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]
-qE-C1YH6~s6iib+pd
看不懂的直接绕过 [3q.Wqz9ei
加密前为:hwy123456bj{i _%G e
加密后为0F9C1C7A2EE76EBB5C1F4D7526CC4F883F4E

============================================LH4Wi0\vd*BP$k
上面是当时自己校内的心情。现在已经解破出来了，分享给大家M T}%c,C
也就不隐瞒大家什么了，这个是国内某知名大学的后台加密算法
以下是RSA算法文件
/WYO'e1N`~}?
文件名RSA.ASP
==============================================="cpe9T?T_~
<% (C0?+p*K:l#\%D,D2~#t
Class clsRSA 3Kktj}l?3j?(OI
Public PrivateKey 0bb,xYU i
Public PublicKey
Public Modulus
Public Function Crypt(pLngMessage, pLngKey)
On Error Resume Next (Y3Y4T5~Q
Dim lLngMod "gy$Zw$Li
Dim lLngResult ]9|mnWr7^XEX
Dim lLngIndex
If pLngKey Mod 2 = 0 Then mdM)\L
lLngResult = 1
For lLngIndex = 1 To pLngKey / 2
lLngMod = (pLngMessage ^ 2) Mod Modulus 6s*]V z*pC^2ht%X"d
' Mod may error on key generation f;T$aN-k(_!v5ms
lLngResult = (lLngMod * lLngResult) Mod Modulus (G?DA.h*_B+C/o%v5[
If Err Then Exit Function ;A]!\ L/m
Next
Else j7Ft5kZ5d
lLngResult = pLngMessage
For lLngIndex = 1 To pLngKey / 2 0djSYfTT9a
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
End If !L*b!KQXP,q1O7}|
Crypt = lLngResult ] XL9[*r
End Function

Public Function Encode(ByVal pStrMessage) 6JUd\k*`
Dim lLngIndex 0[5V Kj#Zx*m
Dim lLngMaxIndex *D7c1{8t7NE
Dim lBytAscii z q2|'L]re
Dim lLngEncrypted s6Dy:ra,SH`
lLngMaxIndex = Len(pStrMessage) {?u8iV3lV&o
If lLngMaxIndex = 0 Then Exit Function
For lLngIndex = 1 To lLngMaxIndex !T2GU;^$K:y&O(C
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1)) Mh `3v%P!h
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4) $o8ov\:n#T I
Next
End Function
Public Function Decode(ByVal pStrMessage)
Dim lBytAscii P.y@q`&H){.d
Dim lLngIndex ]-M{%KX)DA
Dim lLngMaxIndex
Dim lLngEncryptedData &uP^i#RtNo J8od
Decode = "" @*B0pRn0T4Wk
lLngMaxIndex = Len(pStrMessage) 3?"v4J\O!}z
For lLngIndex = 1 To lLngMaxIndex Step 4 9D$hRb!J"w
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4)) I`-V1N+DfUd
lBytAscii = Crypt(lLngEncryptedData, PrivateKey) 9v`s0d4O4\
Decode = Decode & Chr(lBytAscii) "a`@)idh
Next
End Function
Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength) :S2\*h*G*d3B(o%s DTC
End Function
Private Function HexToNumber(ByRef pStrHex) +Bb}0|_zm
HexToNumber = CLng("&h" & pStrHex)
End Function -DKbJn
End Class
function Encryptstr(Message) 9Zxf(H4[)[
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN *t7P2i}]q
Dim StrMessage
Dim ObjRSA N'cyE'J6R B`X
LngKeyE = "32823"
LngKeyD = "20643" %M6d1U*je6`r!sc
LngKeyN = "29893"
StrMessage = Message x`gHZ@k:xM
Set ObjRSA = New clsRSA *@Uc~%cM
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN 8Wt,hI ?
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing /\W:rQ(|N4ZmF
end function2b|G-]XA/b,h
Function Decryptstr(Message) +hO2Y8HPQ(k6HJ
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893" /G#q5AoP iW3G
StrMessage = Message RAnh*x
Set ObjRSA = New clsRSA SU[M e[bo
ObjRSA.PrivateKey =LngKeyD beT]g;\}
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing
end function vx4plcR;QH
%>
===============================================4|(WlR F
(p`1h$B"J#z
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"`f0V'H+x(|h#L'N

<%
function Encryptstr(Message) 0PDr+e,Mo1Ah:Ck
Dim LngKeyE
Dim LngKeyD )Nn&ttl!u
Dim LngKeyN
Dim StrMessage
Dim ObjRSA -zUm!cY*m2v
LngKeyE = "32823"
LngKeyD = "20643" /pu!MI#Nx/cp
LngKeyN = "29893"
StrMessage = Message ^/k kgIm.^
Set ObjRSA = New clsRSA $_4D#jzk,G0M(vo
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing /_o``*t;{
end function V6l.fm qox8Xr l n
function decryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN ^1p8p?q1m
Dim StrMessage
Dim ObjRSA 1B:eU_BKXM
LngKeyE = "32823"
LngKeyD = "20643" 1CP9} I:Wy2T8{&PH/q
LngKeyN = "29893"
StrMessage = Message nv'jh0@td"@_"`
Set ObjRSA = New clsRSA
ObjRSA.PrivateKey =LngKeyD 9RTX[Y'C.D
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage) U:K#p RO{;w8\
Set ObjRSA = Nothing k2KkW/^8O_ Y
end function k"x)m5[$c"P
dim last,first
first="!@#$%^&*()"
Response.Write "加密前为:"&first
last=Encryptstr(first)
Response.Write "加密后为"&last
Response.Write "解密后为" &decryptstr(last)
%> ==============================================&^_cCEyC%c
剩下的就是字符的对照表了
===================字符集================8|m-FX9|{8Fr
1_____6EBB MF_T+Z2V1U?~$U)F
2_____5C1F
3_____4D75
4_____26CC3QD1j$g3s
5_____4F88
6_____3F4Em9\utLzT
7_____0A9D:uq&?io,I
8_____1A1C8\nE}A&s!F9_kV
9_____6D20w3B3vkk
0_____1089`C7O'jG%Ii r:?#C b
a_____0F3E
b_____3159e,@ S|SQ8ya
c_____3517E*?|;~ttH;Uu?L
d_____419C
e_____615CJy5?^n8T){O
f_____556F.?zZhG'X0Peg~8K%c
g_____2B7F.Wy)r2e5b-bPc}o
h_____0F9C
i_____00FA~4e$a ajU3f
j_____5A50
k_____2850
l_____3E7B&i's)CbuG[
m_____71C5$tticO4IR*I
n_____1FC8
o_____74C1 y9vf_G'| C1v
p_____5FB8
q_____6085Sf,tf}U
r_____3AC4&OP qF Il3uS
s_____2F50
t_____36F8
u_____7010,plr5SjI
v_____0B426ni n}o9u4A
w_____1C7A#@aYvr G q
x_____16F8w\)x0]dT)y
y_____2EE7
z_____5CF3
!_____62334Rs{ s+@H j p!V)W
@_____3A45$RL4Gmvk:pUJ(^
#_____2291
$_____5D5C9OJkUz-C Q
%_____09B9+WYt%xB!f ~
^_____43EA-~/Do3p-o3piw2r_
&_____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