Java Ma simple implementation and encryption verification code application, java Ma

Java Ma simple implementation and encryption verification code application, java Ma

I recently watched a movie "Imitation games". In the movie "Imitation games", Alan Turing deciphered Ma and made me curious about Ma. So I began to read the materials to study it, but after all, wisdom is limited, so I use Java to implement a simple Ma.

If anyone is interested in Ma, can you read this blog https://www.0xaa55.com/forum.php? Mod = viewthread & tid = 1208 & extra = page % 3D1 & page = 1.

Ma encrypts each character into an ascii code into another character:

The simple Ma implementation here is just entertainment. The main implementation principle is to enter Scanner cin = new Education (System. in); then read msg = cin. nextLine (); then, each character is cyclically traversed to add 26 to it. To increase the complexity, determine if (msg. charAt (I)-count) <'A ')Encodemsg + = (char) (msg. charAt (I) + 26-count );Else
Encodemsg + = (char) (msg. charAt (I)-count); this achieves a simple Ma, which is actually the ascii code of the variable characters.
// Implementation

Static void encode ()
{
String msg = "";
String encodemsg = "";
Int count = 0;
System. out. printf ("% s", "Input the message :");
Using cin = new Login (System. in); // enter in the console
Msg = cin. nextLine (); // obtain the information entered in the console
For (int I = 0; I <msg. length (); I ++) // cyclically traverse every character of information
{
System. out. println ("msg =" + msg. charAt (I ));
If (Character. isSpace (msg. charAt (I) // determines whether the Character is blank
{
Encodemsg + = "";

}
Else if (msg. charAt (I)-count) <'A') // determines if the ascii code of msg. charAt (I)-count is less than a, + 26

Encodemsg + = (char) (msg. charAt (I) + 26-count );

Else
Encodemsg + = (char) (msg. charAt (I)-count );

Count ++;



}
System. out. println (encodemsg );
}

Note: I have a small idea to encrypt the verification code with this simple Ma, which is conducive to the security protection of the verification code.

The verification code I implemented here is to randomly generate several digits or letters through the random () class, also combined with the ascii code, not to mention the Code directly:

// Generate random numbers and letters,
Public static String getStringRandom (int length ){

String val = "";
Random random = new Random ();

// Parameter length, which indicates the number of random numbers generated
For (int I = 0; I <length; I ++ ){
// Determine whether the output is a number or a letter
String charOrNum = random. nextInt (2) % 2 = 0? "Char": "num ";
// Output letters or numbers
If ("char". equalsIgnoreCase (charOrNum )){
// Whether the output is uppercase or lowercase letters
System. out. println ("n 2 =" + random. nextInt (2 ));
Int temp = random. nextInt (2) % 2 = 0? 65: 97;
Val + = (char) (random. nextInt (26) + temp );
} Else if ("num". equalsIgnoreCase (charOrNum )){
Val + = String. valueOf (random. nextInt (10 ));
}
}
Return val;
}

Then, use the Ma method for encryption. You only need to replace the console input in the previous MA method with the random number generated by getStringRandom (4.

Note: At the beginning of my blog writing, I am not very familiar with this typographical style and feel ugly.