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will namely add a numberic based algorithm, maybe even the stack .to_string() is too slow also might not do it, grialion already optimized tf out of it
67 lines
1.6 KiB
Rust
67 lines
1.6 KiB
Rust
#[unsafe(no_mangle)]
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fn challenge_usize_duple(b: &[u8]) -> (usize, usize) {
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let mut total1 = 0;
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let mut total2 = 0;
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let s = unsafe { str::from_utf8_unchecked(b) };
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let ranges = s.trim_end().split(',');
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for range in ranges {
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let (start, end) = range.split_once('-').unwrap();
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let (start, end): (usize, usize) = (start.parse().unwrap(), end.parse().unwrap());
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for id in start..=end {
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let mut buf = [0; _];
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let id_s = &fast_to_string(id, &mut buf);
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if is_repeating1(id_s) {
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total1 += id;
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}
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if is_repeating2(id_s) {
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total2 += id;
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}
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}
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}
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(total1, total2)
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}
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const MAX_USIZE_LEN: usize = const { usize::MAX.ilog10() + 1 } as usize;
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/// Output is reversed btw
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fn fast_to_string(mut n: usize, into: &mut [u8; MAX_USIZE_LEN]) -> &mut [u8] {
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let mut len = 0;
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while n != 0 {
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into[len] = (n % 10) as u8 + b'0';
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n /= 10;
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len += 1;
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}
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&mut into[0..len]
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}
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fn is_repeating1(s: &[u8]) -> bool {
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if !s.len().is_multiple_of(2) {
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return false;
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}
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let mid = s.len() / 2;
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let (left, right) = s.split_at(mid);
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left == right
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}
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fn is_repeating2(s: &[u8]) -> bool {
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for sublen in 0..(s.len() / 2) {
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let sublen = sublen + 1;
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if !s.len().is_multiple_of(sublen) {
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continue;
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}
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let pref = &s[0..sublen];
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let is_repeating = s[sublen..].chunks_exact(sublen).all(|c| c == pref);
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if is_repeating {
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return true;
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}
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}
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false
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}
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