Ask the user to enter the number of elements in a list of numbers. take in the numbers from the user and then output the list the user entered . followed by the list the user entered with only the 2nd half sorted. note: you may assume the user will enter a list with an even number of elements. you should sort only the 2nd half the first half is to remain as is.

#Section 1

   lst= []

   lstNo=int(input("Enter even number of elements in List: "))

   for i in range(0, lstNo):

       pr=int(input(": " ))

       lst.append(pr)

   print(lst)

#Section 2

   nlst=[]

   dlst=[]

   n =len(lst)

   i=0

   while (i < n/2):

       nlst.append(lst[i])

       i = i+1

   j = n-1

   while j >= n/2:

       dlst.append(lst[j])

       j = j-1

   dlst.sort()

   for a in range(len(dlst)):

       nlst.append(dlst[a])

   print(nlst)

Explanation:

#section 1:

An empty list is declared to hold the list inputs by the user lst= []

The program then prompts the user to enter an even number of elements that will be contained in the list.

The for loop is used to iterate from zero(0) to the number of elements stated, in order to get the input that is appended to the list.

lastly, the list is printed out.

#Section 2:

In this section two new lists are created to hold one half of the value respectively.

The first list collects the first half of the list using a while loop and stores it, it does not perform any sorting on it.

The second list collects the second half using a while loop and sorts the list using the .sort()  method which arranges elements in an ascending order.

Finally, the second list that has been sorted is added to the first and the result is printed to the screen.

A picture of how the code will run has been attached.

answer; /// i do believe that the correct answer is (the area the player will see in the game screen)

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.