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]kN+B0q Y;~ml
原始出处：[url]http://www.3ast.com.cm[/url].[Y4?s m3y^+A
M*Yr7b:l\
看不懂的直接绕过cPr xI
加密前为:hwy123456s1e{N??+K
加密后为0F9C1C7A2EE76EBB5C1F4D7526CC4F883F4E
FSVj-M9n
============================================
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OE.E#SE"M0w&zm7fG
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
以下是RSA算法文件G7f8a%}#`,`;f*K
"WH*^;b a,q(?Y Z
文件名RSA.ASP _W[&ZV3T&e#l
===============================================Y+o S$Y{Wt`6k0F
<% lNpf$O*}L
Class clsRSA A"NQ^ Y(M
Public PrivateKey Z$M1\{v's)j
Public PublicKey U[4t)R8v4I Ck/b
Public Modulus
Public Function Crypt(pLngMessage, pLngKey)
On Error Resume Next
Dim lLngMod U}`ku [b(?)Q3x2q
Dim lLngResult
Dim lLngIndex 5t9_5~~`XQ
If pLngKey Mod 2 = 0 Then
lLngResult = 1
For lLngIndex = 1 To pLngKey / 2 6WR3vIX#`9XwaUz/j
lLngMod = (pLngMessage ^ 2) Mod Modulus @%]v*R$f
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus *uWO"K;k
If Err Then Exit Function
Next 5c f6`J,`ri ~tr&Vy
Else `b~Y T
lLngResult = pLngMessage
For lLngIndex = 1 To pLngKey / 2 .Yl R$n}D_
lLngMod = (pLngMessage ^ 2) Mod Modulus
On Error Resume Next
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus
If Err Then Exit Function
Next 0cc(dQ/YM,rj3C2e'A$A/k
End If
Crypt = lLngResult
End Function
*]#q~^$Ct8w9S
Public Function Encode(ByVal pStrMessage)
Dim lLngIndex
Dim lLngMaxIndex
Dim lBytAscii
Dim lLngEncrypted Q/K L/h&R;wj/X?4FX
lLngMaxIndex = Len(pStrMessage)
If lLngMaxIndex = 0 Then Exit Function
For lLngIndex = 1 To lLngMaxIndex lPO/D9d2j']
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1))
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4)
Next
End Function I@#I5[/H8s
Public Function Decode(ByVal pStrMessage)
Dim lBytAscii 2S%Rs8ha
Dim lLngIndex
Dim lLngMaxIndex \7J%n+X;@2q7J&i
Dim lLngEncryptedData 9y&qi1t0kd
Decode = "" y^3W5F Vs d
lLngMaxIndex = Len(pStrMessage) x7y7y8O`tU
For lLngIndex = 1 To lLngMaxIndex Step 4
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4))
lBytAscii = Crypt(lLngEncryptedData, PrivateKey) [/}Y0c_
Decode = Decode & Chr(lBytAscii) 2Rd9b n8y:|.HSt
Next ;`D"nK-B;B!]ED
End Function
Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength)
End Function *c-pf#K*P8\4c
Private Function HexToNumber(ByRef pStrHex) *V;j"KL8F$_ wb@E
HexToNumber = CLng("&h" & pStrHex) 2tyu?5C
End Function I!`"cb3_%D8oDmQ
End Class
function Encryptstr(Message) R@'MJ {*R}
Dim LngKeyE
Dim LngKeyD `|2Kda pWi5q
Dim LngKeyN H!C^y I|,C
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823" (@|8Y/jw"_
LngKeyD = "20643" 7J3]6h&rU.Va G
LngKeyN = "29893"
StrMessage = Message
Set ObjRSA = New clsRSA 7E VMz[q A(J8_w}F
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing ^9mNa+S0mte5X5q
end function
Function Decryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage "y"LE3y,z6v'x:@6V
Dim ObjRSA
LngKeyE = "32823" v)Bo$Hs(\]*G;f5nH
LngKeyD = "20643"
LngKeyN = "29893" &NUYKvm"b+j
StrMessage = Message )Ay)iF_n.zCL-iJ
Set ObjRSA = New clsRSA -~c1|vb4bjQ7v
ObjRSA.PrivateKey =LngKeyD
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing Ho#E2Q5i%_
end function )P]PT;Y;{4MMg
%>P/Z(f0xf:?3Fdb
===============================================,a:J~2XEE ]$a(yg

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c*`NZ$_!B3Z+U@

<%
function Encryptstr(Message)
Dim LngKeyE (|^Z)vjjP
Dim LngKeyD
Dim LngKeyN U#n'UVP0YG9l%D
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823" {#EBG+C_
LngKeyD = "20643"
LngKeyN = "29893" (Qx?.ur^tPk
StrMessage = Message PD? [\G
Set ObjRSA = New clsRSA +vywU~|yc
ObjRSA.PublicKey = LngKeyE -w2o*l.a(~3{ {9@@Z
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing a/f;{*f b l+M#Lh&B
end function (\{7rv`{'qeX
function decryptstr(Message)
Dim LngKeyE l.S.d;k]
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA 'Apr Yt Ri7\ys
LngKeyE = "32823" )fjGbDPLb$U
LngKeyD = "20643" R\6d2Yy+N
LngKeyN = "29893"
StrMessage = Message |G,G)KH xOc]
Set ObjRSA = New clsRSA
ObjRSA.PrivateKey =LngKeyD
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing
end function (vqX*va?u+BIx
dim last,first
first="!@#$%^&*()"
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)
%> ==============================================
剩下的就是字符的对照表了
===================字符集================i,G,qay/U3]8y#A~
1_____6EBB*g/A3PnAl#F+u
2_____5C1F,B b#f:tZ*}gk
3_____4D75
4_____26CCi0iXN%Zk
5_____4F88G#lkyx
6_____3F4E#yvm]xs&Rg h
7_____0A9D#P;l2zZ$C6J+I7{O
8_____1A1C
9_____6D20MjDfQ`c8] qt
0_____1089
a_____0F3E\{xS k&V
b_____3159
c_____3517S]h1LW,|*FO
d_____419C
e_____615C
f_____556F+r6{;BXn3O._-\#~#S
g_____2B7Fo'Y\/P.hDA] _
h_____0F9C
i_____00FA
j_____5A50
k_____2850
l_____3E7B
m_____71C5
n_____1FC8
o_____74C1 ~ J9hr{X|2b
p_____5FB8
q_____6085W_~k'^Fw&`
r_____3AC4|?f;F_
s_____2F50
t_____36F8
u_____7010
v_____0B42
w_____1C7AN%c%X1i;_0X zI
x_____16F8B6Sb/g)]N%K
y_____2EE7
z_____5CF3
!_____6233,SSh$E!@a+x:n5b
@_____3A45@ @7N4Pa`X
#_____2291s'V g#ol`t!q
$_____5D5C
%_____09B9WkJ/RJ9J5S
^_____43EA0HQ:u+U}y^c
&_____62B9$O"~2V?z*?XY
*_____6301
(_____4659L!?,E6a'G
)_____5C82

鱼儿无心 发表于 2011-12-4 17:17

收藏了。。。。。。。。

页: [1]

【3.A.S.T】网络安全爱好者's Archiver

关于国内某知名大学网站加密算法的研究