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]z }8_'T|
原始出处：[url]http://www.3ast.com.cm[/url]!Y(d&f4l)A z,Doc0y8M1u-t
+x:gOsa5l(v jcG
看不懂的直接绕过d4ZO8K`E S-J"V
加密前为:hwy123456
加密后为0F9C1C7A2EE76EBB5C1F4D7526CC4F883F4E9|s4U`?dm'\-Z
P1C+kT-K
============================================ t%wj0~0bE;c#x"r i
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iu;Pk7h:y [8pY
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3P+aF2a9QuPw1l b[
以下是RSA算法文件:j^k;]/{-IVK

文件名RSA.ASP
===============================================
<% 6i'O:r ~o@q
Class clsRSA o$W ]8mRD
Public PrivateKey
Public PublicKey )jGzL*D;[lZz
Public Modulus
Public Function Crypt(pLngMessage, pLngKey)
On Error Resume Next 0s7jId+L&_XDD
Dim lLngMod
Dim lLngResult
Dim lLngIndex q*}5\(D1y:Nl
If pLngKey Mod 2 = 0 Then *Admy'lv-b_ ?
lLngResult = 1 #\-` hAN9b
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus NdZ.g^k~LP
' Mod may error on key generation 'Ovh"X-jTnx_Y
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function
Next
Else
lLngResult = pLngMessage Ge1N$l(u+^4H
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus
On Error Resume Next
' Mod may error on key generation 5`lA)O5l4B.bC
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function
Next hV\m8bWO}(Xz
End If k/Bxc%~)tN
Crypt = lLngResult
End Functiony4| \9p)]h
zb1{R Ky7E)AZ
Public Function Encode(ByVal pStrMessage) Yw4x r)^UDs+WY
Dim lLngIndex QU3DEIL
Dim lLngMaxIndex !?9M;K&Y1\
Dim lBytAscii
Dim lLngEncrypted
lLngMaxIndex = Len(pStrMessage) s:M6|!X?'@g{
If lLngMaxIndex = 0 Then Exit Function 2jKrlh
For lLngIndex = 1 To lLngMaxIndex
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1)) UG&~V-_(a%zZ
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4) jzrwD/Q"u&W
Next
End Function 8Sa@b#fg*s
Public Function Decode(ByVal pStrMessage)
Dim lBytAscii "F1O[:q5Ak"cz
Dim lLngIndex )^IXv;V0qnYI:D?
Dim lLngMaxIndex wF.otVmL
Dim lLngEncryptedData w'D1lA(?)U`:v j9SSn
Decode = ""
lLngMaxIndex = Len(pStrMessage) 2Dp:\ s%\
For lLngIndex = 1 To lLngMaxIndex Step 4
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4)) }O%mH9}3O;QXW
lBytAscii = Crypt(lLngEncryptedData, PrivateKey) ,t1Q*}Nk%p l1n
Decode = Decode & Chr(lBytAscii)
Next
End Function WLGUgd)h
Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength)
End Function
Private Function HexToNumber(ByRef pStrHex) w^"^!Ma3|C
HexToNumber = CLng("&h" & pStrHex)
End Function
End Class 8x8S^q-XZ
function Encryptstr(Message) k2^`0H2?rT*lr1Jd
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823"
LngKeyD = "20643" .we_+aI
LngKeyN = "29893"
StrMessage = Message
Set ObjRSA = New clsRSA &v@`;zy li-Az(y
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage) ]zhUh:].H
Set ObjRSA = Nothing 7c!n9In+}/pa
end function
Function Decryptstr(Message)
Dim LngKeyE M*Zb0AOs
Dim LngKeyD 0M-h)`l6`$^a
Dim LngKeyN !H!Ia*o$w
Dim StrMessage
Dim ObjRSA Tzj:lP
LngKeyE = "32823" ;y'OaM7^_1k
LngKeyD = "20643"
LngKeyN = "29893" Sb'z4M*E6_,m8K B[
StrMessage = Message
Set ObjRSA = New clsRSA
ObjRSA.PrivateKey =LngKeyD
ObjRSA.Modulus=LngKeyN ,Tv o!d:^[ Qy$Qq
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing 6C7Z"^*o4A'k,_
end function
%>K(]ySF7?8Wv
===============================================B6hu;Y-Jw'?

还有一个用于测试这段代码的test.asp
有兴趣的自己搭建个IIS测试下bqOg,]4p?O

<%
function Encryptstr(Message)
Dim LngKeyE (Fdh H6H+Y,Tw
Dim LngKeyD /kO:|?U"DR
Dim LngKeyN
Dim StrMessage 7Sq v0JD}2{
Dim ObjRSA
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893"
StrMessage = Message
Set ObjRSA = New clsRSA
ObjRSA.PublicKey = LngKeyE ;a5rydW;v
ObjRSA.Modulus = LngKeyN !l Qk&F^L Jf!@
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing 4j}gb6T/~3us!N:^
end function
function decryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN [ O%xZ!P J+u*u
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823" QKOL@5H
LngKeyD = "20643" ViN-rq
LngKeyN = "29893" t0J N6f.Y#Z
StrMessage = Message gAiJ _eA+u6m
Set ObjRSA = New clsRSA Z9{?f I2l ^6nf1V
ObjRSA.PrivateKey =LngKeyD
ObjRSA.Modulus=LngKeyN -}m?-IU
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing
end function 4?(Z]M#hEq-K!L
dim last,first
first="!@#$%^&*()"
Response.Write "加密前为:"&first
last=Encryptstr(first)
Response.Write "加密后为"&last
Response.Write "解密后为" &decryptstr(last) zkM#H*a-E-Q2m
%> ==============================================
剩下的就是字符的对照表了
===================字符集================
1_____6EBB
2_____5C1Fa/c:F!nyT
3_____4D75+NG#DOP3]/[
4_____26CC6g1}[&^UCYD"x
5_____4F88
6_____3F4E
7_____0A9Dy*x'hP:N9Cn DE
8_____1A1CYp;I$FO5FUL
9_____6D20
0_____1089|PCFu(cz p }n6e
a_____0F3E
b_____3159
c_____35174?(D^&]x*uSE f!Z\
d_____419C
e_____615C
f_____556F
g_____2B7F^(^W)K9T1wN
h_____0F9C
i_____00FA
j_____5A50j:Z:a Y*L\
k_____2850
l_____3E7BR&b9u"o*`hhx
m_____71C55rVsG8f8C#e*cb3U
n_____1FC8;q"\:`8~;x{&uaw
o_____74C1.J;W.E,jL*e
p_____5FB8
q_____6085
r_____3AC4
s_____2F50
t_____36F8
u_____7010-}&C.Tvb(nX
v_____0B42J"J&~AW/o
w_____1C7A
x_____16F8
y_____2EE7/Y8`*^vrZ
z_____5CF3
!_____6233
@_____3A45eK`$m9x W2F
#_____2291
$_____5D5C(Le'nQ&Li*O8m#Z
%_____09B9
^_____43EA
&_____62B9
*_____6301
(_____4659
)_____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