题目大意:用1*2的砖头建n*2的墙,问有多少种不同的砖头排列方式?与斐波那契序列相似. #include <cstdio> #define MAXN 60 #define N 50 unsigned int F[MAXN]; int main() { #ifdef LOCAL freopen("in", "r", stdin); #endif F[] = ; F[] = ; ; i <= N; i++) F[i] = F[i-] + F[i-];…

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the leastbricks. The brick wall…

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the leastbricks. The brick wall…

［抄题］: There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the least bricks. The bric…

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the leastbricks. The brick wall…

[LeetCode]554. Brick Wall 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/brick-wall/description/ 题目描述: There is a brick wall in front of you. The wall is rectangular and has…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1325 题意: 从迷宫的S处出发,每次可以往东.南.西.北4个方向之一前进.如果前方有墙壁,游戏者可以把墙壁往前推一格.如果有两堵或者多堵连续的墙,则不能推动.另外,游戏者也不能推动游戏区域边界上的墙.用最少的步数走出迷宫(边界处没有墙的地方就是出口).迷宫总是有4行6列,多解时任意…

称号:背景:brick game有N块,给你一个整数的定数S,两个人轮流木: 的木块数是集合S中存在的随意数字.取走最后木块的人获胜.无法取则对方获胜. 题干:如今让你先取,给你一个你的结果序列串T,当中第k个字符代表有k个木块时你的结果: 可能赢就是T[k] = W.一定输就是T[k] = L. 问题:请你确定一个最小的集合,使得这个序列串T成立.(集合中元素为不超过20的正整数) 分析:博弈,状态压缩+dp验证. 由于集合最多20个元素,利用20个位表示每一个元素(1~20)的选取状态. 枚…

题目传送门 快速的通道I 快速的通道II 题目大意 给定一堆短串,和一个文本串,问哪些短串在文本串中出现的次数最多. 我觉得刘汝佳的做法,时间复杂度有问题.只是似乎这道题短串串长太短不好卡.比如给出的串是一坨$a$.暴力跳$last$会比较gg. 考虑如何计算一个短串在长串中的出现次数. 当短串在长串的某个位置出现的时候,这意味着它的结束位置在fail树上的祖先中某个状态是短串的终止状态. 我们会在长串经过的每个状态都去做这样一个操作来统计每个短串出现的次数. 这个可以看成在fail树上的以根为…

class Solution { public: int leastBricks(vector<vector<int>>& wall) { unordered_map<int,int> m; ; i < wall.size(); i++) , t = ; j < wall[i].size() - ; j++) { t += wall[i][j]; m[t]++; } ; for (auto & p : m) { _max = max(_max…