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]&rWy+h p-N
原始出处：[url]http://www.3ast.com.cm[/url]3]:t%JBb4@Qy

看不懂的直接绕过
加密前为:hwy123456
加密后为0F9C1C7A2EE76EBB5C1F4D7526CC4F883F4E7vP8h{z&O
PB#e Y$k7g%{5\
============================================
上面是当时自己校内的心情。现在已经解破出来了，分享给大家/]9CaW/d"D1i'R\
也就不隐瞒大家什么了，这个是国内某知名大学的后台加密算法\/z y#L]b%r+s3N*j
以下是RSA算法文件
nS_I@7T#m
文件名RSA.ASP
===============================================
<%
Class clsRSA
Public PrivateKey Y9?V!Nb#r&|E
Public PublicKey
Public Modulus
Public Function Crypt(pLngMessage, pLngKey) J s{ O*]J
On Error Resume Next
Dim lLngMod -Xp|n(EL(z%N
Dim lLngResult )m/Nk9sb"n'L c7cRr.q j
Dim lLngIndex S1E+q'E L+lT3^}x#Mw
If pLngKey Mod 2 = 0 Then F"k'p8VrDU'u
lLngResult = 1 ziPIE-tZ
For lLngIndex = 1 To pLngKey / 2 tAIhtm$bZ
lLngMod = (pLngMessage ^ 2) Mod Modulus -d+E5x1~T#{I3?
' Mod may error on key generation
lLngResult = (lLngMod * lLngResult) Mod Modulus #hrp/|2^-C7c
If Err Then Exit Function
Next
Else +r,h7g;SW7J^5CIe/Fy${5v
lLngResult = pLngMessage (|]xw f!h
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 R/{gq6K{?
Next
End If
Crypt = lLngResult @~-Syl;w
End Functiongw:QU%D'A
Gi3PE l0e*lj
Public Function Encode(ByVal pStrMessage)
Dim lLngIndex 5A/q.B{D
Dim lLngMaxIndex
Dim lBytAscii
Dim lLngEncrypted !ppAL'^ d
lLngMaxIndex = Len(pStrMessage) wlnR_-Z/T|&R
If lLngMaxIndex = 0 Then Exit Function
For lLngIndex = 1 To lLngMaxIndex
lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1)) ]'pZ/N9d+u i
lLngEncrypted = Crypt(lBytAscii, PublicKey)
Encode = Encode & NumberToHex(lLngEncrypted, 4)
Next
End Function \s)E-^mqlln)B_q
Public Function Decode(ByVal pStrMessage)
Dim lBytAscii
Dim lLngIndex
Dim lLngMaxIndex 2M7y9RdRJ:R
Dim lLngEncryptedData [ o"\b"N:S%dKsS
Decode = ""
lLngMaxIndex = Len(pStrMessage)
For lLngIndex = 1 To lLngMaxIndex Step 4
lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4))
lBytAscii = Crypt(lLngEncryptedData, PrivateKey) -`phj^
Decode = Decode & Chr(lBytAscii) 2ZShUV1v%T:F'E"{,p9c
Next
End Function #bE&xg3J dy1`
Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength)
End Function QB Pn|(qw
Private Function HexToNumber(ByRef pStrHex)
HexToNumber = CLng("&h" & pStrHex)
End Function (@c;eXdT
End Class %h&n&]t*U3FA4QH n
function Encryptstr(Message) w3Q?c iM0?
Dim LngKeyE
Dim LngKeyD G|#\7S/W+lIsU
Dim LngKeyN 4d ]2M:G ?"L%Ya%l)J
Dim StrMessage
Dim ObjRSA
LngKeyE = "32823" [~:b1f4uGNb:Y7hK/KH
LngKeyD = "20643"
LngKeyN = "29893"
StrMessage = Message P/Q2n%^[8R1fe.r
Set ObjRSA = New clsRSA 5S2s*MQ@;j)C kss0z
ObjRSA.PublicKey = LngKeyE
ObjRSA.Modulus = LngKeyN } lGC~Iw
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing
end function
Function Decryptstr(Message)
Dim LngKeyE
Dim LngKeyD \X`#M W+K'I B;vx Rz
Dim LngKeyN ;N |-g YN
Dim StrMessage EN2?/EJ y"R3i
Dim ObjRSA 5P.L\lCW6h8t.YZM'\
LngKeyE = "32823"
LngKeyD = "20643" Z1Sy1wi7z;N2H
LngKeyN = "29893"
StrMessage = Message 0?}x T'I2Z3L
Set ObjRSA = New clsRSA +d9Gp `"tW/GG d]7J
ObjRSA.PrivateKey =LngKeyD
ObjRSA.Modulus=LngKeyN
decryptstr=ObjRSA.Decode(StrMessage) +v-T1d guy2Lm-O
Set ObjRSA = Nothing
end function
%>
===============================================
|Bq-t+I;G&j
还有一个用于测试这段代码的test.asp9T#r'zS@]x.Q[t%]q
有兴趣的自己搭建个IIS测试下

<%HV6QS*b\
function Encryptstr(Message)
Dim LngKeyE
Dim LngKeyD D N PHp]
Dim LngKeyN ngQRvM-?oG
Dim StrMessage
Dim ObjRSA 9~Xq0J5{ ]6_
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893" $k7w2Rot_UXm:IF
StrMessage = Message
Set ObjRSA = New clsRSA
ObjRSA.PublicKey = LngKeyE UV6Ir~8xs$B+KO
ObjRSA.Modulus = LngKeyN .['HVy8Ro"uhG
Encryptstr = ObjRSA.Encode(StrMessage)
Set ObjRSA = Nothing (UmK B2W%r LtW'OPP
end function 1E-E(@3s_
function decryptstr(Message) 3QC yeJo@&r
Dim LngKeyE LnlMK7W:|
Dim LngKeyD
Dim LngKeyN 'ER"S.^U gCb0P {w
Dim StrMessage 0o!OmL.I$Q{P%tZsT
Dim ObjRSA b*?? F;_ f#B^`6tM
LngKeyE = "32823"
LngKeyD = "20643"
LngKeyN = "29893" A2x_|8v&u5~i~2O6~
StrMessage = Message ,?Yg6O1m
Set ObjRSA = New clsRSA @m4?6|zh&_y
ObjRSA.PrivateKey =LngKeyD ,AV3{Q*_6L8pI
ObjRSA.Modulus=LngKeyN ;F$`"V Zf7as(o
decryptstr=ObjRSA.Decode(StrMessage)
Set ObjRSA = Nothing
end function
dim last,first
first="!@#$%^&*()" %yj(T;]E.yp9H
Response.Write "加密前为:"&first
last=Encryptstr(first) a}`m]e_:mD
Response.Write "加密后为"&last w(MR9[l7L&P0Xp
Response.Write "解密后为" &decryptstr(last) Q N }}1k x^
%> ==============================================
剩下的就是字符的对照表了
===================字符集================
1_____6EBB&x$h4Qoru!_)ql
2_____5C1FS'D1SOToi'q
3_____4D75
4_____26CC
5_____4F88^]7\ Ps!F@I'i
6_____3F4E-Kpdcp+JQ5d
7_____0A9D#DA7R+L6X[;fh A2c$\
8_____1A1C
9_____6D20
0_____1089
a_____0F3E5KDH$c)TfRiq#u
b_____3159_z7F1q.J0ny6d
c_____3517
d_____419Cq*Z.v G(`D ]
e_____615C?'hT*EG?+S?^
f_____556F
g_____2B7Fn7l_GS6a[ kn
h_____0F9C(^$VD"b&su
i_____00FA,|,^!A-w {B m2i
j_____5A50
k_____2850)f q c(UUOh[$iGH
l_____3E7B
m_____71C5
n_____1FC8
o_____74C1
p_____5FB8H4t4| Ul1^A~9I4|6|
q_____6085+^ |!n1|y&Zl
r_____3AC4"vVKu_'zK)a
s_____2F50/^vT%V7d{6@
t_____36F8
u_____70107?ZepNJt
v_____0B42
w_____1C7A_)f \0N&\ ?_t
x_____16F80cM5\X]6O$x
y_____2EE7
z_____5CF3
!_____6233
@_____3A45
#_____2291
$_____5D5C
%_____09B9J)uAL7F:x
^_____43EA
&_____62B92Y,k1x5X5a|B7w
*_____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