Help me solve the following algorithm. In this case, it is a vector of 20 positions. My doubt is what values would be stored in vector A?
for (i=1, A[0]=1; i< A.lenght; i++)
A[i] = A[i-1]*2;
Correct answer:
a - Elements of type 3i for i
b - Elements of type 2i for i
c - Elements of type i2 for i
d - Elements of type ii for i
Could you explain the correct answer to me?
Well, it's easy, let's break it down.
Breaking the for i = 1, A[0] = 1 . This statement initializes i as 0 and the first position of A as 1 .
After that it's just normal. The condition, with% that is "execute this while i < A.lenght is less than the length of the array." And the i which is the% in 1 every loop .
i++ ) code is as follows.
The position i of the array will get the value of the previous position of array multiplied by 2. That is, answers B and C are correct. >