Day 2: Cube Conundrum

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[-] sjmulder 10 points 1 year ago

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);}

this post was submitted on 02 Dec 2023

23 points (96.0% liked)

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