That is correct. Because this crack me has a bug here. It want to read C:\A.txt file, if this file is not exist then exception coming.
So you have two ways:
1. Create a new file named A.txt and put it into root of C: driver .
2. Just click continue button to ignore this message.
Public Shared Function DeCrypt(ByVal txtToCode As String, ByVal txtMeth As String) As String
Dim text1 As String = ""
Dim text2 As String = ""
Dim text3 As String = txtToCode
Dim num1 As Integer = 0
Dim num3 As Long = Crypt.GetSecret(txtMeth)
Dim text4 As String = num3.ToString
If (text4.Length < 3) Then
text4 = ("0" & text4)
End If
Try
Dim num2 As Integer = 0
Do While (num2 <= txtToCode.Length)
text3 = txtToCode.Substring(num2, 3)
num1 = (Integer.Parse(text3) - (Integer.Parse(text4.Substring(2, 1)) * 21))
num1 = (num1 - Integer.Parse(text4))
Dim ch1 As Char = CType(CType(num1, UInt16), Char)
text2 = ch1.ToString
text1 = (text1 & text2)
num2 = (num2 + 3)
Loop
Catch exception1 As Exception
End Try
Return text1
End Function
Public Shared Function EnCrypt(ByVal txtToCode As String, ByVal txtMeth As String) As String
Dim num1 As Long = Crypt.GetSecret(txtMeth)
Dim text1 As String = num1.ToString
Dim text2 As String = ""
Dim text3 As String = ""
Dim chArray1 As Char() = txtToCode.ToCharArray
Dim chArray2 As Char() = chArray1
Dim num2 As Integer = 0
Do While (num2 < chArray2.Length)
Dim ch1 As Char = chArray2(num2)
ch1.ToString
Dim num3 As Integer = (ch1 + Integer.Parse(text1))
text3 = num3.ToString
If (text3.Length < 3) Then
text3 = ("0" & text3)
End If
If (text1.Length < 3) Then
text1 = ("0" & text1)
End If
Dim num4 As Integer = (Integer.Parse(text3) + (Integer.Parse(text1.Substring(2, 1)) * 21))
text3 = num4.ToString
text2 = (text2 & text3)
num2 += 1
Loop
Return text2
End Function