张培超 2008-2009(Seed) Jakarta 2 nd Chengdu 13 th bbsID: failedstar

1. 张培超 2008-2009(Seed) Jakarta 2nd Chengdu 13th bbsID: failedstar starforever00@gmail.com

2. Funny Problems for Beginners Peichao Zhang

3. SGU 238 Uncle Vasya and Bags for Potatoes

4. Problem Statement • N different bags • some bags on the floor • some bags in the other bags • one can do the following operation at each turn

5. Problem Statement • select some bag and open it • put all other bags originally on the floor into the selected bag • pour all bags originally inside the selected bag to the floor

8. Problem Statement • question • if we repeatedly perform the above operation • how many different layouts of bags can we get?

9. A Stragithforward Solution • devise a way to represent the layout of bags • maintain a collection of layouts that have been expanded • simulate the above operation on each layout to expand the collection • count the number of layouts in the collection

10. A Smart Solution • just print N+1 • why?

11. Solution Analysis • tree structure

12. Solution Analysis • 2 operations on the same bag = nothing changed • number of operations = number of bags on the floor • if we choose a bag, we cannot choose other bags originally on the floor after the operation

14. Solution Analysis • relation • layout  tree node • operation  tree edge

16. SGU 251 polymania

17. Problem Statement • N points (N>=3) • each point has a positive special number • at least two points have the same special number

18. Problem Statement • try to arrange the points in a Cartesian coordinate, so that for each triangle formed by 3 different points, the area of the triangle equals the sum of the special number associated with each point

19. Problem Statement 2 2 1 1

20. Solution Analysis • how to utilize the condition “at least two points have the same special number”?

21. X=Y A X B C

22. Solution Analysis • we can conclude from this condition that for any N>4, no solution exists!

23. Solution Analysis • so we need only consider the situation when N=3 and N=4 • when N=3, we can construct a solution easily • when N=4, we can also construct a solution with some calculations

24. Solution Analysis X Y XAB=YAB XYA=XYB A B

25. Solution Analysis X XAB=YAB XYA+XYB=XAB+YAB A B Y

26. Solution Analysis X XAB=YAB XYA-XYB=XAB+YAB A B Y

27. Solution Analysis • for N=4, we can calculate the areas of all the 4 triangulars • we can choose to apply one position layout above that satisfying the corresponding equations

28. SGU 246 Black & White

29. Problem Statement • a necklace with 2N-1 beads • K of them are black, the rest are white • the necklace is “beautiful” if there exists two black beads such that exactly N beads are between them

30. Problem Statement • find the minimal K, such that the necklace will always be “beautiful” • regardless of how the beads are arranged

31. N=4 K=4

32. Solution Analysis • try to make the situation as worst as possible • that is, try to maximize the number of black beads, and keep the necklace “ugly” • two beads with the distance of N+1 in a circular manner cannot be black at once

33. Solution Analysis • construct a graph of 2N-1 nodes, each representing a bead • for any two beads that cannot be black at once, connect the corresponding nodes with an edge in the graph • try to paint as many nodes as possible black, such that no two nodes are adjacent to each other

36. Solution Analysis • with some math analysis • when 2N-1 = 0 (mod 3), the graph is composed of 3 circles • otherwise, the graph is just a circle • the proof is not difficult, so we omit it here

38. Solution Analysis • finally, with some basic calculations, we end up with a fairly simple answer for the problem • when 2N-1 = 0 (mod 3), the answer is N-1, otherwise the answer is N

39. Q&A Thanks!