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Doss [256]
3 years ago
14

A common technique for masking contents of messages or other information traffic so that opponents can not extract the informati

on from the message is: A. integrity B. encryption C. analysis
Computers and Technology
1 answer:
Sladkaya [172]3 years ago
5 0

Answer:

Option B is the correct answer for the above question.

Explanation:

Encryption is techniques that give the features to hide the original text in the back of some other text which has no means for the other person. It is a technique that is followed by the person at the time of message transfer. The encryption changes the original message into a duplicate message on the sender ends and get the original message from the duplicates message on the destination ends with the help of some algorithm.

The above question asked about the techniques which convert the message into duplicate one so that no unauthorized user can read the message. Then the techniques known as Encryption which is stated from the option B. Hence Option B is the correct answer while the other is not because--

  • Option A states about integrity which is not a technique to change the original message into a duplicate message.
  • Option C states about analysis which is also not a technique to change the original message into the duplicate message.

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Read two numbers on the keyboard. Display for each: how many digits it consists of, which is the sum of the digits, and which of
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Answer:

Here it is

Explanation:

#include <iostream>

using namespace std;

int main()

{

   int a, b;

   cin >> a >> b;

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   int sum1 = 0, sum2 = 0;

   int max_digit1 = 0, max_digit2 = 0;

   while (a > 1)

   {

       int digit = a % 10;

       nr_digits1 += 1;

       sum1 += digit;

       if (digit > max_digit1)

       {

           max_digit1 = digit;

       }

       a /= 10;

   }

   while (b > 1)

   {

       int digit = b % 10;

       nr_digits2 += 1;

       sum2 += digit;

       if (digit > max_digit2)

       {

           max_digit2 = digit;

       }

       b /= 10;

   }

   

   cout << "For a: \n" << "   No. of digits: " << nr_digits1 << "\n";

   cout << "   Sum of digits: " << sum1 << "\n";

   cout << "   Max digit: " << max_digit1 << "\n";

   cout << "\n";

   cout << "For b: \n" << "   No. of digits: " << nr_digits2 << "\n";

   cout << "   Sum of digits: " << sum2 << "\n";

   cout << "   Max digit: " << max_digit2 << "\n";

   return 0;

}

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Answer:

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