2013 AIME II 第 9 题

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9.

一个 7×17 \times 1 的板完全由 m×1m \times 1 的砖覆盖且不重叠;每块砖可以覆盖任意数量的连续小方格,并且每块砖都完全位于板上。每块砖要么是红色、蓝色,要么是绿色。设 NN 为铺满 7×17 \times 1 的板且三种颜色都至少使用一次的铺法数。例如,一块 1×11 \times 1 的红砖后接一块 2×12 \times 1 的绿砖、一块 1×11 \times 1 的绿砖、一块 2×12 \times 1 的蓝砖、以及一块 1×11 \times 1 的绿砖,是一种合法铺法。注意,如果那块 2×12 \times 1 的蓝砖换成两块 1×11 \times 1 的蓝砖,会得到一种不同的铺法。求 NN 除以 10001000 的余数。

A 7×17 \times 1 board is completely covered by m×1m \times 1 tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let NN be the number of tilings of the 7×17 \times 1 board in which all three colors are used at least once. For example, a 1×11 \times 1 red tile followed by a 2×12 \times 1 green tile, a 1×11 \times 1 green tile, a 2×12 \times 1 blue tile, and a 1×11 \times 1 green tile is a valid tiling. Note that if the 2×12 \times 1 blue tile is replaced by two 1×11 \times 1 blue tiles, this results in a different tiling. Find the remainder when NN is divided by 1000.1000.

答案:106
知识点:铺砖容斥原理乘法原理
难度评级:2610
解答:

先数有 kk 种颜色可用时的彩色铺法。第一个方格所在的砖有 kk 种颜色可选;之后的 66 个方格中,每个方格要么延续当前砖,要么以 kk 种颜色之一开始一块新砖,所以每个方格有 k+1k + 1 种选择。因此共有 k(k+1)6k(k+1)^6 种铺法。

三种颜色时共有 346=122883 \cdot 4^6 = 12288 种铺法。对未使用的颜色做容斥,使用全部三种颜色的铺法数为 N=3463(236)+3(126)=122884374+192=8106. \begin{aligned} N &= 3 \cdot 4^6 - 3 \cdot (2 \cdot 3^6) \\ &\quad {}+ 3 \cdot (1 \cdot 2^6) \\ &= 12288 - 4374 + 192 \\ &= 8106. \end{aligned}

NN 除以 10001000 的余数是 106106

First count colored tilings when kk colors are available. The first square's tile can be colored in kk ways, and each of the remaining 66 squares either extends the current tile or starts a new tile in one of the kk colors, giving k+1k + 1 choices per square. So there are k(k+1)6k(k+1)^6 tilings.

With three colors that is 346=122883 \cdot 4^6 = 12288 tilings. By inclusion-exclusion over the unused colors, the number using all three colors is N=3463(236)+3(126)=122884374+192=8106. \begin{aligned} N &= 3 \cdot 4^6 - 3 \cdot (2 \cdot 3^6) \\ &\quad {}+ 3 \cdot (1 \cdot 2^6) \\ &= 12288 - 4374 + 192 \\ &= 8106. \end{aligned}

The remainder when NN is divided by 10001000 is 106.106.

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