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Semmy [17]
3 years ago
14

Suppose that a computer can run an algorithm on a problem of size 1,024 in time t. We do not know the complexity of the algorith

m. We note that when we run the same algorithm on a computer 8 times faster, in the same time t, we can solve a problem of size 8,192. We also note that when we run the same algorithm on a computer 8 times slower, in the same time t, we can solve a problem of size 128.
Computers and Technology
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

Time Complexity of Problem - O(n)

Explanation:

When n= 1024 time taken is t. on a particular computer.

When computer is 8 times faster in same time t , n can be equal to 8192. It means on increasing processing speed input grows linearly.

When computer is 8 times slow then with same time t , n will be 128 which is (1/8)th time 1024.

It means with increase in processing speed by x factor time taken will decrease by (1/x) factor. Or input size can be increased by x times. This signifies that time taken by program grows linearly with input size n. Therefore time complexity of problem will be O(n).

If we double the speed of original machine then we can solve problems of size 2n in time t.

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Develop a C program that calculates the final score and the average score for a student from his/her (1)class participation, (2)
Ghella [55]

Answer:

#include <iomanip>

#include<iostream>

using namespace std;

int main(){

char name[100];

float classp, test, assgn, exam, prctscore,ave;

cout<<"Student Name: ";

cin.getline(name,100);

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; }

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){  cout<<"Test: "; cin>>test; }

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; }

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; }

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; }

ave = (int)(classp + test + assgn + exam + prctscore)/5;

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

return 0;}

Explanation:

The required parameters such as cin, cout, etc. implies that the program is to be written in C++ (not C).

So, I answered the program using C++.

Line by line explanation is as follows;

This declares name as character of maximum size of 100 characters

char name[100];

This declares the grading items as float

float classp, test, assgn, exam, prctscore,ave;

This prompts the user for student name

cout<<"Student Name: ";

This gets the student name using getline

cin.getline(name,100);

This prompts the user for class participation. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Class Participation: "; cin>>classp; </em>

<em> while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; } </em>

This prompts the user for test. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Test: "; cin>>test; </em>

<em> while(test <0 || test > 100){  cout<<"Test: "; cin>>test; } </em>

This prompts the user for assignment. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Assignment: "; cin>>assgn; </em>

<em> while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; } </em>

This prompts the user for examination. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Examination: "; cin>>exam; </em>

<em> while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; } </em>

This prompts the user for practice score. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Practice Score: "; cin>>prctscore; </em>

<em> while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; } </em>

This calculates the average of the grading items

ave = (int)(classp + test + assgn + exam + prctscore)/5;

This prints the calculated average

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

8 0
3 years ago
Have you ever used a device that relies solely on the cloud?
Morgarella [4.7K]
Yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeesss i have
8 0
3 years ago
Where are parameters such boot order, processor voltage, motherboard raid, hardware virtualization support, and overclocking con
ololo11 [35]

This is configured on the BIOS Setup. During computer start-up usually for windows system this can be accessed via pressing on ESC or Function keys on immediate start up.

6 0
2 years ago
A technician is buying a PC that will host three VMs running at the same time with the current configuration. The technician bel
raketka [301]

Answer:

B. SSD

Explanation:

VMs or virtual machines are virtually created environment for multiple operating system on a host operating system. The host operating system has an extension called Hyper-V. The hyper-V manager helps to manage the virtual machine's activities.

The VMs can hold applications in their respective containers, which requires a partition of the storage. So when more VMs are configured, more storage memory is required to store their individual data. The VMs in a computer system can share a network interface card.

7 0
3 years ago
Write a recursive method called repeat that accepts a string s and an integer n as parameters and that returns s concatenated to
svp [43]

Answer:

public static String repeat(String text, int repeatCount) {

   if(repeatCount < 0) {

       throw new IllegalArgumentException("repeat count should be either 0 or a positive value");

   }

   if(repeatCount == 0) {

       return "";

   } else {

       return text + repeat(text, repeatCount-1);

   }

}

Explanation:

Here repeatCount is an int value.

at first we will check if repeatCount is non negative number and if it is code will throw exception.

If the value is 0 then we will return ""

If the value is >0 then recursive function is called again untill the repeatCount value is 0.

6 0
3 years ago
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