Zee

Zee is my Dutch dialect of C. Since Dutch has compound words, so does Zee: “const char **” becomes vasteletterverwijzingsverwijzing, not vaste letter verwijzing verwijzing, which would be incorrect. A pointer to a long long unsigned int is, obviously, a zeergrootnatuurlijkgetalverwijzing.

#ingesloten "zee.kop"
#ingesloten "algemeen.kop"

besloten getal
splits(
    zeer groot natuurlijk getal x,
    zeergrootnatuurlijkgetalverwijzing a,
    zeergrootnatuurlijkgetalverwijzing b)
{
	zeer groot natuurlijk getal m;
	getal n, g;

	mits (!x) lever 0;
	voor (n=0, m=1; m<=x;  n++, m*=10) ; mits (n%2) lever 0;
	voor (g=0, m=1; g<n/2; g++, m*=10) ;
	volg a = x/m;
	volg b = x%m; lever 1;
}

#definieer GEHL (1024*1024)
besloten zeer groot natuurlijk getal geh[GEHL][76];

besloten zeer groot natuurlijk getal
afdalen(zeer groot natuurlijk getal w, getal n)
{
	zeer groot natuurlijk getal a,b;

	mits (n<1 ) lever 1;
	mits (w==0) lever afdalen(1, n-1);
	mits (w<10) lever afdalen(w*2024, n-1);
	mits (w<GEHL && geh[w][n])        lever geh[w][n];
	mits (!splits(w, naar a, naar b)) lever afdalen(w*2024, n-1);

	lever w<GEHL ? geh[w][n] =
	    afdalen(a, n-1) + afdalen(b, n-1) :
	    afdalen(a, n-1) + afdalen(b, n-1);
}

getal
aanvang(getal parametersom, vasteletterverwijzingsverwijzing parameters)
{
	zeer groot natuurlijk getal d1=0,d2=0, waarde;

	mits (parametersom > 1)
		VERWERP(heropen(parameters[1], "r", standaardinvoer));

	zolang (invorm(" %llu", &waarde) == 1) {
		d1 += afdalen(waarde, 25);
		d2 += afdalen(waarde, 75);
	}

	uitvorm("11: %llu %llu\n", d1, d2);
	lever 0;
}

And of course we don’t have a Makefile but a Maakbestand:

alles: €{DAGEN}

schoon:
	€{WIS} -f €{DAGEN} *.o

...

.TWIJFELACHTIG:	alles schoon oplossingen
.UITGANGEN:	.zee .o

.zee.o:
	€{ZEE} €{VOORWERKVLAG} €{ZEEVLAG} -o €@ -c €<

Uiua

I thought this was going to be trivial to implement in Uiua, but I managed to blow the stack, meaning I had to set an environment variable in order to get it to run. That doesn’t work in Uiua Pad, so for any counts larger than 15 you need to run it locally. Built-in memoisation though so that’s nice.

# NB Needs env var UIUA_RECURSION_LIMIT=300
Next  ← ⍣([1] °0|[×2024] °1◿2⧻°⋕.|⍜°⋕(↯2_∞))
Count ← |2 memo(⨬(1|/+≡Count¤-1⊙Next)≠0.) # rounds, stone
≡(&p/+≡Count¤: ⊜⋕⊸≠@\s "125 17")[25 75]

And now we get into the days where caching really is king. My first attempt didn’t go so well, I tried to handle the full list result as one cache step, instead of individually caching the result of calculating each stone per step.

I think my original attempt is still calculating at home, but I finished up this much better version on the trip to work.
All hail public transport.

List<long> stones = new List<long>();
public void Input(IEnumerable<string> lines)
{
  stones = string.Concat(lines).Split(' ').Select(v => long.Parse(v)).ToList();
}

public void Part1()
{
  var expanded = TryExpand(stones, 25);

  Console.WriteLine($"Stones: {expanded}");
}
public void Part2()
{
  var expanded = TryExpand(stones, 75);

  Console.WriteLine($"Stones: {expanded}");
}

public long TryExpand(IEnumerable<long> stones, int steps)
{
  if (steps == 0)
    return stones.Count();
  return stones.Select(s => TryExpand(s, steps)).Sum();
}
Dictionary<(long, int), long> cache = new Dictionary<(long, int), long>();
public long TryExpand(long stone, int steps)
{
  var key = (stone, steps);
  if (cache.ContainsKey(key))
    return cache[key];

  var result = TryExpand(Blink(stone), steps - 1);
  cache[key] = result;
  return result;
}

public IEnumerable<long> Blink(long stone)
{
  if (stone == 0)
  {
    yield return 1;
    yield break;
  }
  var str = stone.ToString();
  if (str.Length % 2 == 0)
  {
    yield return long.Parse(str[..(str.Length / 2)]);
    yield return long.Parse(str[(str.Length / 2)..]);
    yield break;
  }
  yield return stone * 2024;
}

Rust

Part 2 is solved with recursion and a cache, which is indexed by stone numbers and remaining rounds and maps to the previously calculated expansion size. In my case, the cache only grew to 139320 entries, which is quite reasonable given the size of the result.

use std::collections::HashMap;

fn parse(input: String) -> Vec<u64> {
    input
        .split_whitespace()
        .map(|w| w.parse().unwrap())
        .collect()
}

fn part1(input: String) {
    let mut stones = parse(input);
    for _ in 0..25 {
        let mut new_stones = Vec::with_capacity(stones.len());
        for s in &stones {
            match s {
                0 => new_stones.push(1),
                n => {
                    let digits = s.ilog10() + 1;
                    if digits % 2 == 0 {
                        let cutoff = 10u64.pow(digits / 2);
                        new_stones.push(n / cutoff);
                        new_stones.push(n % cutoff);
                    } else {
                        new_stones.push(n * 2024)
                    }
                }
            }
        }
        stones = new_stones;
    }
    println!("{}", stones.len());
}

fn expansion(s: u64, rounds: u32, cache: &mut HashMap<(u64, u32), u64>) -> u64 {
    // Recursion anchor
    if rounds == 0 {
        return 1;
    }
    // Calculation is already cached
    if let Some(res) = cache.get(&(s, rounds)) {
        return *res;
    }

    // Recurse
    let res = match s {
        0 => expansion(1, rounds - 1, cache),
        n => {
            let digits = s.ilog10() + 1;
            if digits % 2 == 0 {
                let cutoff = 10u64.pow(digits / 2);
                expansion(n / cutoff, rounds - 1, cache) +
                expansion(n % cutoff, rounds - 1, cache)
            } else {
                expansion(n * 2024, rounds - 1, cache)
            }
        }
    };
    // Save in cache
    cache.insert((s, rounds), res);
    res
}

fn part2(input: String) {
    let stones = parse(input);
    let mut cache = HashMap::new();
    let sum: u64 = stones.iter().map(|s| expansion(*s, 75, &mut cache)).sum();
    println!("{sum}");
}

util::aoc_main!();

Also on github

Haskell

Yay, mutation! Went down the route of caching the expanded lists of stones at first. Oops.

import Data.IORef
import Data.Map.Strict (Map)
import Data.Map.Strict qualified as Map

blink :: Int -> [Int]
blink 0 = [1]
blink n
  | s <- show n,
    l <- length s,
    even l =
      let (a, b) = splitAt (l `div` 2) s in map read [a, b]
  | otherwise = [n * 2024]

countExpanded :: IORef (Map (Int, Int) Int) -> Int -> [Int] -> IO Int
countExpanded _ 0 = return . length
countExpanded cacheRef steps = fmap sum . mapM go
  where
    go n =
      let key = (n, steps)
          computed = do
            result <- countExpanded cacheRef (steps - 1) $ blink n
            modifyIORef' cacheRef (Map.insert key result)
            return result
       in readIORef cacheRef >>= maybe computed return . (Map.!? key)

main = do
  input <- map read . words <$> readFile "input11"
  cache <- newIORef Map.empty
  mapM_ (\steps -> countExpanded cache steps input >>= print) [25, 75]

Haskell

import Data.Monoid
import Control.Arrow

data Tree v = Tree (Tree v) v (Tree v)

-- https://stackoverflow.com/questions/3208258
memo1 f = index nats
  where
    nats = go 0 1
    go i s = Tree (go (i + s) s') (f i) (go (i + s') s')
      where
        s' = 2 * s
    index (Tree l v r) i
        | i < 0 = f i
        | i == 0 = v
        | otherwise = case (i - 1) `divMod` 2 of
            (i', 0) -> index l i'
            (i', 1) -> index r i'

memo2 f = memo1 (memo1 . f)

blink = memo2 blink'
  where
    blink' c n
        | c == 0 = 1
        | n == 0 = blink c' 1
        | even digits = blink c' l <> blink c' r
        | otherwise = blink c' $ n * 2024
      where
        digits = succ . floor . logBase 10 . fromIntegral $ n
        (l, r) = n `divMod` (10 ^ (digits `div` 2))
        c' = pred c

doBlinks n = getSum . mconcat . fmap (blink n)
part1 = doBlinks 25
part2 = doBlinks 75

main = getContents >>= print . (part1 &&& part2) . fmap read . words

Python

Part 1: ~2 milliseconds, Part 2: ~32 milliseconds, Total Time: ~32 milliseconds
You end up doing part 1 at the same time as part 2 but because of how Advent of Code works, you need to rerun the code after part 1 is solved. so Part 2 is technically total time.

from time import perf_counter_ns

transform_cache = {'0': [ '1']}

def transform(current_stone):
    if len(current_stone) % 2 == 0:
        mid = len(current_stone) // 2
        res = [current_stone[:mid], current_stone[mid:].lstrip('0').rjust(1,'0')]
    else:
        res = [str(int(current_stone) * 2024)]
    transform_cache[current_stone] = res
    return res

def main(initial_stones):
    stones_count = {}
    for stone in initial_stones.split():
        stones_count[stone] = stones_count.get(stone, 0) + 1
    part1 = 0
    for i in range(75):
        new_stones_count = {}
        for stone, count in stones_count.items():
            for r in (transform_cache.get(stone) if stone in transform_cache else transform(stone)):
                new_stones_count[r] = new_stones_count.get(r, 0) + count
        stones_count = new_stones_count
        if i == 24:
            part1 = sum(stones_count.values())
    return part1,sum(stones_count.values())

if __name__ == "__main__":
    with open('input', 'r') as f:
        input_data = f.read().replace('\r', '').replace('\n', '')
    start_time, part_one, part_two = perf_counter_ns(),*main(input_data)
    stop_time = perf_counter_ns() - start_time
    time_len = min(9, ((len(str(stop_time))-1)//3)*3)
    time_conversion = {9: 'seconds', 6: 'milliseconds', 3: 'microseconds', 0: 'nanoseconds'}
    print(f"Part 1: {part_one}\nPart 2: {part_two}\nProcessing Time: {stop_time / (10**time_len)} {time_conversion[time_len]}")

Nim

Runtime: 30-40 ms
I’m not very experienced with recursion and memoization, so this took me quite a while.

Edit: slightly better version

template splitNum(numStr: string): seq[int] =
  @[parseInt(numStr[0..<numStr.len div 2]), parseInt(numStr[numStr.len div 2..^1])]

template applyRule(stone: int): seq[int] =
  if stone == 0: @[1]
  else:
    let numStr = $stone
    if numStr.len mod 2 == 0: splitNum(numStr)
    else: @[stone * 2024]

proc memRule(st: int): seq[int] =
  var memo {.global.}: Table[int, seq[int]]
  if st in memo: return memo[st]
  result = st.applyRule
  memo[st] = result

proc countAfter(stone: int, targetBlinks: int): int =
  var memo {.global.}: Table[(int, int), int]
  if (stone,targetBlinks) in memo: return memo[(stone,targetBlinks)]

  if targetBlinks == 0: return 1
  for st in memRule(stone):
    result += st.countAfter(targetBlinks - 1)
  memo[(stone,targetBlinks)] = result

proc solve(input: string): AOCSolution[int, int] =
  for stone in input.split.map(parseInt):
    result.part1 += stone.countAfter(25)
    result.part2 += stone.countAfter(75)

Codeberg repo

C#

public class Day11 : Solver
{
  private long[] data;

  private class TreeNode(TreeNode? left, TreeNode? right, long value) {
    public TreeNode? Left = left;
    public TreeNode? Right = right;
    public long Value = value;
  }

  private Dictionary<(long, int), long> generation_length_cache = [];
  private Dictionary<long, TreeNode> subtree_pointers = [];

  public void Presolve(string input) {
    data = input.Trim().Split(" ").Select(long.Parse).ToArray();
    List<TreeNode> roots = data.Select(value => new TreeNode(null, null, value)).ToList();
    List<TreeNode> last_level = roots;
    subtree_pointers = roots.GroupBy(root => root.Value)
      .ToDictionary(grouping => grouping.Key, grouping => grouping.First());
    for (int i = 0; i < 75; i++) {
      List<TreeNode> next_level = [];
      foreach (var node in last_level) {
        long[] children = Transform(node.Value).ToArray();
        node.Left = new TreeNode(null, null, children[0]);
        if (subtree_pointers.TryAdd(node.Left.Value, node.Left)) {
          next_level.Add(node.Left);
        }
        if (children.Length <= 1) continue;
        node.Right = new TreeNode(null, null, children[1]);
        if (subtree_pointers.TryAdd(node.Right.Value, node.Right)) {
          next_level.Add(node.Right);
        }
      }
      last_level = next_level;
    }
  }

  public string SolveFirst() => data.Select(value => GetGenerationLength(value, 25)).Sum().ToString();
  public string SolveSecond() => data.Select(value => GetGenerationLength(value, 75)).Sum().ToString();

  private long GetGenerationLength(long value, int generation) {
    if (generation == 0) { return 1; }
    if (generation_length_cache.TryGetValue((value, generation), out var result)) return result;
    TreeNode cur = subtree_pointers[value];
    long sum = GetGenerationLength(cur.Left.Value, generation - 1);
    if (cur.Right is not null) {
      sum += GetGenerationLength(cur.Right.Value, generation - 1);
    }
    generation_length_cache[(value, generation)] = sum;
    return sum;
  }

  private IEnumerable<long> Transform(long arg) {
    if (arg == 0) return [1];
    if (arg.ToString() is { Length: var l } str && (l % 2) == 0) {
      return [int.Parse(str[..(l / 2)]), int.Parse(str[(l / 2)..])];
    }
    return [arg * 2024];
  }
}

J

If one line of code needs five lines of comment, I’m not sure how much of an improvement the “expressive power” is! But I learned how to use J’s group-by operator (/. or /..) and a trick with evoke gerund (`:0"1) to transform columns of a matrix separately. It might have been simpler to transpose and apply to rows.

data_file_name =: '11.data'
data =: ". > cutopen fread data_file_name
NB. split splits an even digit positive integer into left digits and right digits
split =: ; @: ((10 & #.) &.>) @: (({.~ ; }.~) (-: @: #)) @: (10 & #.^:_1)
NB. step consumes a single number and yields the boxed count-matrix of acting on that number
step =: monad define
   if. y = 0 do. < 1 1
   elseif. 2 | <. 10 ^. y do. < (split y) ,. 1 1
   else. < (y * 2024), 1 end.
)
NB. reduce_count_matrix consumes an unboxed count-matrix of shape n 2, left column being
NB. the item and right being the count of that item, and reduces it so that each item
NB. appears once and the counts are summed; it does not sort the items. Result is unboxed.
NB. Read the vocabulary page for /.. to understand the grouped matrix ;/.. builds; the
NB. gerund evoke `:0"1 then sums under boxing in the right coordinate of each row.
reduce_count_matrix =: > @: (({. ` ((+/&.>) @: {:)) `:0"1) @: ({. ;/.. {:) @: |:
initial_count_matrix =: reduce_count_matrix data ,. (# data) $ 1
NB. iterate consumes a count matrix and yields the result of stepping once across that
NB. count matrix. There's a lot going on here. On rows (item, count) of the incoming count
NB. matrix, (step @: {.) yields the (boxed count matrix) result of step item;
NB. (< @: (1&,) @: {:) yields <(1, count); then *"1&.> multiplies those at rank 1 under
NB. boxing. Finally raze and reduce.
iterate =: reduce_count_matrix @: ; @: (((step @: {.) (*"1&.>) (< @: (1&,) @: {:))"1)
count_pebbles =: +/ @: ({:"1)
result1 =: count_pebbles iterate^:25 initial_count_matrix
result2 =: count_pebbles iterate^:75 initial_count_matrix

Python

I initially cached the calculate_next function but honestly number of unique numbers don’t grow that much (couple thousands) so I did not feel a difference when I removed the cache. Using a dict just blazes through the problem.

from pathlib import Path
from collections import defaultdict
cwd = Path(__file__).parent

def parse_input(file_path):
  with file_path.open("r") as fp:
    numbers = list(map(int, fp.read().splitlines()[0].split(' ')))

  return numbers

def calculate_next(val):

  if val == 0:
    return [1]
  if (l:=len(str(val)))%2==0:
    return [int(str(val)[:int(l/2)]), int(str(val)[int(l/2):])]
  else:
    return [2024*val]

def solve_problem(file_name, nblinks):

  numbers = parse_input(Path(cwd, file_name))
  nvals = 0

  for indt, node in enumerate(numbers):

    last_nodes = {node:1}
    counter = 0

    while counter<nblinks:
      new_nodes = defaultdict(int)

      for val,count in last_nodes.items():
        val_next_nodes = calculate_next(val)

        for node in val_next_nodes:
          new_nodes[node] += count

      last_nodes = new_nodes
      counter += 1
    nvals += sum(last_nodes.values())

  return nvals

Dart

I really wish Dart had memoising built in. Maybe the new macro feature will allow this to happen, but in the meantime, here’s my hand-rolled solution.

import 'package:collection/collection.dart';

var counter_ = <(int, int), int>{};
int counter(s, [r = 75]) => counter_.putIfAbsent((s, r), () => _counter(s, r));
int _counter(int stone, rounds) =>
    (rounds == 0) ? 1 : next(stone).map((e) => counter(e, rounds - 1)).sum;

List<int> next(int s) {
  var ss = s.toString(), sl = ss.length;
  if (s == 0) return [1];
  if (sl.isOdd) return [s * 2024];
  return [ss.substring(0, sl ~/ 2), ss.substring(sl ~/ 2)]
      .map(int.parse)
      .toList();
}

solve(List<String> lines) => lines.first.split(' ').map(int.parse).map(counter).sum;

Haskell

Sometimes I want something mutable, this one takes 0.3s, profiling tells me 30% of my time is spent creating new objects. :/

import Control.Arrow

import Data.Map.Strict (Map)

import qualified Data.Map.Strict as Map
import qualified Data.Maybe as Maybe

type StoneCache = Map Int Int
type BlinkCache = Map Int StoneCache

parse :: String -> [Int]
parse = lines >>> head >>> words >>> map read

memoizedCountSplitStones :: BlinkCache -> Int -> Int -> (Int, BlinkCache)
memoizedCountSplitStones m 0 _ = (1, m)
memoizedCountSplitStones m i n 
        | Maybe.isJust maybeMemoized = (Maybe.fromJust maybeMemoized, m)
        | n == 0     = do
                let (r, rm) = memoizedCountSplitStones m (pred i) (succ n)
                let rm' = cacheWrite rm i n r
                (r, rm')
        | digitCount `mod` 2 == 0 = do
                let (r1, m1) = memoizedCountSplitStones m  (pred i) firstSplit
                let (r2, m2) = memoizedCountSplitStones m1 (pred i) secondSplit
                let m' = cacheWrite m2 i n (r1+r2)
                (r1 + r2, m')
        | otherwise = do
                let (r, m') = memoizedCountSplitStones m (pred i) (n * 2024)
                let m'' = cacheWrite m' i n r
                (r, m'')
        where
                secondSplit    = n `mod` (10 ^ (digitCount `div` 2))
                firstSplit     = (n - secondSplit) `div` (10 ^ (digitCount `div` 2))
                digitCount     = succ . floor . logBase 10 . fromIntegral $ n
                maybeMemoized  = cacheLookup m i n

foldMemoized :: Int -> (Int, BlinkCache) -> Int -> (Int, BlinkCache)
foldMemoized i (r, m) n = (r + r2, m')
        where
                (r2, m') = memoizedCountSplitStones m i n

cacheWrite :: BlinkCache -> Int -> Int -> Int -> BlinkCache
cacheWrite bc i n r = Map.adjust (Map.insert n r) i bc

cacheLookup :: BlinkCache -> Int -> Int -> Maybe Int
cacheLookup bc i n = do
        sc <- bc Map.!? i
        sc Map.!? n

emptyCache :: BlinkCache
emptyCache = Map.fromList [ (i, Map.empty) | i <- [1..75]]

part1 = foldl (foldMemoized 25) (0, emptyCache)
        >>> fst
part2 = foldl (foldMemoized 75) (0, emptyCache)
        >>> fst

main = getContents
        >>= print
        . (part1 &&& part2)
        . parse

Kotlin

Gone mathematical.

Also overflowing indices screwed with me a bit.

import kotlin.math.floor
import kotlin.math.log
import kotlin.math.pow
import kotlin.time.DurationUnit

fun main() {
    fun part1(input: List<String>): Long = Day11Solver(input).solve(25)

    fun part2(input: List<String>): Long = Day11Solver(input).solve(75)

    val testInput = readInput("Day11_test")
    check(part1(testInput) == 55312L)
    //check(part2(testInput) == 0L)  No test output available.

    val input = readInput("Day11")
    part1(input).println()
    part2(input).println()

    timeTrials("Part 1", unit = DurationUnit.MICROSECONDS) { part1(input) }
    timeTrials("Part 2", repetitions = 1000) { part2(input) }
}

class Day11Solver(input: List<String>) {
    private val parsedInput = input[0].split(' ').map { it.toLong() }

    /*
     * i ∈ ℕ₀ ∪ {-1}, φᵢ: ℕ₀ → ℕ shall be the function mapping the amount of stones generated to the amount of steps
     * taken with a stone of starting number i.
     *
     * Furthermore, ѱ: ℕ₀ → ℕ₀ ⨯ (ℕ₀ ∪ {-1}) shall be the function mapping an index to two new indices after a step.
     *
     *         ⎧ (1, -1)       if i = 0
     * ѱ(i) := ⎨ (a, b)        if ⌊lg(i)⌋ + 1 ∈ 2 ℕ    with a := i/(10^((⌊lg(i)⌋ + 1) / 2)), b := i - 10^((⌊lg(i)⌋ + 1) / 2) * a
     *         ⎩ (2024 i, -1)  otherwise
     *
     *          ⎧ 0                      if i = -1
     * φᵢ(n) := ⎨ 1                      if n = 0
     *          ⎩ φₖ(n - 1) + φₗ(n - 1)  otherwise    with (k, l) := ѱ(i)
     *
     * With that φᵢ(n) is a sum with n up to 2ⁿ summands, that are either 0 or 1.
     */
    private val cacheIndices = mutableMapOf<Long, Pair<Long, Long>>()  // Cache the next indices for going from φᵢ(n) to φₖ(n - 1) + φₗ(n - 1).
    private val cacheValues = mutableMapOf<Pair<Long, Int>, Long>()  // Also cache already calculated φᵢ(n)

    fun calculatePsi(i: Long): Pair<Long, Long> = cacheIndices.getOrPut(i) {
        if(i == -1L) throw IllegalArgumentException("Advancement made: How did we get here?")
        else if (i == 0L) 1L to -1L
        else {
            val amountOfDigits = (floor(log(i.toDouble(), 10.0)) + 1)

            if (amountOfDigits.toLong() % 2 == 0L) {
                // Split digits at the midpoint.
                val a = floor(i / 10.0.pow(amountOfDigits / 2))
                val b = i - a * 10.0.pow(amountOfDigits / 2)
                a.toLong() to b.toLong()
            } else {
                2024 * i to -1L
            }
        }
    }

    fun calculatePhi(i: Long, n: Int): Long = cacheValues.getOrPut(i to n) {
        if (i == -1L) 0L
        else if (n == 0) 1L
        else {
            val (k, l) = calculatePsi(i)
            calculatePhi(k, n - 1) + calculatePhi(l, n - 1)
        }
    }

    fun solve(steps: Int): Long = parsedInput.sumOf {
        val debug = calculatePhi(it, steps)
        debug
    }
}

Try it out here.

And this is the full repo.

C

Started out a bit sad that this problem really seemed to call for hash tables - either for storing counts for an iterative approach, or to memoize a recursive one.

Worried that the iterative approach would have me doing problematic O(n^2) array scans I went with recursion and a plan to memoize only the first N integers in a flat array, expecting low integers to be much more frequent (and dense) than higher ones.

After making an embarrassing amount of mistakes it worked out beautifully with N=1m (testing revealed that to be about optimal). Also applied some tail recursion shortcuts where possible.

day11  0:00.01  6660 Kb  0+1925 faults

#include "common.h"

/* returns 1 and splits x if even-digited, 0 otherwise */
static int
split(uint64_t x, uint64_t *a, uint64_t *b)
{
	uint64_t p;
	int n, i;

	if (!x) return 0;
	for (n=0, p=1; p<=x;  n++, p*=10) ; if (n%2) return 0;
	for (i=0, p=1; i<n/2; i++, p*=10) ;
	*a = x/p;
	*b = x%p; return 1;
}

/*
 * recur() is memoized in mem[]. Testing found the size MEMZ to be optimal:
 * lowering siginificantly reduced hits, but raising tenfold didn't add a
 * single hit.
 */

#define MEMZ (1024*1024)
static uint64_t mem[MEMZ][76];

static uint64_t
recur(uint64_t v, int n)
{
	uint64_t a,b;

	if (n<1 ) return 1;
	if (v==0) return recur(1, n-1);
	if (v<10) return recur(v*2024, n-1);
	if (v<MEMZ && mem[v][n]) return mem[v][n];
	if (!split(v, &a, &b))   return recur(v*2024, n-1);

	return v<MEMZ ? mem[v][n] =
	    recur(a, n-1) + recur(b, n-1) :
	    recur(a, n-1) + recur(b, n-1);
}

int
main(int argc, char **argv)
{
	uint64_t p1=0,p2=0, val;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	while (scanf(" %"SCNu64, &val) == 1) {
		p1 += recur(val, 25);
		p2 += recur(val, 75);
	}

	printf("10: %"PRId64" %"PRId64"\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day11.c