Question

PART 2: BLUE BOOK I TO VI ANSWER THESE IN THE BLUE BOOK SHOW ALL WORK for full credit IF NECESSARY (10 pts): Show the program trace IN A TABLE OF VALUES AND OUTPUT of the following using the input stream 5 2-1 10 int A, B, N; A 3; B-6; cout<<"Enter a sequence of values" do cin >> N A=A+N+B; J while (N< 10) B *= 2; cout << "(NAB) = " << "(" << N << ", " << A << ", " << B << ")"くくendl; return 0;

Answer 1

NOTE: I have completed the program for your assignment. Please check and let me know if you have any questions. I will acknowledge back with a response within 24 hours. Thanks for your patience.

I have pasted below a table containing program statements and change in variable values based on statement execution. The bolded statements are copied for easy understanding of execution of the code but really code will not duplicate and extend lines of code.

Output of the program is:
Input:- Enter a sequence of values 5 2 -1 10   
Output:- (N, A, B) = (10,49,20)

Table of Statement and change in Variable values during program execution:-

Statement	N	A	B	Comments
int A, B, N;
A = 3;		3
B = 6;		3	6
cout << "Enter a sequence of values ";
do{
cin >> N;	5	3	6	for input 5
A = A + N + B;	5	14	6
B++;	5	14	7
cin >> N;	2	14	7	for input 2
A = A + N + B;	2	23	7
B++;	2	23	8
cin >> N;	-1	23	8	for input -1
A = A + N + B;	-1	30	8
B++;	-1	30	9
cin >> N;	10	30	9	for input 10
A = A + N + B;	10	49	9
B++;	10	49	10
}while (N < 10);	10	49	10
B *= 2;	10	49	20
cout << "(N, A, B) = "<<"("<< N << "," << A << "," << B << ")" << endl;	10	49	20

Explanation:-
1) The initial values of A is 3 and B is 6
2) The do-while loop reads a sequence of values 5 2 -1 10 in the loop
3) In each iteration the value of A and B changes and is depicted clearly in the above table
4) Once the program comes out from do-while loop the value of (N, A, B ) is (10, 49, 10)
5) With the execution of statement B*=2 the value of B changes from 10 to 20
6) Hence the final values of (N, A, B) is (10, 49, 20)