giles
ce013fa138
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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.
14 lines
169 B
OCaml
14 lines
169 B
OCaml
let fib_mod n m =
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let a = ref 0 in
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let b = ref 1 in
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for _ = 1 to n do
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let c = (!a + !b) mod m in
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a := !b;
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b := c
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done;
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!a
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;;
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fib_mod 100 1000003
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