day 14 part 1 (stupid algorithm) + init part 2
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@@ -24,7 +24,7 @@ LDLIB := -l$(LIB)
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export LD_LIBRARY_PATH = $(LIBDIR)
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CFLAGS += -std=gnu99
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#CFLAGS += -O2
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CFLAGS += -O2
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CFLAGS += -g
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CFLAGS += -Wall
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CFLAGS += -Wextra
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@@ -50,3 +50,14 @@ After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
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This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After step 10, B occurs 1749 times, C occurs 298 times, H occurs 161 times, and N occurs 865 times; taking the quantity of the most common element (B, 1749) and subtracting the quantity of the least common element (H, 161) produces 1749 - 161 = 1588.
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Apply 10 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
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Your puzzle answer was 2712.
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The first half of this puzzle is complete! It provides one gold star: *
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--- Part Two ---
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The resulting polymer isn't nearly strong enough to reinforce the submarine. You'll need to run more steps of the pair insertion process; a total of 40 steps should do it.
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In the above example, the most common element is B (occurring 2192039569602 times) and the least common element is H (occurring 3849876073 times); subtracting these produces 2188189693529.
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Apply 40 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
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196
2021/day14/aoc-c.c
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196
2021/day14/aoc-c.c
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@@ -0,0 +1,196 @@
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/* aoc-c.c: Advent of Code 2021, day 14 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 "debug.h"
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#include "bits.h"
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//#include "list.h"
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/* possible 10 letters: B, C, F, H, K, N, O, P, S, V */
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int weight[] = { ['B'] = 0, ['C'] = 1, ['F'] = 2,
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['H'] = 3, ['K'] = 4, ['N'] = 5, ['O'] = 6,
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['P'] = 7, ['S'] = 8, ['V'] = 9
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};
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int count[26];
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/*struct weigth {
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char key;
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int val;
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} weight1[] = {
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{ 'A', -1 },
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{ 'B', 0 },
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{ 'C', 1 },
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{ 'D', -1 },
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{ 'E', -1 },
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{ 'F', 2 },
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{ 'G', -1 },
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{ 'H', 3 },
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{ 'I', -1 },
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{ 'J', -1 },
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{ 'K', 4 },
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{ 'L', -1 },
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{ 'M', -1 },
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{ 'N', 5 },
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{ 'O', 6 },
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{ 'P', 7 },
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{ 'Q', -1 },
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{ 'R', -1 },
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{ 'S', 8 },
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{ 'T', -1 },
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{ 'U', -1 },
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{ 'V', 9 },
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{ 'W', -1 },
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{ 'X', -1 },
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{ 'Y', -1 },
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{ 'Z', -1 }
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};
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*/
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#define MAX_RUN 10
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#define MAX_SIZE 1024 * 10 * 2 /* max resulting string size */
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char seq[MAX_SIZE], seq2[MAX_SIZE];
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int nseq;
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int rules[10 * 10];
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int nrules;
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/* read data and create graph.
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*/
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static int read_input()
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{
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ssize_t len;
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size_t alloc = 0;
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char *buf;
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char val[10] = { 0 };
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int pos;//, ins;
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nseq = getline(&buf, &alloc, stdin) - 1;
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buf[nseq] = 0;
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strcpy(seq, buf);
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printf("str = %d [%s]\n", nseq, seq);
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printf("empt = %ld\n", getline(&buf, &alloc, stdin));
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/* get rules */
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while ((len = getline(&buf, &alloc, stdin)) > 1) {
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//log(3, "len = %d [%s]\n", len, buf);
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sscanf(buf, "%c%c -> %c", val, val+1, val+2);
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//*val -= 'A';
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//*(val + 1) -= 'A';
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//*(val + 2) -= 'A';
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pos = weight[(int)*val] * 10 + weight[(int)*(val + 1)];
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//ins = weight[(int)*(val + 2)];
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rules[pos] = *(val + 2);
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//log(3, val);
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//log(3, "%s : %c %d %c %d %d %d pos=%d\n", val, *val, (int) *val - 'A', val[1], val[1] - 'A',
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// weight[(int)*val] * 10, weight[(int)*(val + 1)], pos);
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log(3, "%.2s : array[%d]=%c\n", val, pos, rules[pos]);
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nrules++;
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}
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free(buf);
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return nrules;
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}
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static int part1()
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{
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char *src, *dst;
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int key;//, last;
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unsigned int count[26] = {0};
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unsigned min, max;
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read_input();
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for (int i = 0; i < 10; ++i) {
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int j, k;
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src = i % 2? seq2: seq;
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dst = i % 2? seq: seq2;
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//savedst = dst;
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//savesrc = dst;
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//last = strlen(src) - 1;
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for (j = 0, k = 0; src[j+1]; ++j, k += 2) {
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key = weight[(int) src[j]] * 10 + weight[(int) src[j+1]];
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//(*src - 'A') * 10 + *(src+1) -'A';
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//log(3, "%d/%d: src=%s dst=%s s1=%c s2=%c key=%d\n", i, j, src, dst, *src, *(src + 1), key);
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dst[k] = src[j];
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dst[k+1] = rules[key];
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dst[k+2] = src[j+1];
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//src++;
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//dst += 3;
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//log_i(3, "j=%d k=%d dst=%s\n", j, k, dst);
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}
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log(3, "i = %d len=%d dst = %s\n", i, strlen(dst), dst);
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//log(3, "%d: src=%p dst=%p seq=%p seq2=%p\n", src, dst, seq, seq2);
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}
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for (int i=0; dst[i]; ++i)
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count[dst[i] - 'A']++;
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min = ~0;
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max = 0;
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for (unsigned i = 0; i < sizeof(count)/sizeof(*count); ++i)
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printf("count[%c]=%d\n", i + 'A', count[i]);
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for (unsigned i = 0; i < sizeof(count)/sizeof(*count); ++i) {
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if (count[i]) {
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if (count[i] < min)
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min = count[i];
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if (count[i] > max)
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max = count[i];
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}
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}
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printf("min = %i max=%u\n", min, max);
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return max - min;
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}
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static int part2()
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{
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return 2;
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}
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static int doit(int part)
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{
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//read_input();
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return part == 1? part1(): part2();
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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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printf("%s : res=%d\n", *av, doit(part));
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exit (0);
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}
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