Version de la solution F optimisé

contest/f_goodies/input/input31.txt

@@ -1,3 +1,3 @@
-20
-30
-3 5 11 14 12 6 10 7 5 6 5 3 3 5 11 6 4 3 13 12 2 12 3 12 4 2 8 12 9 10
+50
+27
+1 1 2 2 3 3 3 4 4 5 5 6 6 7 8 9 9 10 10 11 12 12 13 13 14 14 14 

contest/f_goodies/sol/f_goodies_optimized.py

@@ -0,0 +1,52 @@
+# Solution by ArnaudF
+# Largely inspired by the solution from Sebastien Goll & Louis Hasenfratz
+import collections
+from collections import defaultdict
+
+Q = int(input())
+N = int(input())
+# The goodies are sorted to make sure that two goodies of the same price are grouped together.
+# This is useful when constructing combinations, line 27-29.
+goodies = sorted(list(map(int, input().split())))
+
+#Sac definition
+bag = defaultdict(set)
+bag[0].add(tuple())
+
+#List substration. Does the same as the - operator with sets.
+# list1 = [1,2,3,4,5]; list2 = [4,5]
+# Return: [1,2,3]
+def diff(list1, list2):
+	for element in list2:
+		list1.remove(element)
+	return list1
+
+#Build all the possible combinations that give a bag size from 0 to Q.
+for card in goodies:
+	for size in range(Q+1, -1 , -1):
+		if size + card <= Q and size in bag:
+			for combinaison in bag[size]:
+				combinaison = combinaison + (card,)
+				bag[size+card].add(combinaison)
+
+#For each combination that gives the desired sum Q, 
+# the combination is removed from the goodies list. 
+# Then we check whether it is still possible to build a bag with a price Q.
+# If not: We win, print "OUI"
+weWin = False
+for combinaison in bag[Q]:
+	newgoodiesList = diff(goodies.copy(), combinaison)
+	newbag = defaultdict(bool)
+	newbag[0] = True
+	for card in newgoodiesList:
+		for size in range(Q+1, -1 , -1):
+			if size + card <= Q and size in newbag:
+				newbag[size+card] = True
+	if Q not in newbag:
+		weWin = True
+		break
+
+if weWin:
+	print("OUI")
+else:
+	print("NON")