Answer:
The correct answer to the following question will be "Θ(n2)
". The further explanation is given below.
Explanation:
If we're to show all the objects that exist from either the first as well as the second vector, though not all of them, so we'll have to cycle around the first vector, so we'll have to match all the objects with the second one.
So,
This one takes:
= 
And then the same manner compared again first with the second one, this takes.
=
Therefore the total complexity,
= Θ(n2)
Answer:
In C++:
int PrintInBinary(int num){
if (num == 0)
return 0;
else
return (num % 2 + 10 * PrintInBinary(num / 2));
}
Explanation:
This defines the PrintInBinary function
int PrintInBinary(int num){
This returns 0 is num is 0 or num has been reduced to 0
<em> if (num == 0) </em>
<em> return 0; </em>
If otherwise, see below for further explanation
<em> else
</em>
<em> return (num % 2 + 10 * PrintInBinary(num / 2));
</em>
}
----------------------------------------------------------------------------------------
num % 2 + 10 * PrintInBinary(num / 2)
The above can be split into:
num % 2 and + 10 * PrintInBinary(num / 2)
Assume num is 35.
num % 2 = 1
10 * PrintInBinary(num / 2) => 10 * PrintInBinary(17)
17 will be passed to the function (recursively).
This process will continue until num is 0
Answer: Hi im Sergeant von im here to help you in any way ma'm;)
I would think it would be the last one ( A SCANNER ) i hope this maybe the right answer if not im sorry at least i gave it a shot
Explanation:
The last one " a scanner " is the one i would choose ma'm it seems logic the the Question and seems too fit in with the rest ;)
Have a good day ma'm
Rubbing two sticks together will cause friction
