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+#advent of code 2024
+#day 13
+#similar problem was last year
+#rewritten to include both parts, could get rid of second for loop but w/e
+import re
+
+def formula(machinevalues,p2):
+	xa,ya,xb,yb,XP,YP = machinevalues;
+	XP += p2;
+	YP += p2;
+	pusha = (yb*XP - xb*YP)/(yb*xa - xb*ya); #number of pushes - button A
+	pushb = (XP - xa*pusha)/xb; 			 #number of pushes - button B
+	Valid = (pusha%1 == 0)*(pushb%1==0); #check if solution is an integer
+	cost = 3*int(pusha) + int(pushb);
+	return Valid*cost;
+
+machines = [];
+f = open("13.in","r");
+for l in f.read().split("\n\n"):
+	values = re.findall(r'\d+',l);
+	machines.append([int(v) for v in values]);
+f.close();
+
+part1, part2 = 0, 0;
+for machine in machines:
+	part1 += formula(machine,0);
+	part2 += formula(machine,10000000000000);
+	
+print("part 1 =",part1);
+print("part 2 =",part2);
+