2026 AIME II 第 13 题

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13.

若两个有限整数集合 SSTT 满足以下条件,就称它们为表亲

SSTT 的元素个数相同,

SSTT 不相交,并且

• 可以把 SS 的元素与 TT 的元素配对,使每一对中的两个元素恰好相差 11

例如,{1,2,5}\{1, 2, 5\}{0,3,4}\{0, 3, 4\} 是表亲。已知集合 SS 恰好有 40404040 个表亲。 求集合 SS 可能拥有的最少元素个数。

Call finite sets of integers SS and TT cousins if

SS and TT have the same number of elements,

SS and TT are disjoint, and

• the elements of SS can be paired with the elements of TT so that the elements in each pair differ by exactly 1.1.

For example, {1,2,5}\{1, 2, 5\} and {0,3,4}\{0, 3, 4\} are cousins. Suppose that the set SS has exactly 40404040 cousins. Find the least number of elements the set SS can have.

答案:107
知识点:双射质因数分解最优化
难度评级:3370
解答:

一个表亲 TT 是某个映射的像,该映射把每个 xSx \in S 送到 x1x - 1x+1x + 1,落在 SS 外,并且是单射。若 x1,x,x+1Sx - 1, x, x + 1 \in S,那么 xx 无处可去,所以 SS 的每个最大连续块 长度只能是 1122。一个双点块 {a,a+1}\{a, a+1\} 被迫映到 {a1,a+2}\{a - 1, a + 2\},而一个单点块 {a}\{a\} 可选择 a1a - 1a+1a + 1。只有当两个块之间恰好隔一个整数时,它们才可能争抢同一个值, 因此把块分成若干链:链内相邻块之间的间隔恰好为一。在一条链中,唯一一致的模式是“前 ii 个块向左移, 后面的块向右移”,因为某块向右而下一块向左会发生碰撞;双点块同时起到左移与右移的作用,迫使切换恰好在那里发生。 因此一条含 kk 个单点块的链产生 k+1k + 1 个不同的像;含一个双点块的链恰好产生 11 个;含两个双点块的链产生 00 个。不同模式给出不同的集合 TT,而不同链的选择彼此独立,所以表亲个数等于所有纯单点链的 (ki+1)(k_i + 1) 的乘积。

需要 (ki+1)=4040=235101\prod (k_i + 1) = 4040 = 2^3 \cdot 5 \cdot 101,同时最小化元素个数 ki\sum k_i (含双点块的链只会浪费元素)。把一个合数因子 f=ghf = gh 替换成两个因子 g,h2g, h \ge 2 会严格降低代价, 因为 (g1)+(h1)<gh1(g - 1) + (h - 1) \lt gh - 1。所以最优方案使用质因数分解: 可由五条分别含 1,1,1,4,1001, 1, 1, 4, 100 个单点块的链实现,也就是若干段相隔一个整数的单点块,彼此放得足够远。 (f1)=1+1+1+4+100=107, \begin{aligned} &\sum (f - 1) \\ &= 1 + 1 + 1 + 4 + 100 \\ &= 107, \end{aligned}

最少元素个数为 107107

A cousin TT is the image of an injection sending each xSx \in S to x1x - 1 or x+1,x + 1, landing outside S.S. If x1,x,x+1Sx - 1, x, x + 1 \in S then xx has nowhere to go, so every maximal block of consecutive elements of SS has size 11 or 2.2. A double block {a,a+1}\{a, a+1\} is forced to map to {a1,a+2},\{a - 1, a + 2\}, while a singleton {a}\{a\} chooses a1a - 1 or a+1.a + 1. Two blocks can fight over a value only when exactly one integer separates them, so group blocks into chains: consecutive blocks with gaps of exactly one. Within a chain the only consistent patterns are "the first ii blocks shift left and the rest shift right," since a block choosing right and its successor choosing left would collide; a double block acts as both left and right, forcing the switch to happen exactly at it. Hence a chain of kk singletons produces k+1k + 1 distinct images, a chain containing one double produces exactly 1,1, and a chain with two doubles produces 0.0. Distinct patterns give distinct sets T,T, and choices in different chains are independent, so the number of cousins is the product of (ki+1)(k_i + 1) over the all-singleton chains.

We need (ki+1)=4040=235101\prod (k_i + 1) = 4040 = 2^3 \cdot 5 \cdot 101 while minimizing the element count ki\sum k_i (chains with doubles only waste elements). Replacing a composite factor f=ghf = gh with the two factors g,h2g, h \ge 2 strictly lowers the cost, because (g1)+(h1)<gh1.(g - 1) + (h - 1) \lt gh - 1. So the optimum uses the prime factorization: (f1)=1+1+1+4+100=107, \begin{aligned} &\sum (f - 1) \\ &= 1 + 1 + 1 + 4 + 100 \\ &= 107, \end{aligned} realized by five chains of 1,1,1,4,1001, 1, 1, 4, 100 singletons — runs of every-other integer — placed far apart.

The least possible number of elements is 107.107.

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