Iterative two-ref Fibonacci with modular reduction every step:
let fib_mod n m =
let a = ref 0 in
let b = ref 1 in
for _ = 1 to n do
let c = (!a + !b) mod m in
a := !b;
b := c
done;
!a
The 100th Fibonacci is 354_224_848_179_261_915_075, well past JS
safe-int (2^53). Modular reduction every step keeps intermediate
values within int53 precision so the answer is exact in our
runtime. fib(100) mod 1000003 = 391360.
76 baseline programs total.
