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Viefleur [7K]
3 years ago
6

What is an accessory?

Computers and Technology
1 answer:
frosja888 [35]3 years ago
3 0

Answer:

An accessory is a person who assists in the commission of a crime, but who does not actually participate in the commission of the crime.Explanation:

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​A(n) ________ database makes it possible to store information across millions of machines in hundreds of data centers around th
Kisachek [45]

Answer:

distributed

Explanation:

According to my expertise in information technology, it seems that the type of database being described is a distributed database. Like mentioned in the question this is a database that works by saving information in data centers from various locations and information is processed through multiple database node. This allows information to be more secured, faster, and also acts as a fail-safe in case of any malfunction where data may otherwise be lost.

3 0
3 years ago
Write a C++ program that searches for anagrams in a dictionary. An anagram is a word obtained by scrambling the letters of some
Semmy [17]

Answer:

d, avt, car, feet, more, pitch, rome, tac, teef,

Anagrams of b in dictionary

Anagrams of cat in dictionary

avt, cat, tac,

Anagrams of room in dictionary

more, rome,

Anagrams of hello in dictionary

Anagrams of  in dictionary

Explanation:

// FindAnagrams.cpp : it is the starting point of application console.

//

#include <vector>

#include <hash_map>

#include <iostream>

#include <string>

#include <algorithm>

#include <map>

#include <set>

using namespace std;

using namespace stdext;

bool IsAnagrams(string& strA, string& strB)

{

   list<char> listA(strA.begin(), strA.end());

   list<char> listB(strB.begin(), strB.end());

   listA.sort();

   listB.sort();

   return listA == listB;

   // return equal(listA.begin(), listA.end(), listB.begin());

}

string SortChars(string str)

{

   list<char> l(str.begin(), str.end());

   l.sort();

   return string(l.begin(), l.end());

}

set<string> FindAnagrams(list<string>& dict, string findStr)

{

   map<string, set<string>> d;

   typedef pair<string, set<string>> MyPair;

   for(list<string>::const_iterator it = dict.begin(); it != dict.end(); ++it){

       string str(SortChars(*it));

       if(d.find(str) == d.end()){

           set<string> s;

           s.insert(*it);

           d.insert(MyPair(str, s));

       }

       else{

           d[str].insert(*it);

       }

   }

   string sortedStr(SortChars(findStr));

   return d[sortedStr];

}

int main(int argc, char* argv[])

{

   list<string> dict;

   dict.push_back("c");

   dict.push_back("car");

   dict.push_back("avt");

   dict.push_back("taac");

   dict.push_back("feet");

   dict.push_back("teef");

   dict.push_back("rom");

   dict.push_back("more");

   dict.push_back("pit");

   dict.sort();

   cout << "The dictionary: " << endl;

   copy(dict.begin(), dict.end(), ostream_iterator<string>(cout, ", "));

   cout << endl;

   list<string> testCases;

   testCases.push_back("d");

   testCases.push_back("car");

   testCases.push_back("rome");

   testCases.push_back("hell");

   testCases.push_back("");

   for(list<string>::iterator it = testCases.begin(); it != testCases.end(); ++it)

   {

       cout << endl << "Anagrams of " << *it << " in dictionary" << endl;

       set<string> output = FindAnagrams(dict, *it);

       copy(output.begin(), output.end(), ostream_iterator<string>(cout, ", "));

       cout << endl;

   }

   return 0;

}

8 0
3 years ago
With Voice over Internet Protocol (VoIP), _____. a. voicemails cannot be received on the computer b. call quality is significant
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c. Users can have calls forwarded from anywhere in the world

Explanation:

As all you need is the internet, there would be no need to try to sort out roaming as you would on a regular phone line

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I’ll give a you a good guess because I’ve only deleted a whole conversation like 3 times. I honestly feel like eventually there will be a warning on deleting the full conversation or on the other hand nothing may happen; or that it will automatically stop you from deleting the conversation.
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