Day 6: Trash Compactor ## Megathread guidelines - Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever) - You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL ## FAQ - What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268 - Where do I participate?: https://adventofcode.com/ - Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465

  • janAkali
    2·
    8 个月前

    Nim

    The hardest part was reading the part 2 description. I literally looked at it for minutes trying to understand where the problem numbers come from and how they’re related to the example input. But then it clicked.

    The next roadblock was that my template was stripping whitespace at the end of the last line, making parsing a lot harder. I’ve replaced strip() with strip(chars={'\n'}) to keep the trailing space intact.

    Runtime: 1.4 ms 618 μs

    view code 
    type
  AOCSolution[T,U] = tuple[part1: T, part2: U]

proc solve(input: string): AOCSolution[int, int] =
  let lines = input.splitLines()
  let numbers = lines[0..^2]
  let ops = lines[^1]

  block p1:
    let numbers = numbers.mapIt(it.splitWhiteSpace().mapIt(parseInt it))
    let ops = ops.splitWhitespace()
    for x in 0 .. numbers[0].high:
      var res = numbers[0][x]
      for y in 1 .. numbers.high:
        case ops[x]
        of "*": res *= numbers[y][x]
        of "+": res += numbers[y][x]
      result.part1 += res

  block p2:
    var problems: seq[(char, Slice[int])]
    var ind = 0
    while ind < ops.len:
      let len = ops.skipWhile({' '}, ind+1)
      problems.add (ops[ind], ind .. ind + len - (if ind+len < ops.high: 1 else: 0))
      ind += len + 1

    for (op, cols) in problems:
      var res = 0
      for x in cols:
        var num = ""
        for y in 0 .. numbers.high:
          num &= numbers[y][x]

        if res == 0:
          res = parseInt num.strip
        else:
          case op
          of '*': res *= parseInt num.strip
          of '+': res += parseInt num.strip
          else: discard

      result.part2 += res

    Full solution at Codeberg: solution.nim

  • strlcpyEnglish
    2·
    8 个月前

    C

    Well so much for reading a grid of ints in part 1! For part 2, initially I reworked the parsing to read into a big buffer, but then thought it would be fun to try and use memory-mapped I/O as not to use any more memory than strictly necessary for the final version:

    Code 
    #include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <assert.h>
#include <sys/mman.h>
#include <unistd.h>
#include <err.h>

#define GH	5

int
main()
{
	char *data, *g[GH], *p;
	uint64_t p1=0,p2=0, acc;
	int len, h=0, i, x,y, val;
	char op;

	if ((len = (int)lseek(0, 0, SEEK_END)) == -1)
		err(1, "<stdin>");
	if (!(data = mmap(NULL, len, PROT_READ, MAP_SHARED, 0, 0)))
		err(1, "<stdin>");
	for (i=0; i<len; i++)
		if (!i || data[i-1]=='\n') {
			assert(h < GH);
			g[h++] = data+i;
		}

	for (x=0; g[h-1]+x < data+len; x++) {
		if ((op = g[h-1][x]) != '+' && op != '*')
			continue;

		for (acc = op=='*', y=0; y<h-1; y++) {
			val = atoi(&g[y][x]);
			acc = op=='+' ? acc+val : acc*val;
		}

		p1 += acc;

		for (acc = op=='*', i=0; ; i++) {
			for (val=0, y=0; y<h-1; y++) {
				p = &g[y][x+i];
				if (p < g[y+1] && isdigit(*p))
					val = val*10 + *p-'0';
			}
			if (!val)
				break;
			acc = op=='+' ? acc+val : acc*val;
		}

		p2 += acc;
	}

	printf("06: %"PRIu64" %"PRIu64"\n", p1, p2);
}