Day 9: Mirage Maintenance
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
🔒 Thread is locked until there’s at least 100 2 star entries on the global leaderboard
🔓 Unlocked after 5 mins
C
Thought there must be some clever way to infer the formula without storing those derived rows but implemented it like the example to get started. Turns out that was fine for part 2 as well so yay!
GitHub link
(Slightly abridged) code
int main() { char line[128], *tok,*rest; size_t n,d,i; int a[24][24], p1=0,p2=0, nz, acc1,acc2; while ((rest = fgets(line, sizeof(line), stdin))) { for (n=0; (tok = strsep(&rest, " ")); n++) a[0][n] = atoi(tok); /* generate rows until all 0, 'd' is depth */ for (d=1, nz=1; nz && d<n; d++) for (i=0, nz=0; i<n-d; i++) nz |= (a[d][i] = a[d-1][i+1] - a[d-1][i]); /* extrapolate forward and backwards */ for (i=0, acc1=acc2=0; i<d; i++) { acc1 += a[d-i-1][n-d+i]; acc2 = a[d-i-1][0] - acc2; } p1 += acc1; p2 += acc2; } printf("09: %d %d\n", p1, p2); return 0; }