Fastest Fibonacci Sequence/Number Computation

The fastest way (in O(n)) of calculating Fibonacci sequence is by using matrix multiplication approach using following relation.
\[\begin{bmatrix}0 & 1 \\ 1 & 1 \end{bmatrix}^n = \begin{bmatrix} F_{n – 1} & F_n \\ F_n & F_{n + 1}\end{bmatrix}\]
Calculating $F_{34}$ is, therefore, multiplying the matrix $\begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix}$ 34 times. The $a_{01}$ or $a_{10}$ gives the right fibonacci number. In fact $F_{34}$ can be calculated in less than 34 multiplication in following away.
\[ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^2 = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix} X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix} \\
\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^4 = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^2 X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^2 \\
\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^8 = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^4 X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^4 \\
\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{16} = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^8 X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^8 \\
\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{32} = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{16} X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{16} \\
\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{34} = \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{32} X \ \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}^{2} \\\]

This means, the multiplication can be done in logarithmic time. The C code for this computation is given below (also available on GitHub).

#include 
#include 
 
void mat_product(unsigned long long a[], unsigned long long b[], unsigned long long result[]) {
 // assuming matrix is always 2 x 2
 result[0] = a[0] * b[0] + a[1] * b[2];
 result[1] = a[0] * b[1] + a[1] * b[3];
 result[2] = a[2] * b[0] + a[3] * b[2];
 result[3] = a[2] * b[1] + a[3] * b[3];
}
 
int main(int argc, char* argv[]) {
 int n, i, temp, len;
 unsigned long long bin[1000], **fib;
 unsigned long long result[4];
 if (argc != 2) {
  printf("Usage: outputfile n\n");
  exit(1);
 }
 
 n = atoi(argv[1]);
 temp = n;
 len = 1;
 while(n / 2 > 0) {
  bin[len - 1] = n % 2;
  n = n/2;
  len++;
 }
 if (n == 1) {
  bin[len - 1] = 1;
}
 
 fib = (unsigned long long**) malloc(sizeof(unsigned long long*) * len);
 for (i = 0; i  %llu\n", temp, result[1]);
 
 return 0;
}