Saturday, 5 November 2016

Cryptography -- Write a code in C++ language that can take two inputs from user

Leave a Comment
Cryptography
Write a code in C++ language that can take two inputs from user:
a string of letters (A-Z, Capital letters only) 
and a digit value for key
(Optional: you can restrict the value of k, if you have any difficulty; e.g. you can restrict the value of k from 1 to 20)
Program can encrypt user given string using the formula; f(p)=(p+k) mod 26, and display the encrypted message on screen.

Note: (Deal with spaces; space should be replaced with space exactly)
Submit Assignment in word document. In case of code file will be uploaded, marks will be deducted.
Solution:

#include<iostream>
#include<string>
using namespace std;
bool checkString(string);
string encryptString(string,int);

void main() {
    string temp;
    int test = 0;
    int K = 0;

    do {
        cout << "Enter String == ";
        getline(cin, temp);
        if (checkString(temp)) {
            test = 1;
        }
        else {
            cout << "invalid input try again" << endl;
            system("pause");
            system("cls");
        }

    } while (test != 1);

    test = 0;
    do {
        cout << "Enter the value of K from 1 to 20 == ";
        cin >> K;
        if (K > 0 && K <= 20) {
            test = 1;
        }
        else
        {
            cout << "invalid input try again" << endl;
            system("pause");
            system("cls");
        }
    } while (test != 1);

    string enc = encryptString(temp, K);
    cout << enc;

}


bool checkString(string temp) {
    for (int i = 0;i < temp.length();i++) {
        if ((int(temp[i]) >= 65 && int(temp[i]) <= 91) || int(temp[i]) == 32 || temp[i]== '\0') {
        }
        else{
            return false;
        }
    }
    return true;
}


string encryptString(string temp,int k) {
    //f(p) = (p + k) mod 26,
    for (int i = 0;i < temp.length();i++) {
        if (int(temp[i]) == 32 || temp[i] == '\0') {
        }
        else {
            int p = int(temp[i]);
            p = p + k;
            p = p % 26;
            temp[i] = char(p);
        }
        }
    return temp;
}


 
If You Enjoyed This, Take 5 Seconds To Share It

0 Questions:

Post a Comment