关于国内某知名大学网站加密算法的研究(页 1) - 原创专区 - 【3.A.S.T】网络安全爱好者 -- 网络安全，势在必行

柔肠寸断 发表于 2009-4-19 21:52

关于国内某知名大学网站加密算法的研究

原创作者：柔肠寸断 [3.A.S.T] [url]http://www.3ast.com.cm[/url]
原始出处：[url]http://www.3ast.com.cm[/url]
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看不懂的直接绕过bJvD(z"y8C
加密前为:hwy1234565OB_8J*C?N@
加密后为0F9C1C7A2EE76EBB5C1F4D7526CC4F883F4E;Los V*ve"~+B#a
v"t^*RT5bx;Mja
============================================'u5`p `{\:pH;P
上面是当时自己校内的心情。现在已经解破出来了，分享给大家,E!v({L6A;l
也就不隐瞒大家什么了，这个是国内某知名大学的后台加密算法8O0{,H(khR J\
以下是RSA算法文件{1P*d Y:uz$R

文件名RSA.ASPgCRS;]R@(@
===============================================+\~W%nu7C2Y(q
<%
Class clsRSA
Public PrivateKey :Ft)BLt0f
Public PublicKey "wS#hh3_c@)R
Public Modulus Pr(n8[m|B _&K
Public Function Crypt(pLngMessage, pLngKey) 2A-m#sq_O/JFZ
On Error Resume Next }$Z QS c0XD G
Dim lLngMod
Dim lLngResult
Dim lLngIndex
If pLngKey Mod 2 = 0 Then
lLngResult = 1 l)Kg6h&OpMnO
For lLngIndex = 1 To pLngKey / 2 y)t6nqSc/b
lLngMod = (pLngMessage ^ 2) Mod Modulus GmW^Y4q
' Mod may error on key generation A?lih5u
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function !|+gT1N3WwKY
Next mvqp;w6B5S_
Else 5]@)Z8Bhc
lLngResult = pLngMessage (kmVe nb
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus
On Error Resume Next
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function i4y'd;C$^L^
Next
End If PZC`Dx&hcC`J
Crypt = lLngResult
End Function?m)r`}

Public Function Encode(ByVal pStrMessage) "SC NbZ0rg,}.KP3?
Dim lLngIndex
Dim lLngMaxIndex |9rkTG.shDv
Dim lBytAscii C)~l\8l
Dim lLngEncrypted
lLngMaxIndex = Len(pStrMessage)
If lLngMaxIndex = 0 Then Exit Function )n%Y&gSe:M N y:i
For lLngIndex = 1 To lLngMaxIndex !i(xFq6Z;Jum
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1))
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4)
Next
End Function
Public Function Decode(ByVal pStrMessage)
Dim lBytAscii 1SD q/C2iX-k:C
Dim lLngIndex
Dim lLngMaxIndex N7P W uW{)GH
Dim lLngEncryptedData 'T|.Xy2C2y1l!fT
Decode = ""
lLngMaxIndex = Len(pStrMessage)
For lLngIndex = 1 To lLngMaxIndex Step 4
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4))
lBytAscii = Crypt(lLngEncryptedData, PrivateKey) @2Y.n.\b
Decode = Decode & Chr(lBytAscii)
Next !~ha#pS,x \)z3_
End Function DTnU WZ
Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength) dvuNzb-K.D-V
End Function +IPC)C.R [r*^
Private Function HexToNumber(ByRef pStrHex)
HexToNumber = CLng("&h" & pStrHex) 8LY _:m,ee.A i]
End Function
End Class &mm3WY8S5s.~'`
function Encryptstr(Message)
Dim LngKeyE )\7K9uC\r(a,?
Dim LngKeyD :T7t#fs_s$nY
Dim LngKeyN
Dim StrMessage ,L"s7v$[vm
Dim ObjRSA 5_p{(asL9T8TU
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893"
StrMessage = Message
Set ObjRSA = New clsRSA xYR4Lnp1ea
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage) i{o^_(a]3i
Set ObjRSA = Nothing +M%[K6b:sP5X
end function
Function Decryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN -J j Rh&Dz&xR
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893" O:KR+~:j+H9b1U;\x
StrMessage = Message
Set ObjRSA = New clsRSA B-ay VCm}
ObjRSA.PrivateKey =LngKeyD s.I1R?;\&?"T Yk
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage) ,gg bof
Set ObjRSA = Nothing
end function -~(Olf1C!d$u
%>
===============================================7C { ] yPWRQ

还有一个用于测试这段代码的test.asp;@Q'C.A1LF
有兴趣的自己搭建个IIS测试下 ADW!O0cb

<%/c)O"A U&_I(N
function Encryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN ,m%uk}lR#x
Dim StrMessage 8n3X"Jr3{
Dim ObjRSA 7V&F2CY |fD
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893"
StrMessage = Message c{/ai3m&b$U
Set ObjRSA = New clsRSA Ee7@-J;t6KYQWu
ObjRSA.PublicKey = LngKeyE !]G LI(\2` B}0n
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage) .IM l3QG&c9K"I
Set ObjRSA = Nothing
end function
function decryptstr(Message) x]F)R?F}\G
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA +C/|T3Y,C9wN T_
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893"
StrMessage = Message ,eMX3[Cs-x[-rP
Set ObjRSA = New clsRSA TL/`iHRs!I
ObjRSA.PrivateKey =LngKeyD J? R&rB
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing m/\-gR t\O#e;ZF
end function
dim last,first
first="!@#$%^&*()"
Response.Write "加密前为:"&first $MwBs%_ZSGR
last=Encryptstr(first) +W3io6{G J
Response.Write "加密后为"&last
Response.Write "解密后为" &decryptstr(last) 5R.qMmZ-YD
%> ==============================================
剩下的就是字符的对照表了
===================字符集================
1_____6EBB.pP3k2n-{DRu5T4M
2_____5C1FF'Qs{6_8R+H4gV0D4H
3_____4D75
4_____26CCuF9M0kq?B
5_____4F88
6_____3F4E
7_____0A9D,Q(Yik/k4Nq)SL0R'`
8_____1A1Cd}_1~X|@
9_____6D20
0_____1089
a_____0F3E@w7E[CrJp
b_____3159
c_____3517MPOUH,oT
d_____419C
e_____615C
f_____556F&|;[zEpk`%o
g_____2B7F
h_____0F9C&WF#LH4|V m+z
i_____00FA1PD*I'c7z-Q]
j_____5A50V^{Qw1F,Z
k_____2850(V3J#c(jw] w
l_____3E7B
m_____71C5
n_____1FC8
o_____74C1
p_____5FB8
q_____6085;Y"b(GY-c,fy{
r_____3AC4
s_____2F50
t_____36F85@M_Z8uG
u_____7010
v_____0B42|#p5g ^&lW%M
w_____1C7A
x_____16F87\tL,CA_c
y_____2EE7b^aJf6?nn3}8M0lK
z_____5CF3
!_____6233
@_____3A45
#_____2291w7z-FY0H1]
$_____5D5C
%_____09B9
^_____43EA4NK7he"JP+D1b)`
&_____62B9
*_____6301
(_____4659J1[jrio!l
)_____5C82

鱼儿无心 发表于 2011-12-4 17:17

收藏了。。。。。。。。

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关于国内某知名大学网站加密算法的研究