张培超 2008-2009(seed) jakarta 2 nd chengdu 13 th bbsid: failedstar

张培超

2008-2009(Seed)Jakarta 2nd

Chengdu 13th

bbsID: failedstarstarforever00@gmail.com

Funny Problems for Beginners

Peichao Zhang

UNCLE VASYA AND BAGS FOR POTATOES

SGU 238

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

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

Problem Statement

• question– if we repeatedly perform the above operation– how many different layouts of bags can we get?

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

A Smart Solution

• just print N+1• why?

Solution Analysis

• tree structure

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

Solution Analysis

• relation– layout tree node– operation tree edge

POLYMANIASGU 251

Problem Statement

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

number

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

Problem Statement

1 1

22

Solution Analysis

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

B

A

C

X

X=Y

Solution Analysis

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

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

XAB=YABXYA=XYB

Solution Analysis

A

X

B

Y

XAB=YABXYA+XYB=XAB+YAB

Solution Analysis

A

X

B

Y

XAB=YABXYA-XYB=XAB+YAB

Solution Analysis

A

X

B

Y

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

BLACK & WHITESGU 246

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

Problem Statement

• find the minimal K, such that the necklace will always be “beautiful”

• regardless of how the beads are arranged

N=4K=4

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

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

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

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

Q&A

Thanks!