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- ⚡ Fun fact: Did you know that you can calculate the modulo of two numbers $n$ and $d$ by doing modulo reductions by $b$, which you can simply set to a power of two. This allows you to calculate modulo with only bitshifting, additions and a multiplication. Anyways, here is an example:
fn fmod(n: u32, d: u32) -> u32
{
// Check if n is less than d first
if n < d { return n; }
// Find a suitable p for our base
for p in 2..32 {
if 1 << p > n {
// Will always find a p.
return fmod_p(n, d, p)
}
}
unreachable!(); // Mark as unreachable
}
fn fmod_p(n: u32, d: u32, p: u32) -> u32
{
if n < 1 << p {
if n < d { return n; }
if n < d << 1 { return n - d; }
return fmod_p(n, d, p-1);
}
fmod_p(
((1 << p) - d) * (n >> p) + (n & ((1 << p) - 1)),
d,
p
)
}
fn main() {
// Simple fizzbuzz example
for i in 1..101 {
let mut s = String::new();
if fmod(i, 3) == 0 { s.push_str("Fizz"); }
if fmod(i, 5) == 0 { s.push_str("Buzz"); }
if s.is_empty() { s.push_str(&i.to_string()); }
println!("{}", s);
}
}