Compute the classic Fibonacci running times quantitatively via the third variable

//Util.h
static unsigned long long factorialLoops;
long double Fibonacci35(int len);

//Util.cpp
unsigned long long Util::factorialLoops=0;

long double Util::Fibonacci35(int i)
{
    Util::factorialLoops++;
    if(i==0 || i==1)
    {
        return 1;
    }
    else
    {
        return Fibonacci35(i-1)+Fibonacci35(i-2);
    }
}

//main.cpp
void recursion16(int i)
{
    Util ul;
    long double fib = 0;
    for (int x = 0; x < i; x++)
    {
        fib = ul.Fibonacci35(x);
        cout <<fixed<< "Fib of " << x << " is " << fib <<",fibonacci loops is "<<Util::factorialLoops<<endl;
    }
    cout<<endl<<"The recursion loops is "<<Util::factorialLoops<<endl;
}


int main(int args, char **argv)
{
    recursion16(atoi(argv[1]));
    return 0;
}

Compile and run as below command which will cost more time definitely

./h1 100

The following snapshot can demonstrate everything

 

 When run the 54 it  call the classic Fibonacci method more than 730 billion times.