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Leni [432]
3 years ago
11

An individual is first with the network before they are authorized to access resources on the network A. countermeasure B. vulne

rability C. adversary D. risk
Computers and Technology
1 answer:
ozzi3 years ago
8 0

Answer:A) Countmeasure

Explanation: Countermeasure is the activity that is used for the reduction or avoidance of the threats or malfunction that occur in the operating system.The tool consists of the major components like firewalls , antiviruses etc to secure the system. It protects the servers, data systems , networks etc.

Other options are incorrect because vulnerability is the capability of a system being in exposed situation, adversary is a conflict situation and risk is the event that can cause loss in the operating system.Thus , the correct option is option(A) for checking the authorization of the individual and then only letting him access the network.

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Write a program that takes paragraph from the user and prints all unique words in that paragraph using strings in c++
Bezzdna [24]

Answer:

void printUniquedWords(char filename[])

void printUniquedWords(char filename[]) {

void printUniquedWords(char filename[]) {     // Whatever your t.text file is

    fstream fs(filename);

    fstream fs(filename);   

    fstream fs(filename);       

    fstream fs(filename);           map<string, int> mp;

    fstream fs(filename);           map<string, int> mp;   

    fstream fs(filename);           map<string, int> mp;       // This serves to keep a record of the words

    string word;

    string word;     while (fs >> word)

    string word;     while (fs >> word)     {

    string word;     while (fs >> word)     {         // Verifies if this is the first time the word appears hence "unique word"

        if (!mp.count(word))

        if (!mp.count(word))             mp.insert(make_pair(word, 1));

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }   

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();   

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;     }

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;     } }

5 0
3 years ago
Which type of evidence should victims collect to help officials catch cyber bullies
iren [92.7K]

Answer:

How you get cyberbullied just look away from the screen -Tyler The Creator

Explanation:

3 0
4 years ago
Read 2 more answers
in an advertisement is made of a wireless keyboard and a mouse.Name two commonly used methods of connecting devices wirelessly​
DENIUS [597]

Answer:

Wireless LAN

Wireless MAN

Explanation:

I use these methods to connect my wireless devices all of the time and most of my friends do it as well so I am pretty sure it is a commonly used method.

3 0
3 years ago
Question 6 options: This is a category or partitioned group of the small units of a programming language include its numeric lit
Wittaler [7]

Lexeme is a category or partitioned group of the small units of a programming language include its numeric literals, operators, and special words.

<h3>What is a lexeme in programming?</h3>

Lexemes are known to be those character strings that are put together from the character group of a program, and the token stands for what aspect of the program's grammar they are made of.

Hence, Lexeme is a category or partitioned group of the small units of a programming language include its numeric literals, operators, and special words.

Learn more about Lexeme from

brainly.com/question/14125370

#SPJ1

6 0
2 years ago
Fill in the blank: In a Word chart, text that describes the data represented in a chart and that is typically displayed on the r
Triss [41]

Answer:

Data bar

Explanation:

8 0
3 years ago
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