Outer loop executes $\log{n}$ times ...
For $i=n$, inner loop will execute $n$ times.
For $i=\frac{n}{2}$, inner loop executes $\frac{n}{2}$ times. And so on...
count+=1 statement is in the inner loop, and will be executed $n + \frac{n}{2} + \frac{n}{4} + \ldots + 1$ times.
Solving,
$\underbrace{n + \frac{n}{2} + \frac{n}{4} + \ldots + 1}_{\log{n} \text{ terms}}$
$= n \cdot \frac{1- (1/2)^{\log{n}}}{1/2}$ (using GP, first term $=1$ and common ratio $=1/2$)
$=2n(1-\frac1 n) = 2n - 2$
complexity $=O(n)$