161 lines
3.7 KiB
C
161 lines
3.7 KiB
C
/* aoc-c.c: Advent of Code 2021, day 16 parts 1 & 2
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*
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* Copyright (C) 2021 Bruno Raoult ("br")
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* Licensed under the GNU General Public License v3.0 or later.
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* Some rights reserved. See COPYING.
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*
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* You should have received a copy of the GNU General Public License along with this
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* program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
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*
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* SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <unistd.h>
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#include <string.h>
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#include <malloc.h>
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#include <ctype.h>
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#include <stdint.h>
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#include <math.h>
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#include "debug.h"
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#include "pool.h"
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#include "bits.h"
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#include "list.h"
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typedef struct vector {
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s64 x, y;
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} vector_t;
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#define TMIN 0
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#define TMAX 1
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struct target {
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s64 x1, y1;
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s64 x2, y2;
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} target;
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struct bounds {
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s64 dx_min, dx_max;
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s64 dy_min, dy_max;
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s64 steps_min;
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} bounds;
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inline static int hit(s64 x, s64 y)
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{
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return x >= target.x1 && y >= target.y1 &&
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x <= target.x2 && y <= target.y2;
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}
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/* if we take hypothesis target y is always negative, we can exclude steps
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* which are still over 0. We reach back zero after 2 x d1 +1 steps for
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* initial d1 positive (0 otherwise).
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*/
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static s64 nth(s64 x, s64 n)
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{
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return ((2 * x) + -1 * (n - 1)) * n / 2;
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}
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static s64 nth_x(s64 x0, s64 n)
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{
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return n >= x0? nth(x0, x0): nth(x0, n);
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}
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/* determine max possible initial dx :
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* ( dx )(dx + 1) <= minx * 2
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* <=> dx² + dx - minx * 2 <= 0
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* solution is x = (-b ± sqrt(b² - 4 * a * c)) / 2 * b, with a = 1, b = 1, c = -2 * minx
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* => (-1 ± sqrt(1 + 8 * minx)) / 2 <= 0
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*/
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static s64 dx_min(s64 xmin)
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{
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s64 res = (s64) ((-1) + sqrt((double)(1 + 8 * xmin))) / 2;
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bounds.steps_min = 1;
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bounds.dx_min = res;
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//printf("xmin=%ld res=%ld nmin=%ld\n", xmin, res, nmin);
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return res;
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}
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/* The highest solution is the solution of:
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* y1 * (y1 +1) / 2
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* (we can ignore x velocity, as we can reach any target x1 with x velocity
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* becoming zero at x1).
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*/
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static s64 part1()
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{
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return target.y1 * (target.y1 + 1) / 2;
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}
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static s64 part2()
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{
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s64 count = 0;
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for (s64 dx = bounds.dx_min; dx <= bounds.dx_max ; ++dx) {
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for (s64 dy = bounds.dy_min; dy <= bounds.dy_max; ++dy) {
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for (s64 step = 1; ; step++) {
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s64 newx = nth_x(dx, step);
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s64 newy = nth(dy, step);
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if (newx > target.x2 || newy < target.y1)
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goto nexty;
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if (hit(newx, newy)) {
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count++;
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goto nexty;
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}
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}
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nexty:
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}
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}
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return count;
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}
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/* read input
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*/
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static int read_input()
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{
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if (scanf("target area: x=%ld..%ld, y=%ld..%ld",
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&target.x1, &target.x2,
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&target.y1, &target.y2))
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return 0;
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return 1;
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}
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static int usage(char *prg)
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{
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fprintf(stderr, "Usage: %s [-d debug_level] [-p part]\n", prg);
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return 1;
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}
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int main(int ac, char **av)
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{
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int opt, part = 1;
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while ((opt = getopt(ac, av, "d:p:")) != -1) {
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switch (opt) {
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case 'd':
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debug_level_set(atoi(optarg));
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break;
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case 'p': /* 1 or 2 */
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part = atoi(optarg);
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if (part < 1 || part > 2)
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return usage(*av);
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break;
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default:
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return usage(*av);
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}
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}
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if (optind < ac)
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return usage(*av);
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read_input();
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dx_min(target.x1);
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bounds.dx_max = target.x2;
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bounds.dy_min = target.y1;
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bounds.dy_max = -target.y1;
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printf("%s : res=%ld\n", *av, part == 1? part1(): part2(&target));
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exit (0);
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}
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