递（dong）推（tai）数（gui）列（hua）染色问题

递（dong）推（tai）数（gui）列（hua）染色问题

对于这个问题由于相邻的两个格子可以涂一样的颜色，也就是说相邻的格子可以分类讨论为一样的颜色和不一样的颜色两种情况。如果从左到右一个格子一个格子地涂进行计数统计的话，12个格子要分很多种情况。所以我们要转变一下思维方式，考察一下涂第i个格子时有多少种选择。

由于相邻的两个格子可以涂为一样的样色，所以我们应该研究对象优化为第i-1个格子和第i个格子的整体。又由题意可知，涂色是分步进行，前i-2个格子的涂法与之后两个格子的涂法保持独立。即i-1与i可分为两种情况（count_i表示涂前i个格子的涂法数）：

1.颜色一样,但需与i-2不一样：

\[COUNT_{i - 2} \]

2.样色不一样，此时将i作为研究对象：

\[COUNT_{i - 1} \]

将两者合并：

\[COUNT_i = COUNT_{i - 1} + COUNT_{i -2} \]

这样我们就得到了递推式。又可计算只有1个格子和2个格子时，涂法分别为2,4.

\[COUNT_3 = COUNT_{1} + COUNT_{2} = 2 + 4 = 6 \]

\[COUNT_4 = COUNT_{2} + COUNT_{3} = 4 + 6 = 10 \]

\[COUNT_5 = COUNT_{3} + COUNT_{4} = 6 + 10 = 16 \]

\[COUNT_6 = COUNT_{4} + COUNT_{5} = 10 + 16 = 26 \]

\[COUNT_7 = COUNT_{5} + COUNT_{6} = 16 + 26 = 42 \]

\[COUNT_8 = COUNT_{6} + COUNT_{7} = 26 + 42 = 68 \]

\[COUNT_9 = COUNT_{7} + COUNT_{8} = 42 + 68 = 110 \]

\[COUNT_{10} = COUNT_{8} + COUNT_{9} = 68 + 110 = 178 \]

\[COUNT_{11} = COUNT_{9} + COUNT_{10} = 110 + 178 = 288 \]

\[COUNT_{12} = COUNT_{10} + COUNT_{11} = 178 + 288 = 466 \]

综上共有466种涂法