Consider the following fragment of C++ code, assuming that the constructor Seque

ID: 3761686 • Letter: C

Question

Consider the following fragment of C++ code, assuming that the constructor Sequence creates an empty sequence of integer objects. Recall that division between integers performs truncation (for example, 7/2 = 3). Sequence seq; for (int i = 0; i < n; i++) seq.insertAtRank(i/2, i); a. Assume that the for loop is executed 10 times, that is, n = 10, and show the sequence after each iteration of the loop. b. Draw a schematic illustration of the sequence at the end of the for loop, for a generic number n of iterations.

Explanation / Answer

Hi,

Below is the solution to your problem:

Consider the following code:

Sequence seq;

for (int i = 0; i < n; i++) seq.insertAtRank(i/2, i);

1.Assume that the for loop is executed 10 times, that is, n = 10, and show the sequence after each iteration of the loop.

Solution:If the above for loop is executed 10 times,with n=10,Below is the sequence after each iteration:

operation

The Sequence

Seq.insertAtRank(1, 0)

(-,0)

Seq.insertAtRank(1, 1)

(-,1,0)

Seq.insertAtRank(2, 2)

(-,1,2,0)

Seq.insertAtRank(2, 3)

(-,1,3,2,0)

Seq.insertAtRank(3, 4)

(-,1,3,4,2,0)

Seq.insertAtRank(3, 5)

(-,1,3,5,4,2,0)

Seq.insertAtRank(4, 6)

(-,1,3,5,6,4,2,0)

Seq.insertAtRank(4, 7)

(-,1,3,5,7,6,4,2,0)

Seq.insertAtRank(5, 8)

(-,1,3,5,7,8,6,4,2,0)

Seq.insertAtRank(5, 9)

(-,1,3,5,7,9,8,6,4,2,0)

2.Draw a schematic illustration of the sequence at the end of the for loop, for a generic number n of iterations.

Solution:In order to prove this lets consider the odd and even case seperately,

When n is odd

(-, 1, 3, …., n-6, n-4, n-2, n , n-1, n-3, n-5, …, 2, 0)

When n is even

(-, 1, 3, …., n-5, n-3, n-1, n, n-2, n-4, n-6, …, 2, 0)

Hope that helps...HAPPY ANSWERING!!!!!!!!