Puzzle 4: Consolidate 4a/4b programs into `4` with `-p` option

Like the later puzzles

1 files changed, 81 insertions, 0 deletions

diff --git a/4.c b/4.c
new file mode 100644
index 0000000..1f183b7
--- /dev/null
+++ b/4.c
@@ -0,0 +1,81 @@
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <unistd.h>
+
+
+char *
+getsect(int * dest, char * pch, char expect)
+{
+	char * end;
+
+	if (!pch)
+		return NULL;
+	*dest = strtol(pch, &end, 10);
+	if (*dest == 0 && end == pch)
+		return NULL;
+	if (*end != expect)
+		return NULL;
+	if (expect)
+		++end;
+	return end;
+}
+
+
+int
+main(int argc, char ** argv)
+{
+	char buf[BUFSIZ];
+	int part = 1, i, overlaps = 0;
+
+	while ((i = getopt(argc, argv, "p:")) != -1) {
+		switch (i) {
+		case 'p':
+			part = atoi(optarg);
+			break;
+
+		default:
+			return -1;
+		}
+	}
+
+	while (fgets(buf, sizeof(buf), stdin)) {
+		int n[4];
+		char * pbuf;
+
+		pbuf = getsect(&n[0], buf, '-');
+		pbuf = getsect(&n[1], pbuf, ',');
+		pbuf = getsect(&n[2], pbuf, '-');
+		pbuf = getsect(&n[3], pbuf, '\n');
+		if (!pbuf) {
+			fprintf(stderr, "Unexpected line: %s\n", buf);
+			return -1;
+		}
+		switch (part) {
+		case 1:
+			if ((n[2] >= n[0] && n[3] <= n[1])
+					|| (n[0] >= n[2] && n[1] <= n[3]))
+			{
+				++overlaps;
+			}
+			break;
+
+		case 2:
+			if (!(n[3] < n[0] || n[2] > n[1])) {
+				++overlaps;
+			}
+			break;
+
+		default:
+			fprintf(stderr, "Unexpected puzzle part %d\n", part);
+			return -1;
+		}
+	}
+
+	printf("%s overlaps: %d\n",
+			part == 1 ? "Full" : "Partial",
+			overlaps);
+
+	return 0;
+}
+