time complexity of recursive program

Question

What is the time complexity of the following code snippet? Assume "statement" takes $O(1)$ time.

int x=0;
int A(n)
{

   statement;
   if(n==1)return 1;
   
   else
   {
     x + = 4A(n/2)+n^{2}
      
      return (x);
    }

}

$A)\theta(n^{2}logn)$             $B)\theta(logn)$                $C)\theta(n^{2})$                $D)\theta(nlogn)$

Comments

Recursion relation is T(n)=T(n/2) +O(1)

This will give O(logn)

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yes thanks

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@shubhanshu @ soumya

Can you please give one example where we will write as 4 × T(n/2) + O(1)?

So that I can understand the difference when to multiply recurrence relation and when it is a only constant.

Thanks

Answer 1

4A(n/2) numerical multiplication takes O(1) time

so recurrence relation 

A(n) = A(n/2) + O(1) + O(1)

a = 1 , b = 2 , k = 0 

b^k = 2⁰ = 1

a = b

p = 0 > -1

A(n) = θ(n logb_a* logn)

        = θ(logn)

Correct answer = B

Comments

But in usual scenarios, log is taken of aT(n/b) + f(n)

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log_a b. How did you know its constant