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