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Viefleur [7K]
3 years ago
15

In this lab, you complete a C++ program that swaps values stored in three int variables and determines maximum and minimum value

s. The C++ file provided for this lab contains the necessary variable declarations, as well as the input and output statements. You want to end up with the smallest value stored in the variable named first and the largest value stored in the variable named third. You need to write the statements that compare the values and swap them if appropriate. Comments included in the code tell you where to write your statements.InstructionsEnsure the Swap.cpp file is open in your editor.Write the statements that test the first two integers, and swap them if necessary.Write the statements that test the second and third integer, and swap them if necessary.Write the statements that test the first and second integers again, and swap them if necessary.Execute the program by clicking the "Run Code" button at the bottom of the screen using the following sets of input values.101 22 -23630 1500 921 2 2Provided code:// Swap.cpp - This program determines the minimum and maximum of three values input by// the user and performs necessary swaps.// Input: Three int values.// Output: The numbers in numerical order.#include using namespace std;int main(){ // Declare variables int first = 0; // First number int second = 0; // Second number int third = 0; // Third number int temp; // Used to swap numbers const string SENTINEL = "done"; // Named constant for sentinel value string repeat; bool notDone = true; //loop control // Get user input cout << "Enter first number: "; cin >> first; cout << "Enter second number: "; cin >> second; cout << "Enter third number: "; cin >> third; while(notDone == true){ // Test to see if the first number is greater than the second number // Test to see if the second number is greater than the third number // Test to see if the first number is greater than the second number again // Print numbers in numerical order cout << "Smallest: " << first << endl; cout << "Next smallest: " << second << endl; cout << "Largest: " << third << endl; cout << "Enter any letter to continue or done to quit: "; cin >> repeat; if (repeat == SENTINEL){ notDone = false; } else { cout << "Enter first number: "; cin >> first; cout << "Enter second number: "; cin >> second; cout << "Enter third number: "; cin >> third; } return 0;} // End of main function
Computers and Technology
1 answer:
Sergio039 [100]3 years ago
6 0

Answer:

Following are the code to the given question:

#include <iostream>//header file

using namespace std;

int main()//main method

{

int first = 0,second = 0,third = 0;//defining integer variable  

int temp; //defining integer variable

const string SENTINEL = "done"; // defining a string variable as constant  

string repeat;// defining a string variable  

bool notDone = true; //defining bool variable

cout << "Enter first number: ";//print message

cin >> first;//input value

cout << "Enter second number: ";//print message

cin >> second;//input value

cout << "Enter third number: ";//print message

cin >> third;//input value

while(notDone == true)//defining a loop to check the value  

{

if(first > second)//use if to compare first and second value

{

int temp = first;//defining temp to hold first value

first = second;//holding second value in first variable

second = temp;//holding temp value in second variable

}

if(second > third)//use if to compare second and third value

{

int temp = second;//defining temp to hold second value

second = third;//holding second value in third variable

third = temp;//holding temp value in third variable

}

cout << "Smallest: " << first << endl;//print smallest value

cout << "Next smallest: " << second << endl;//print Next smallest value

cout << "Largest: " << third << endl;////print Largest value

cout << "Enter any letter to continue or done to quit: ";//print message

cin >> repeat;//holding string value

if (repeat == SENTINEL)

{

notDone = false;//holding bool value

}  

else //else block

{

cout << "Enter first number: ";//print message

cin >> first;//input value

cout << "Enter second number: ";//print message

cin >> second;//input value

cout << "Enter third number: ";//print message

cin >> third;//input value

}

return 0;

}

}

Output:

Please find the attached file.

Explanation:

  • Inside the main method Four integer variable "first, second, third, and temp" is declared in which first three variable is used for input value and temp is used to compare value.
  • In thew next step, two string variable "SENTINEL and repeat" is declared in which "SENTINEL" is constant and a bool variable "notDone" is declared.
  • After input the value from the user-end a loop is declared that compare and swap value and print its value.

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Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
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       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
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So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
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So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
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So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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2 years ago
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Explanation:

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I will provide you 1 example:-

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Timely

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