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melamori03 [73]
3 years ago
12

Given that the variable named Boy = "Joey" and the variable named Age = 6, create statements that will output the following mess

age:
Congratulations, Joey! Today you are 6 years old.

First, write one statement that will use a variable named Message to store the message. (You will need to concatenate all of the strings and variables together into the one variable named Message. If you don't know what that means, read the section in Chapter 1 about concatenation.)
Then write a second statement that will simply output the Message variable.
Computers and Technology
1 answer:
stepan [7]3 years ago
5 0

Answer:

isn't it already showing it if not put text box

Explanation:

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Data Structure in C++
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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
) Consider the array called myArray declared below that contains and negative integers. Write number of positive integers in the
grigory [225]

Answer:

#include <iostream>

using namespace std;

int main()

{

 

  int arr[]={3,-9,9,33,-4,-5, 100,4,-23};

  int pos;

  int n=sizeof(arr)/sizeof(arr[0]);

  for(int i=0;i<n;i++){

      if(arr[i]>=0){

          pos++;

      }

  }

  cout<<"Number of positive integers is "<<pos<<endl;

  return 0;

}

Explanation:

create the main function in the c++ programming and declare the array with the element. Then, store the size of array by using the formula:

int n=sizeof(arr)/sizeof(arr[0]);

after that, take a for loop for traversing the array and then check condition for positive element using if statement,

condition is array element greater than or equal to zero.

if condition true then increment the count by 1.

this process happen until the condition true

and finally print the count.

6 0
3 years ago
What is an XML-based open standard for exchanging authentication and authorization information and is commonly used for web appl
nalin [4]

Answer:

SAML.

Explanation:

SAML seems to be an accessible standardized XML-based for some of the sharing of validation and verification details and has been generally implemented for the web apps and it seems to be a design that would be SSO. Verification data is shared via XML documentation which are securely signed. So, the following answer is correct according to the given scenario.

4 0
3 years ago
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