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aliya0001 [1]
3 years ago
13

Using OOP, write a C++ program that will read in a file of names. The file is called Names.txt and should be located in the curr

ent directory of your program. Read in and store the names into an array of 30 names. Sort the array using the selection sort or the bubblesort code found in your textbook. List the roster of students in ascending alphabetical order. Projects using global variables or all code is in main() will result in a grade submission of 0. List of names in Names.txt: Jackie Sam Bill Tom Mary Paul Zev Barb John Sharon Dana Dominic Steven Padro Katey Kathy Darius Angela Mimi Jesse Kenny Lynn Hector Brittany Jenn Joe Chloe Geena Sylvia Dean
Engineering
2 answers:
Katen [24]3 years ago
7 0

Answer: This doesn't work fully, but it's a start. Good Luck

#include <iostream>

#include <fstream>

#include <string>

#include <cstdlib>

using namespace std;

class People

{

private:

const static int SIZE = 30;  

string names[SIZE];  

int birth_years[SIZE];  

int count;  

void sort();  

void display();  

public:

People();

void simulate();

};

People::People()

{

count = 0;

// open both files

ifstream namesFile, birthyearsFile;

namesFile.open("Names.txt");

birthyearsFile.open("BirthYear.txt");

while (!namesFile.eof() && !birthyearsFile.eof() && count < SIZE)

{

 getline(namesFile, names[count]);  

 birthyearsFile >> birth_years[count];  

 count++;  

}

// files open failed, exit the program

if (namesFile.fail() || birthyearsFile.fail())

{

 cout << "Unable to open input file(s). Terminating" << endl;

 exit(1);

}

//close the files

namesFile.close();

birthyearsFile.close();

sort();

display();

}

void People::sort()

{

for (int i = 0; i < count - 1; i++)

{

 for (int j = 0; j < count - 1 - i; j++)

 {

  if (names[j] > names[j + 1])

  {

   string tempName = names[j];

   names[j] = names[j + 1];

   names[j + 1] = tempName;

   int tempYear = birth_years[j];

   birth_years[j] = birth_years[j + 1];

   birth_years[j + 1] = tempYear;

  }

 }

}

}

void People::display()

{

cout << "Alphabetical Roster of Names: " << endl;

for (int i = 0; i < count; i++)

{

 cout << names[i] << "\t" << birth_years[i] << endl;

}

cout << endl;

}

void People::simulate()

{

int year;

cout << endl << "Names by Birth Year" << endl;

// input the birth year

cout << "Please enter the birth year: ";

cin >> year;

// loop that continues until valid input has been read

while (cin.fail() || year < 1995 || year > 2005)

{

 cin.clear();  

 cin.ignore(100, '\n');  

 cout << "Invalid birth year entered, try again: ";  

 cin >> year;

}

bool found = false;  

for (int i = 0; i < count; i++)

{

 if (birth_years[i] == year)  

 {

  if (!found)  

  {

   cout << endl << "For the birth year of " << year << ":" << endl;

   found = true;

  }

  // display the name

  cout << names[i] << endl;

 }

}

// no name with birth year found

if (!found)

 cout << endl << "No names with the birth year " << year << "." << endl;

cout << "End of results" << endl;

}

int main()

{

People people;  

people.simulate();  

return 0;

}

Explanation:

Tanzania [10]3 years ago
5 0
Not really sure sorry for not being able to help
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Please answer fast. With full step by step solution.​
lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

Now condense <em>f(z)</em> into one series:

\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}

\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}

So, the inverse <em>Z</em> transform of <em>f(z)</em> is \boxed{6+2^{2-n}}.

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Answer:

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Explanation:

Data:

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T₂ = 75 °C

Calculations:

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ΔT = 75 °C - 25 °C = 50 °C  = 50 K

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2. Energy in British thermal units

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Algorithm of the Nios II assembly program.

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

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  • DE2-115_SW11
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  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

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