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我有一个加密方法,欢迎大家指正
时间:2022-07-02 23:58:09 编辑:袖梨 来源:一聚教程网
<%
IF Request("Action")=1 then
Dim String,StringLen,i,StringNumTmp,StringRndNum,ResultString,j
String = Request("Pass") '密码字符串
'开始计算字符数据
StringLen = Len(String)
For i = 1 to StringLen
StringNumTmp = Asc(Mid(String,i,1))
Randomize
StringRndNum=Int((18-1)*Rnd+1)
if Len(StringNumTmp + StringRndNum) < 3 then
For j = 1 to 3 - Int(Len(StringNumTmp + StringRndNum))
StringNumber = "0" & (StringNumTmp + StringRndNum)
Next
Else
StringNumber = StringNumTmp + StringRndNum
End if
ResultString = ResultString & Chr(StringNumTmp - StringRndNum) & StringNumber
Next
Response.write "加密后结果:" & ResultString '输出结果
%>
<%
'开始破解
For i = 1 to Len(ResultString) Step 4
PString = Left(Mid(ResultString,i,4),1)
PStringNum1 = Asc(PString)
PStringNum2 = Int(Right(Mid(ResultString,i,4),3))
PStringNum = (PStringNum1 + PStringNum2) / 2
PString1 = PString1 & Chr(PStringNum)
Next
Response.write "破解后结果:" & PString1
End if
%>
IF Request("Action")=1 then
Dim String,StringLen,i,StringNumTmp,StringRndNum,ResultString,j
String = Request("Pass") '密码字符串
'开始计算字符数据
StringLen = Len(String)
For i = 1 to StringLen
StringNumTmp = Asc(Mid(String,i,1))
Randomize
StringRndNum=Int((18-1)*Rnd+1)
if Len(StringNumTmp + StringRndNum) < 3 then
For j = 1 to 3 - Int(Len(StringNumTmp + StringRndNum))
StringNumber = "0" & (StringNumTmp + StringRndNum)
Next
Else
StringNumber = StringNumTmp + StringRndNum
End if
ResultString = ResultString & Chr(StringNumTmp - StringRndNum) & StringNumber
Next
Response.write "加密后结果:" & ResultString '输出结果
%>
<%
'开始破解
For i = 1 to Len(ResultString) Step 4
PString = Left(Mid(ResultString,i,4),1)
PStringNum1 = Asc(PString)
PStringNum2 = Int(Right(Mid(ResultString,i,4),3))
PStringNum = (PStringNum1 + PStringNum2) / 2
PString1 = PString1 & Chr(PStringNum)
Next
Response.write "破解后结果:" & PString1
End if
%>
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