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Mariana [72]
3 years ago
13

Uter 110 Marks )

Computers and Technology
1 answer:
Alexxandr [17]3 years ago
4 0

Answer:

Sent me a message that I was paying R319 for the

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You want to get information about DNS request types and DNS query packet content. What kind of DNS logging do you need to enable
ololo11 [35]

Answer:

Option C i.e., Analytic event logging is the correct answer.

Explanation:

The user required to enable Analytic event logging Type of DNS logging to take data or information related to the type of DNS request and also the content packet of the DNS query. By Analytic event logging, Users get data about the forms of DNS requests and the size of the DNS query packets. That's why the following option is correct.

8 0
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Which describes a design theme in PowerPoint? a set of elements that unify the slides and complement the topic a printed handout
chubhunter [2.5K]

a set of elements that unify the slides and complement the topic

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True or false you have no control over who views your social network information
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Data Structure in C++
agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

  int s1 = 0;

  int mid_sz = sz / 2;

  int s2 = mid_sz;

  int e2 = sz;

  int s3 = 0;

  int end3 = sz;

  int i, j;

   

  if (opt_asc==true)

  {

      i = s1;

      j = e2 - 1;

      while (i < mid_sz && j >= s2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (i != mid_sz)

      {

          while (i < mid_sz)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (j >= s2)

      {

          while (j >= s2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

      }

  }

  else

  {

      i = mid_sz - 1;

      j = s2;

      while (i >= s1 && j <e2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

      if (i >= s1)

      {

          while (i >= s1)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

      }

      if (j != e2)

      {

          while (j < e2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

  }

   

  for (i = 0; i < sz; i++)

      *(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

  int* opt = new int[sz];

   

  mergesort(ip, sz, opt, true);

  return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

  if (sz > 1)

  {

      int q = sz / 2;

      mergesort(ip, sz / 2, opt, true);

      mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

      merge(ip, sz, opt, opt_asc);

  }

}

int main()

{

  int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

  int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

  int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

  int *opt;

  cout << "Arays after sorting:\n";

  cout << "Array 1 : ";

  opt = mergesort(arr1, 12);

  for (int i = 0; i < 12; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 2 : ";

  opt = mergesort(arr2, 14);

  for (int i = 0; i < 14; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 3 : ";

  opt = mergesort(arr3, 10);

  for (int i = 0; i < 10; i++)

      cout << opt[i] << " ";

  cout << endl;

  return 0;

}

4 0
4 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
3 years ago
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