2015 AIME II 第 10 题

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10.

若整数 1,2,,n1, 2, \ldots, n 的一个排列 a1,a2,,ana_1, a_2, \ldots, a_n 满足对每个 1kn11 \le k \le n - 1 都有 akak+1+2a_k \le a_{k+1} + 2,则称它为准递增排列。例如,53421534211425314253 是整数 1,2,3,4,51, 2, 3, 4, 5 的准递增排列,但 4512345123 不是。求整数 1,2,,71, 2, \ldots, 7 的准递增排列个数。

Call a permutation a1,a2,,ana_1, a_2, \ldots, a_n of the integers 1,2,,n1, 2, \ldots, n quasi-increasing if akak+1+2a_k \le a_{k+1} + 2 for each 1kn1.1 \le k \le n - 1. For example, 5342153421 and 1425314253 are quasi-increasing permutations of the integers 1,2,3,4,5,1, 2, 3, 4, 5, but 4512345123 is not. Find the number of quasi-increasing permutations of the integers 1,2,,7.1, 2, \ldots, 7.

答案:486
知识点:排列递推计数
难度评级:2890
解答:

SnS_n1,,n1, \ldots, n 的准递增排列个数。把 nn 插入一个 1,,n11, \ldots, n - 1 的准递增排列中:紧跟在 nn 后面的数必须至少为 n2n - 2,所以 nn 可以放在 n1n - 1 正前方、n2n - 2 正前方,或最后面,恰好有 33 个位置;每种插入都会保持其他相邻条件不变。

反过来,从一个 1,,n1, \ldots, n 的准递增排列中删除 nn 会留下一个 1,,n11, \ldots, n - 1 的准递增排列,因为当 n3n \ge 3 时,被删去的 nn 两侧的数满足 ak1n1ak+1+2a_{k-1} \le n - 1 \le a_{k+1} + 2。所以对 n3n \ge 3Sn=3Sn1S_n = 3S_{n-1}

因为 S2=2S_2 = 2,所以 S7=235=486S_7 = 2 \cdot 3^5 = 486

Let SnS_n be the number of quasi-increasing permutations of 1,,n.1, \ldots, n. Insert nn into a quasi-increasing permutation of 1,,n1:1, \ldots, n - 1: the entry following nn must be at least n2,n - 2, so nn can go immediately before n1,n - 1, immediately before n2,n - 2, or at the very end — exactly 33 positions, and each insertion keeps every other adjacent condition intact.

Conversely, deleting nn from a quasi-increasing permutation of 1,,n1, \ldots, n leaves a quasi-increasing permutation of 1,,n1,1, \ldots, n - 1, since the entries around the deleted nn satisfy ak1n1ak+1+2a_{k-1} \le n - 1 \le a_{k+1} + 2 when n3.n \ge 3. So Sn=3Sn1S_n = 3S_{n-1} for n3.n \ge 3.

Since S2=2,S_2 = 2, we get S7=235=486.S_7 = 2 \cdot 3^5 = 486.

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