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HMMT 二月 2016 · COMB 赛 · 第 8 题

HMMT February 2016 — COMB Round — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Let X be the collection of all functions f : { 0 , 1 , . . . , 2016 } → { 0 , 1 , . . . , 2016 } . Compute the numberof functions f ∈ X such that
    ( )
    ( ) ( )
    max min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i )) = 2015 .
    g ∈ X 0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
解析

英文解析

  1. Let X be the collection of all functions f : { 0 , 1 , . . . , 2016 } → { 0 , 1 , . . . , 2016 } . Compute the numberof functions f ∈ X such that
    ( )
    ( ) ( )
    max min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i )) = 2015 .
    g ∈ X 0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
    Proposed by:
    2017 2017
    Answer: 2 · (3 − 2 )
    For each f, g ∈ X , we define
    ( ) ( )
    d ( f, g ) := min max( f ( i ) , g ( i )) − max min( f ( i ) , g ( i ))
    0 ≤ i ≤ 2016 0 ≤ i ≤ 2016
    Thus we desire max d ( f, g ) = 2015.
    g ∈ X
    First, we count the number of functions f ∈ X such that
    ∃ g : min max { f ( i ) , g ( i ) } ≥ 2015 and ∃ g : min max { f ( i ) , g ( i ) } = 0 .
    i i
    That means for every value of i , either f ( i ) = 0 (then we pick g ( i ) = 2015) or f ( i ) ≥ 2015 (then we
    2017
    pick g ( i ) = 0). So there are A = 3 functions in this case.
    Similarly, the number of functions such that
    ∃ g : min max { f ( i ) , g ( i ) } = 2016 and ∃ g : min max { f ( i ) , g ( i ) } ≤ 1
    i i
    2017
    is also B = 3 .
    Finally, the number of functions such that
    ∃ g : min max { f ( i ) , g ( i ) } = 2016 and ∃ g : min max { f ( i ) , g ( i ) } = 0
    i i
    2017
    is C = 2 .
    Now A + B − C counts the number of functions with max d ( f, g ) ≥ 2015 and C counts the numberg ∈ X
    2017 2017
    of functions with max d ( f, g ) ≥ 2016, so the answer is A + B − 2 C = 2 · (3 − 2 ).
    g ∈ X