aoc22-13.jq

# Usage: jq -s -f aoc22-13.jq input.txt

def cmp_list($l; $r):
	def cmp_elem($l; $r):
		($l | type) as $lt
	|	($r | type) as $rt
	|	if $lt == $rt then
			if $lt == "null" then
				0
			elif $lt == "number" then
				if $l < $r then
					-1
				elif $l == $r then
					0
				else
					1
				end
			else
				cmp_list($l; $r)
			end
		elif $lt == "null" then
			-1
		elif $rt == "null" then
			1
		elif $lt == "number" then
			cmp_list([$l]; $r)
		elif $rt == "number" then
			cmp_list($l; [$r])
		else
			"bad input" | halt_error
		end;

	first([$l, $r] | transpose[]
		| cmp_elem(.[0]; .[1])
		| select(. != 0)
	);

def silver:
	. as $in
|	reduce range(length / 2) as $i (0;
		if $in[$i * 2:] | cmp_list(.[0]; .[1]) == -1 then
			. + ($i + 1)
		else
			.
		end
	);

def gold:
	reduce .[] as $lst ([1, 2];
		if cmp_list($lst; [[2]]) == -1 then
			map(. + 1)
		elif cmp_list($lst; [[6]]) == -1 then
			.[1] += 1
		else
			.
		end
	) | .[0] * .[1];

silver,
gold

aoc22-20.c

#include <assert.h>
#include <stdint.h>
#include <stdio.h>

struct num {
	intmax_t val;
	size_t prev;
	size_t next;
} nums[5000];
size_t len;

size_t
nth(size_t i, size_t n) {
	size_t j;

	n %= len - 1;
	for (j = 0; j < n; j++)
		i = nums[i].next;

	return i;
}

int
main(void) {
	size_t i, j, k;
	intmax_t val;
	size_t zero;
	
	i = 0;
	while (scanf("%jd ", &val) != EOF) {
#if GOLD
		nums[i].val = val * 811589153;
#else
		nums[i].val = val;
#endif
		nums[i].prev = i - 1;
		nums[i].next = i + 1;

		if (val == 0)
			zero = i;

		i++;
	}

	len = i;
	assert(len > 0);
	nums[0].prev = len - 1;
	nums[len - 1].next = 0;

#if GOLD
	for (k = 0; k < 10; k++)
#endif
	for (i = 0; i < len; i++) {
		if (i == zero)
			continue;

		if (nums[i].val < 0)
			j = nth(i, (len - 1) - (-nums[i].val % (len - 1)));
		else
			j = nth(i, nums[i].val);

		if (j == i || j == nums[i].prev)
			continue;

		nums[nums[i].prev].next = nums[i].next;
		nums[nums[i].next].prev = nums[i].prev;

		nums[i].prev = j;
		nums[i].next = nums[j].next;

		nums[nums[j].next].prev = i;
		nums[j].next = i;
	}

	printf("%jd\n", nums[nth(zero, 1000)].val
			+ nums[nth(zero, 2000)].val
			+ nums[nth(zero, 3000)].val);
}