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solong [7]
3 years ago
5

JAVA

Computers and Technology
1 answer:
ivolga24 [154]3 years ago
6 0

import java.util.Scanner;

public class JavaApplication57 {

   

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("Enter two numbers:");

       int num1 = scan.nextInt();

       int num2 = scan.nextInt();

       while (num1 <= num2){

          if (num1 %2 == 0){

              System.out.print(num1+" ");

          }

          num1+=1;

       }

   }

   

}

I hope this helps!

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fenix001 [56]
If it’s computer files, that would be System Software and Application software.

“The System Software is the programs that allow the computer to function and access the functionality of the hardware. Systems software sole function is the control of the operation of the computer.

Applications software is the term used for programs that enable the user to achieve specific tasks such as create a document, use a database or produce a spreadsheet.”
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41. All the following software are examples of operating systems EXCEPT
Hunter-Best [27]

Answer:

vista

Explanation:

the other three I have used before

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Define a function typeHistogram that takes an iterator ""it"" (representing a sequence of values of different types) and builds
Alexeev081 [22]

Answer:

def typeHistogram(it,n):

   d = dict()

   for i in it:

       n -=1

       if n>=0:

           if str(type(i).__name__) not in d.keys():

               d.setdefault(type(i).__name__,1)

           else:

               d[str(type(i).__name__)] += 1

       else:

           break

   return list(d.items())

it = iter([1,2,'a','b','c',4,5])

print(typeHistogram(it,7))

Explanation:

  • Create a typeHistogram function that has 2 parameters namely "it" and "n" where "it" is an iterator used to represent a sequence of values of different types while "n" is the total number of elements in the sequence.
  • Initialize an empty dictionary and loop through the iterator "it".
  • Check if n is greater than 0 and current string is not present in the dictionary, then set default type as 1 otherwise increment by 1.
  • At the end return the list of items.
  • Finally initialize the iterator and display the histogram by calling the typeHistogram.
3 0
3 years ago
A computing company is running a set of processes every day. Each process has its own start time and finish time, during which i
Helga [31]

Answer:

Above all else, let us show that this issue has the greedy choice property.

This implies that worldwide ideal arrangement can be gotten by choosing local ideal arrangements. Presently notice the following:

The global optimal solution of the issue requests us to track down the minimum number of times process_check module should be run.

It is likewise referenced that during each running interaction, process_check module should be called atleast once. Along these lines, the minimum possible number of process_check calls happens when process_check is made close to the quantity of the processes that are run.

Along these lines, we see that the global optimal solution is shaped by choosing optimal answer for the nearby advances.

Along these lines, the issue shows greedy choice property. Thus, we can utilize a greedy algorithm for this issue.

The greedy step in this calculation is to postpone the process_check as far as might be feasible. Thus, it should be run towards the finish of each interaction. So the following algorithm can be utilized:

Sort the cycles by utilizing the completion times.

Towards the finish of the 1st process in the arranged list, run the process_check module. Right now any process that is now running is removed the arranged rundown. The main interaction is additionally removed from the sorted list.

Now repeat stage 2, till all processes are finished.

In the above mentioned algorithm, the costliest advance is the step of sorting. By utilizing the optimal sorting algorithm, for example, merge sort, we can achieve it in O(n log n) asymptotic time.

Along these lines, this greedy algorithm additionally takes O(n log n) asymptotic time complexity.

5 0
3 years ago
Using recursion, write a program that asks a user to enter the starting coordinates (row, then column), the ending coordinates (
Luden [163]

Answer:

The program in Python is as follows

def Paths(row,col):

if row ==0 or col==0:

 return 1

return (Paths(row-1, col) + Paths(row, col-1))

row = int(input("Row: "))

col = int(input("Column: "))

print("Paths: ", Paths(row,col))

Explanation:

This defines the function

def Paths(row,col):

If row or column is 0, the function returns 1

<em> if row ==0 or col==0: </em>

<em>  return 1 </em>

This calls the function recursively, as long as row and col are greater than 1

<em> return (Paths(row-1, col) + Paths(row, col-1))</em>

The main begins here

This prompts the user for rows

row = int(input("Row: "))

This prompts the user for columns

col = int(input("Column: "))

This calls the Paths function and prints the number of paths

print("Paths: ", Paths(row,col))

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