2022 day 4: C, parts 1 & 2 (lazy way: same algo as Bash)

2022/day04/aoc-c.c

@@ -0,0 +1,43 @@
+/* aoc-c.c: Advent of Code 2022, day 3
+ *
+ * Copyright (C) 2022 Bruno Raoult ("br")
+ * Licensed under the GNU General Public License v3.0 or later.
+ * Some rights reserved. See COPYING.
+ *
+ * You should have received a copy of the GNU General Public License along with this
+ * program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
+ *
+ * SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#include "aoc.h"
+
+static int parse(int part)
+{
+    int res = 0, val[4];
+
+    while (scanf("%d-%d,%d-%d", val, val+1, val+2, val+3) == 4) {
+        if (part == 1) {
+            if ( (val[0] >= val[2] && val[1] <= val[3] ) ||
+                 (val[0] <= val[2] && val[1] >= val[3] ) ) {
+                res++;
+            }
+        } else {
+            if ( (val[0] >= val[2] && val[0] <= val[3] ) ||
+                 (val[0] <= val[2] && val[1] >= val[2] ) ) {
+                res++;
+            }
+        }
+    }
+    return res;
+}
+
+int main(int ac, char **av)
+{
+    printf("%s: res=%d\n", *av, parse(parseargs(ac, av)));
+    exit(0);
+}