CareerCup Arrange the 2 x 1 boards on the 2 x n board

本文探讨了如何计算使用2x1板砖完全填充2xN大小的板砖的不同方式的数量。通过将问题简化为寻找第N个斐波那契数的方法,提供了一个高效的解决方案,并进一步介绍了如何在O(log N)时间内找到该数值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Given a board made of 2 x n squares, and boards made of 2 x 1 squares, write a function that will calculate the number of possible ways to arrange the 2 x 1 boards on the 2 x n board, in a way that will fill it completely. 

(Asked to refine the solution to be more efficient)

------------------------------------------------------------------------------------------------------

At first it seems like a dynamic programming type of problem but looking more closely it reduces to the simple problem of finding the nth Fibonacci number. Denote by f(n) the number of possibilities to tile a 2xn board. 

Consider tiling a 2xn board for some n>2. Let us sort all the possibilities according to the last 2 columns. There are f(n-1) possibilities where the last column includes a single tile and there are f(n-2) possibilities where the last 2 columns include 2 horizontally placed tiles. This covers all possibilities and hence: f(n) = f(n-1) + f(n-2) which is exactly the nth Fibonacci number. 

AND there is way to get f(n) in log(n) time. 


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值