Answer:
PART ONE
- import java.util.Scanner;
- public class CountToLimit {
- public static void main(String[] args) {
- Scanner scnr = new Scanner(System.in);
- int countLimit = 0;
- int printVal = 0;
- // Get user input
- System.out.println("Enter Count Limit");
- countLimit = scnr.nextInt();
- do {
- System.out.print(printVal + " ");
- printVal = printVal + 1;
- } while ( printVal<=countLimit );
- System.out.println("");
- return;
- }
- }
PART TWO
- import java.util.Scanner;
- public class NumberPrompt {
- public static void main (String [] args) {
- Scanner scnr = new Scanner(System.in);
- System.out.print("Your number < 100: ");
- int userInput = scnr.nextInt();
- do {
- System.out.print("Your number < 100: ");
- userInput = scnr.nextInt();
- }while (userInput>=100);
- System.out.println("Your number < 100 is: " + userInput);
- return;
- }
- }
Explanation:
In Part one of the question, The condition for the do...while loop had to be stated this is stated on line 14
In part 2, A do....while loop that will repeatedly prompt user to enter a number less than 100 is created. from line 7 to line 10
I think I am not sure that it is probably b
Answer:
p(x,n)
1. if(n==0) [if power is 0]
2. then result =1.
3.else
4. { result=1.
5. for i=1 to n.
6. { result = result * x. } [each time we multiply x once]
7. return result.
8. }
Let's count p(3,3)
3
0, so come to else part.
i=1: result = result *3 = 3
i=2: result = result *3 = 9
i=2: result = result *3 = 27
Explanation:
here the for loop at step 4 takes O(n) time and other steps take constant time. So overall time complexity = O(n)
Answer:
select Control Panel, then Network Protocols, then Internet
