Question

Questions on C++, I am not 100% sure on answers, so need to clarify. Simple explanation next to the question would be nice as well.

In mathematic notation, which is equivalent to the following statement when writing a

pseudocode?

x = 4y

A. x == 4y

B. x := 4y

C. x -> 4y

D. 4y = x

5. Given the following pseudocode, the final value of x is __.

Start

int x = 5;

for i in 1 to 3 do

x = x + i

End

A. 5

B. 7

C. 9

D. 11

E. 13

6. Given the following pseudocode, what will be the final output if m=24 and n=3?

r←m%n

while r !=0 do

m←n

n←r

r←m%n

return n

A. 3

B. 12

C. 6

D. 15

E. 8

7. Given the following code segment, what is the total number of executions?

for (int i=1; i<17; i++)

{

cout << (5 / i) * 3;

}

A. 17

B. 18

C. 34

D. 36

8. The following is an example of __ time complexity.

int findRemainder(int x, int y)

{

int r = x % y;

return r;

}

A. constant

B. linear

C. quadratic

D. exponential

E. logarithmic

10. Given the following code segment, the total number of executions is __.

int x = 3;

int y = 4;

for i = 1 to n do

x = x + i

y = y - i

print x

print y

A. 2n

B. 4n + 2

C. 6n + 4

D. 8n + 6

E. 5n + 6

Answer 1

Q1 Ans) B
   In pseudocode format, you can use = or := for assignment operation.

   Here in Given Options.
       A) x == 4y ; == is used for equality checking.
       B) x:= 4y ; := is an assignment operation(correct)
       C) x-> 4y ; -> denotes Replaced by
       D) 4y = x ; 4y is assigned to x

Q2 Ans) D
   Initially x is assigned to 5
   for i = 1 => x = x + 1 ; x= 5 + 1 = 6
   i = 2 => x = x + 2 ; x= 6 + 2 = 8
   i = 3 => x = x + 3 ; x= 8 + 3 = 11

Q3 Ans)A
  
   Initially, r = 24 % 3 = 0
   Since r == 0 it doesn’t enter into loop and returns simply n value. i.e,, 3

Q4 Ans) None of Them.
   The loop executes for i=1 to i=16 only. Since i<17 doesn’t iterate for i=17. The answer is 16.

Q5 Ans) A
   It just requires constant amount i.e,, O(1) time complexity as it just calculates the remainder.
  
   int findRemainder(int x, int y)
   {
       int r = x % y; -> O(1)
       return r; -> O(1)
   }

Q6 Ans) A
   The total number of statements are 2 and The loop executes for n times.
   So, it executes 2n times.