加密数字C ++

Question

So I am trying to encrypt a four digit integer by adding seven to the digit then dividing the whole digit by ten. In my program I am taking each single digit separately and then I need to divide the whole digit by ten. How can I combine all the separate int into one four digit number?

#include "stdafx.h"
using namespace std;

int main()
{
    //Define all variables needed
    int a,b,c,d,enc,ext;

    //Print dialog and input each single digit for the four number digit
    cout << "Enter Your First Digit:" << endl;
    cin >> a;
    cout << "Enter Your Second Digit:" << endl;
    cin >> b;
    cout << "Enter Your Third Digit:" << endl;
    cin >> c;
    cout << "Enter Your Fourth Digit:" << endl;
    cin >> d;
    //Add seven to each digit
    a += 7;
    b += 7;
    c += 7;
    d += 7;

    a /= 10;
    b /= 10;
    c /= 10;
    d /= 10;

    cout << "Your encrpyted digits:" << c << d << a << b <<endl;

    cout << "Enter 1 to exit:" <<endl;
    cin >> ext;

    if (ext = 1) {
        exit(EXIT_SUCCESS);
    }
}

As you probably noticed I am dividing each number separately. I need to do them together. Then I am also creating a decrypting which I will get me back to the original number in a separate program.

Answer 1

Based on your comment you are trying to do a variation on the Caesar Cipher , in which case you should be using the modulus operator ( % ) not the integer division operator ( / ). Using integer division loses information which will prevent you from decrypting. When your digit is in {0, 1, 2} your division results in a 0. When it is in {3, 4, 5, 6, 7, 8, 9}, the division results in a 1. You can't decrypt {0, 1} back into the original number without some additional information (which you have discarded).

If you want to encrypt on a digit by digit basis using the Caesar Cipher approach, you should be using modulo arithmetic so that each digit has a unique encrypted value which can be retrieved during decryption. If that's really what you are looking for then you should be doing something like the following to encrypt with a 7:

    a = (a + 7) % 10;
    b = (b + 7) % 10;
    c = (c + 7) % 10;
    d = (d + 7) % 10;

To decrpyt, you subtract 7, which in mod 10 arithmetic is an addition by 3, so that would be:

    a = (a + 3) % 10;
    b = (b + 3) % 10;
    c = (c + 3) % 10;
    d = (d + 3) % 10;

This of course presupposes you've properly validated your input (which isn't the case in your example above).

Answer 2

This is what youd'd probably be looking for :

int e = (a*1000)+(b*100)+(c*10)+d;
e=e/10;

Answer 3

Combining the individual digits into one four-digit number is simple; just multiple the first digit by 1000, add the second multiplied by 100, and so on.

But this is a one-way algorithm; you will never be able to retrieve the original four-digit number from this.

Answer 4

It's not clear from your description whether the addition should be modulo 10 or not; if so

((((((a % 10) * 10) + (b % 10)) * 10) + (c % 10)) * 10) + (d % 10) 

if you don't want the modulo 10

(((((a * 10) + b) * 10) + c) * 10) + d 

Answer 5

Stepping aside the fact that you almost certainly want mod instead of divide (as @Andand has said), there's more than one way to turn the digits into a number!

A lot of people using interpreted languages these days would probably want to do it symbolically. C++ can do that too, fairly neatly in fact:

// create a string stream that you can write to, just like
// writing to cout, except the results will be stored
// in a string

stringstream ss (stringstream::in | stringstream::out);

// write the digits to the string stream
ss << a << b << c << d;

cout << "The value stored as a string is " << ss.str() << endl;

// you can also read from a string stream like you're reading
// from cin.  in this case we are reading the integer value
// that we just symbolically stored as a character string

int value;
ss >> value;

cout << "The value stored as an integer is " << value << endl;

It won't be as efficient as multiplications in this narrow case of a 4 digit number, because of the round trip to a string and back. But good to know the technique. Also it's a style of coding that can be a lot easier to maintain and adapt.

You'll get stringstream if you #include <sstream> .