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Illusion [34]
3 years ago
14

Explain the differences between apple and android

Computers and Technology
2 answers:
marta [7]3 years ago
7 0

Answer:

Apple had better camera and now Android is taking over apple is smoth there is more options on it

Mumz [18]3 years ago
4 0

Answer:

you will think android is better but apple is better

Explanation:

android has a wierd camera

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A programmer must add 1 to the value stored in the int variable sum. One way is given to get you started. Write two more distinc
olganol [36]

Answer:

option 1: sum++

option 2: ++sum

Explanation:

In programming, a variable a container or memory location assigned to a value. The C++ syntax specifies the memory location type of value on the declaration. The major types of variables in C++ are; integer, float, double, char, string, and boolean.

To add 1 to an integer variable, the variable value could be summed with one and assigned to that same variable or '++before' and 'after++' with adds one to the variable before and after the execution of a statement.

4 0
3 years ago
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
If num1 and num2 are the same, print equal.
boyakko [2]

num1 = some value

num2 = some value

if num1 == num2:

   print("equal")

You just need to provide the values for the numbers. I hope this helps!

4 0
3 years ago
Louis has two sets of two gears (Set A and Set B) that he is using to build two different machines. He has all the gears laying
soldi70 [24.7K]

Answer: The larger bottom gear will be cooler than the smaller bottom gear, because the energy that transferred to it was spread out over more molecules.

Explanation: Thermale Energy

7 0
3 years ago
Read 2 more answers
In a traditional systems development environment, security issues usually are less complex than with web-based systems, because
NikAS [45]

Answer:

The answer is "True".

Explanation:

Traditional techniques of application development were focused on pre-development stages. These control system flows through claims to design and upgrades and will then tests and repair are directional there.

  • It enables the user to create and run a machine-based data system.  
  • It uses program and device, that specifically the solutions, that enable manual information to be processed.
6 0
4 years ago
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