，則左端也必須為奇，而(2n-1)^4是奇，所以得出q必須為偶，故素?cái)?shù)q=2，上式成為：
(2n-1)^4+2=p×(2n-1)^3，整理成：
p(2n-1)^3-(2n-1)^4=(2n-1)^3×[p-(2n-1)]=1×2
由于(2n-1)^3為奇，所以必有：(2n-1)^3=1，解得：n=1；
則：[p-(2n-1)]=2，解得：p=3；
綜上，對(duì)于素?cái)?shù)p、q，方程x^4-px^3+q=0有整數(shù)解,則p、q分別為3和2。
暈，再補(bǔ)一個(gè)!
1：編程實(shí)現(xiàn) 自動(dòng)生成兩個(gè)素?cái)?shù)p,q
2：編程實(shí)現(xiàn) 計(jì)算n=p*q
f(n)=(p-1)(q-1)
3: 隨機(jī)數(shù)e滿足:0ef(n) and f(n)與e互為素?cái)?shù)
4: 編程實(shí)現(xiàn) 計(jì)算d:d=e mod f(n)=1
如要加密m
5:編程實(shí)現(xiàn)計(jì)算c:
加密公式：c=m^e mod n
6:編程實(shí)現(xiàn)解密m:
解密公式：m=c^d mod n
Private Sub Command10_Click()
End
End Sub
Private Sub Command9_Click()
Text1.Text = ""
Text2.Text = ""
Text3.Text = ""
Text4.Text = ""
Text5.Text = ""
Text6.Text = ""
Text7.Text = ""
Text8.Text = ""
Text9.Text = ""
End Sub
Private Sub Command2_Click()
p = Val(Text1.Text)
q = Val(Text2.Text)
Text3.Text = Str$(p * q)
End Sub
Private Sub Command3_Click()
p = Val(Text1.Text)
q = Val(Text2.Text)
Text4.Text = Str$(p - 1) * (q - 1)
End Sub
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