Day 2: Cube Conundrum
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/ or pastebin (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
🔒This post will be unlocked when there is a decent amount of submissions on the leaderboard to avoid cheating for top spots
🔓 Edit: Post has been unlocked after 6 minutes
Found a C per-char solution, that is, no lines, splitting, lookahead, etc. It wasn’t even necessary to keep match lengths for the color names because they all have unique characters, e.g. ‘b’ only occurs in “blue” so then you can attribute the count to that color.
int main() { int p1=0,p2=0, id=1,num=0, r=0,g=0,b=0, c; while ((c = getchar()) != EOF) if (c==',' || c==';' || c==':') num = 0; else if (c>='0' && c<='9') num = num*10 + c-'0'; else if (c=='d') r = MAX(r, num); else if (c=='g') g = MAX(g, num); else if (c=='b') b = MAX(b, num); else if (c=='\n') { p1 += (r<=12 && g<=13 && b<=14) * id; p2 += r*g*b; r=g=b=0; id++; } printf("%d %d\n", p1, p2); return 0; }Golfed:
c,p,P,i,n,r,g,b;main(){while(~ (c=getchar()))c==44|c==58|59== c?n=0:c>47&c<58?n=n*10+c-48:98 ==c?b=b>n?b:n:c=='d'?r=r>n?r:n :c=='g'?g=g>n?g:n:10==c?p+=++i *(r<13&g<14&b<15),P+=r*g*b,r=g =b=0:0;printf("%d %d\n",p,P);}