2018-09-01 22:38:19 问题描述: 问题求解: 本题如果是第一看到,应该还是非常棘手的,基本没有什么思路. 不妨先从一种简化的版本来考虑.如果仅有一种砖块,那么,填充的方式如下.…

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX <- domino XX <- "L" tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7. (…

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX <- domino XX <- "L" tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7. (…

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX <- domino XX <- "L" tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7. (…

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX <- domino XX <- "L" tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7. (…

We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX <- domino XX <- "L" tromino X Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7. (…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/domino-and-tromino-tiling/description/ 题目描述: We have two types of tiles: a 2x1 domino shape, and an "L" tromino shape. These shapes may be rotated. XX &l…

思路: dp.没有像discuss中的那样优化递推式. 实现: class Solution { public: ; int numTilings(int N) { vector<vector<, vector<, )); dp[][] = ; dp[][] = dp[][] = dp[][] = ; ; i <= N; i++) { dp[i][] = (((dp[i - ][] + dp[i - ][]) % MOD + dp[i - ][]) % MOD + dp[i - ]…

填格子 Time Limit: 2000/1000ms (Java/Others) Problem Description: 在一个n*n格子里边已经填了部分大写字母,现在给你个任务:把剩下的格子也填满大写字母,要求任意两个相邻的格子中的字母不一样. Input: 输入有多组数据.第一行输入T,表示有T组数据.以下每组第一行输入为整数 n (n<=10).下面接着输入有大写字母‘A~Z’或者'.'组成的 n * n 的方格. Output: 对于每一组数据,输出填满字母后的网格.若答案有多组,输…

题意 \(3 * 3\)的网格,给出左上角的数字\(m\)和右下角的数字\(m\),如果当前格子有数字\(x\),格子左边有个数字\(y\),格子上面有个数字\(z\),则\(y|x, z|x\).格子中不存在相同的两个数.问是否存在填满格子的方案. 分析 最优放法肯定是每一个格子放的数相是上面和左边的数的最小公倍数或乘上了一个质因子. 首先我们质因子肯定是\(p=m/n\)的质因子(如果\(m \% n \neq 0\)则显然无解)而且可以把左上角看成\(1\),右下角看成\(p\). (下面…