2026 AIME II 第 13 题

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13.

若两个有限整数集合 SS 和 TT 满足以下条件,就称它们为表亲:

• SS 和 TT 的元素个数相同,

• SS 和 TT 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\} 与 {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
小提示:

每个元素都必须移动 ±1\pm 1 到 SS 外的值,并且映射要一一对应。若某个元素的两个邻居都在 SS 中,它就无法移动,所以 SS 由长度为 11 或 22 的连续整数块组成。

Each element must move by ±1\pm 1 to a value outside S,S, injectively. An element with both neighbors in SS is stuck, so SS consists of blocks of 11 or 22 consecutive integers.

大提示:

只隔一个缺失整数的块会连成链:一条含 kk 个单点块的链恰好有 k+1k + 1 个像,不同链的选择相乘。分解 40404040,使总大小最小。

Blocks separated by a single missing integer chain together: a chain of kk singleton blocks has exactly k+1k + 1 images, and separate chains multiply. Factor 40404040 to minimize total size.

解答:

一个表亲 TT 是某个映射的像,该映射把每个 x∈Sx \in S 送到 x−1x - 1 或 x+1x + 1,落在 SS 外,并且是单射。若 x−1,x,x+1∈Sx - 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 11 或 22。一个双点块 {a,a+1}\{a, a+1\} 被迫映到 {a−1,a+2}\{a - 1, a + 2\},而一个单点块 {a}\{a\} 可选择 a−1a - 1 或 a+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=23⋅5⋅101\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,h≥2g, h \ge 2 会严格降低代价,因为 (g−1)+(h−1)<gh−1(g - 1) + (h - 1) \lt gh - 1。所以最优方案使用质因数分解: ∑(f−1)=1+1+1+4+100=107 \begin{aligned} &\sum (f - 1) \\ &= 1 + 1 + 1 + 4 + 100 \\ &= 107 \end{aligned} 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。

最少元素个数为 107107。

A cousin TT is the image of an injection sending each x∈Sx \in S to x−1x - 1 or x+1,x + 1, landing outside S.S. If x−1,x,x+1∈Sx - 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 {a−1,a+2},\{a - 1, a + 2\}, while a singleton {a}\{a\} chooses a−1a - 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=23⋅5⋅101\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,h≥2g, h \ge 2 strictly lowers the cost, because (g−1)+(h−1)<gh−1.(g - 1) + (h - 1) \lt gh - 1. So the optimum uses the prime factorization: ∑(f−1)=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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