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2 years ago

. Reorder the following efficiencies from smallest to largest:

Computers and Technology

1 answer:

alexdok [17]2 years ago

3 0

Answer:

The order of the efficiencies is as following:-

10,000 < 2n < nlog(n) < n5 < n!.

Explanation:

10,000 is constant time whatever will be the size of the problem the efficiency will remain the same.

2n this efficiency is linear it will grow proportionally as the size of the problem increases.

nlog(n) this efficiency is is a bit greater than 2n though it will grow faster than 2n but slower than n2 as the size of the problem increases.

n5 this efficiency is very poor.It is growing very rapidly as the size of the problem increases.

n! is the worst efficiency of them all.

n!=n*(n-1)*(n-2)*(n-3)*(n-4)*.......2*1.

It will grow beanstalk in jack and the beanstalk.

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2 years ago

Which of the following devices are least likely to deny a connection inline when an attack is detected? (select 2)

user100 [1]

Answer: Option (A) & (D) are correct.

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2 years ago

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Zielflug [23.3K]

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3 years ago

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3 years ago

Write a program that converts a number entered in Roman numerals to decimal. your program should consist of a class, say, Roman.

arlik [135]

Answer:

The code is given below in Java with appropriate comments

Explanation:

import java.util.*;

public class RomantoDecimal {

public static void main(String[] args)

{

Scanner SC = new Scanner(System.in);

RomantoDecimal r = new RomantoDecimal();

System.out.println("Enter a Roman number :");

// INPUT A ROMAN NUMBER

String rNum = SC.next();

// CALL convertToDecimal FOR CONVERSION OF ROMAN TO DECIMAL

r.convertToDecimal(rNum);

}

// M=1000, D=500, C=100, L=50, X=10, V=5, I=1

public void convertToDecimal(String roamNo)

{

int number = 0;

// TEACK EACH DIGIT IN THE GIVEN NUMBER IN REVERSE ORDER

for (int i = roamNo.length() - 1; i >= 0; i--)

{

// FIND OUT WHETHER IT IS 'M' OR NOT

if (roamNo.charAt(i) == 'M')

{

if (i != 0)

{ // CHECK WHETHER THERE IS C BEFORE M

if (roamNo.charAt(i - 1) == 'C')

{

number = number + 900;

i--;

continue;

}

number = number + 1000;

}

// FIND OUT WHETHER IT IS 'D' OR NOT

else if (roamNo.charAt(i) == 'D')

{

if (i != 0)

{

// CHECK WHETHER THERE IS C BEFORE D

if (roamNo.charAt(i - 1) == 'C')

{

number = number + 400;

i--;

continue;

}

number = number + 500;

}

// FIND OUT WHETHER IT IS 'C' OR NOT

else if (roamNo.charAt(i) == 'C')

{

if (i != 0)

{

if (roamNo.charAt(i - 1) == 'X')

{

number = number + 90;

i--;

continue;

}

number = number + 100;

}

else if (roamNo.charAt(i) == 'L')

{

if (i != 0)

{

if (roamNo.charAt(i) == 'X')

{

number = number + 40;

i--;

continue;

}

number = number + 50;

}

// FIND OUT WHETHER IT IS 'X' OR NOT

else if (roamNo.charAt(i) == 'X')

{

if (i != 0)

{

if (roamNo.charAt(i - 1) == 'I')

{

number = number + 9;

i--;

continue;

}

number = number + 10;

}

// FIND OUT WHETHER IT IS 'V' OR NOT

else if (roamNo.charAt(i) == 'V')

{

if (i != 0)

{

if (roamNo.charAt(i - 1) == 'I')

{

number = number + 4;

i--;

continue;

}

number = number + 5;

}

else // FIND OUT WHETHER IT IS 'I' OR NOT

{

number = number + 1;

}

}// end for loop

System.out.println("Roman number: " + roamNo + "\nIts Decimal Equivalent: " + number);

}

3 0

3 years ago