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HMMT 十一月 2012 · 冲刺赛 · 第 36 题

HMMT November 2012 — Guts Round — Problem 36

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 20 ] Let π and π be permutations of the numbers from 1 through 7. Call π superior to π if the
    1 2 1 2
    sum of all i such that π ( i ) > π ( i ) exceeds the sum of all i such that π ( i ) > π ( i ). Write down a
    1 2 2 1
    permutation of the integers from 1 through 7. Let N be the total number of answers submitted forthis problem, and let n be the number of submitted answers your answer is superior to. Your scorewill be ⌈ 20 ⌉ .n
解析

这是 Guts Round 的互动题，无固定答案——得分取决于所有队伍的提交。
题目： 提交 {1,…,7}\{1,\ldots,7\} 的一个排列。你的排列比另一个排列"优越"如果它的值在更多高索引处更大。得分 ⌈20⋅n/N⌉\lceil 20 \cdot n/N \rceil。
策略： 最优提交取决于其他队伍的分布。选择在高索引处放较大值的"中心型"排列通常表现较好。

英文解析

This is a Guts Round interactive problem. There is no single fixed answer — the score depends on submissions from all teams.
Problem: Submit a permutation of {1,…,7}\{1,\ldots,7\}. Your answer is "superior" to another submitted answer if the sum of indices ii where your permutation has a larger value exceeds the sum where the other does. Your score is ⌈20⋅n/N⌉\lceil 20 \cdot n/N \rceil where nn is the number of submitted answers your answer is superior to, and NN is the total number of submissions.
Strategy: The optimal submission depends on the distribution of other teams' answers. In practice, choosing a "central" permutation that dominates many others (e.g., one with large values at high indices) tends to perform well.