By the for loops ==> 10.$\frac{n}{2}$ ===> C$_1$.$\frac{n}{2}$.
$\text{return 4}*\color{red}{\text{algorithm}(\frac{n}{2})}*\color{green}{\text{algorithm}(\frac{n}{2})}+\color{magenta}{\text{algorithm}(\frac{n}{2})}*\color{blue}{\text{algorithm}(\frac{n}{2})} $
Total four times(red,blue,green,magenta) you are recalling the original problem with the size of half of input.
Note that, the return value is different, and the calls are different.
Totally ===> T(n) = C$_1$.$\frac{n}{2}$ + 4.T($\frac{n}{2}$) = 4.T($\frac{n}{2}$) + n