洛谷题目页面传送门 & CodeForces题目页面传送门 题意见洛谷里的翻译. 首先我们可以用区间DP算出对于每个子01串,能表示的字母串的个数. 设\(dp_{i,j}\)表示长度为\(i\),起点为\(j\)(下标从\(1\)开始)的子01串能表示的字母串的个数(也许我设的DP状态有点奇怪).显然,\(dp_{0,i}=1\),因为空子01串能且仅能表示空字母串. 那么转移怎么转移呢?可以从串的首部转移,从首部挖出\(1\sim4\)个字符组成字母,累加上剩下的串的DP值.那么状态转移方程…

morse code,摩斯电码,是一种时通时断的信号代码,通过不同的排列顺序来表达不同的英文字母.数字和标点符号. 摩斯电码,是一种早期的数字化通信形式,但是它不同于现代只使用0和1两种状态的二进制代码,它的代码包括5种: 点.划.点和划之间的停顿.每个字符间短的停顿(在点和划之间).每个词之间中等的停顿以及句子之间长的停顿. 下图为morse code的密码本 通过密码本,我们可以写1个测试程序 懂了这些,morse code基本入门了. 接着,说一下电报机原理 首先,电报机是由发报机与收报机…

摩尔斯电码,在早期的"生产斗争"生活中,扮演了很重要的角色,作为一种信息编码标准,摩尔斯电码拥有其他编码方案无法超越的长久生命.摩尔斯电码在海事通讯中被作为国际标准一直使用到1999年.不过随着科技的发展,最终还是倒在了历史的车轮下(PS:等等,为何这么熟悉??观众:好一片根正苗红的*文啊,啊啊啊啊),1997年,当法国海军停止使用摩尔斯电码时,发送的最后一条消息是:“所有人注意,这是我们在永远沉寂之前最后的一声呐喊!”. 任何关于Morse code可拓展的咨询,请移步维基百科:摩尔…

参考:https://blog.csdn.net/yuweiming70/article/details/79684433 题目描述: International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to &q…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on. F…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on. F…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b"maps to "-...", "c" maps to "-.-.", and so on. Fo…

Description International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.",…

Algorithm [leetcode]Unique Morse Code Words https://leetcode.com/problems/unique-morse-code-words/ 1)problem International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c"maps to "-.-.", and so on. Fo…

题目要求 International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a"maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on. Fo…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a"maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on. Fo…

Unique Morse Code Words Description International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" ma…

题目标签:String 题目给了我们 对应每一个 字母的 morse 密码,让我们从words 中 找出 有几个不同的 morse code 组合. 然后只要遍历 words,把每一个word 转换成 morse code,把 唯一的 存入 HashSet 就可以了,最后返回 set 的 size. Java Solution: Runtime beats  99.77% 完成日期:07/02/2018 关键词:HashSet 关键点:把唯一morse code 存入 set class Sol…

html水波纹样式,源码直接下载,代码有注释教程,小白可以看懂. 动画啥的都做好了,效果我觉得还不错 网上文章看到xbox 工程师使用隐写术,在界面的右下角放上了含有用户激活码的水波纹样式,一般人还真看不出来. 于是仿制了一个 先把莫尔兹转码部分放上来吧,莫尔兹转换部分是网上某位大佬写的,照抄. 完整代码下载在最后边 var utils = utils || {}; utils.isArray = function(value) { return Object.prototype.toStrin…

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a"maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on. Fo…

problem 804. Unique Morse Code Words solution1: class Solution { public: int uniqueMorseRepresentations(vector<string>& words) { vector<string> morse{".-", "-...","-.-.","-..",".","..-…

1675: Morse Code 时间限制: 2 Sec  内存限制: 128 MB提交: 73  解决: 33[提交][状态][讨论版] 题目描述 摩尔斯电码(英语:Morse Code)是一种时通时断的信号代码,通过不同的排列顺序来表达不同的英文字母.数字和标点符号.是由美国人萨缪尔·摩尔斯在1836年发明.                                                                                            …

[Codeforces 1197E]Culture Code(线段树优化建图+DAG上最短路) 题面 有n个空心物品,每个物品有外部体积\(out_i\)和内部体积\(in_i\),如果\(in_i> out_j\),那么j就可以套在i里面.现在我们要选出n个物品的一个子集,这个子集内的k个物品全部套在一起,且剩下的物品都无法添加到这个子集中(没有空间塞进去).定义浪费的空间为子集中空心的部分,即\(in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} -…

链接:https://ac.nowcoder.com/acm/contest/3665/E来源:牛客网 题目描述 Morse code is a character encoding scheme used in telecommunication that encodes text characters as standardized sequences of two different signal durations called dots and dashes or dits and d…

804. Unique Morse Code Words (Easy)# International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" m…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述: 题目大意 解题方法 set + map set + 字典 日期 题目地址:https://leetcode.com/problems/unique-morse-code-words/description/ 题目描述: International Morse Code defines a standard encoding where each letter is m…

Question 804. Unique Morse Code Words [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-…

Codeforces 1129 C 题意:给一个0/1串,问它的每一个前缀中的每一个子串能解析成莫尔斯电码的串的种数. 思路:首先对于这个串构造后缀自动机,那么从起点走到每一个节点的每一条路径都代表了这个串的一个子串,所以考虑以后缀自动机上的节点作为dp的对象,即\(dp(i)\)表示考虑第i个节点所表示的子串们,能够解释成多少种串. 转移方程就考虑现在在u节点,\(dp(go(u,i))\)可以加上\(dp(u)\)的值. 求答案的时候需要注意.我们需要在插入字符的时候就将每一个节点所对应的前…

题意:给你n个01字符,每次问你前缀的所有本质不同的子串,由摩斯密码组成的方案数和. 题解:离线处理,把字符建sam,通过topo序来dp计算每个节点表示的子串方案数的和.统计答案时,把n个字符挨个匹配过去,把所有经过的节点加上方案数,还有每个节点的fa链,也要统计(因为是当前串的后缀),用一个vis标记访问过没有.复杂度O(n*32),于是这题n能出到1e6 //#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC op…

Discription Arkady's code contains nn variables. Each variable has a unique name consisting of lowercase English letters only. One day Arkady decided to shorten his code. He wants to replace each variable name with its non-empty prefix so that these…

http://codeforces.com/problemset/problem/543/A 题目大意:n个人,一共要写m行程序,每个程序员每行出现的bug数为ai,要求整个程序出现的bug数不超过b的方案数. 思路:f[i][j]代表第m行,j个bug的方案数,n^3转移 #include<cstdio> #include<cmath> #include<iostream> #include<algorithm> #include<cstring&g…

有n个程序,这n个程序运作产生m行代码,但是每个程序产生的BUG总和不能超过b, 给出每个程序产生的代码,每行会产生ai个BUG,问在总BUG不超过b的情况下, 我们有几种选择方法思路:看懂了题意之后就是一个完全背包题了 定义dp[ i ][ j ][ k ] 表示前 i 个程序员,已经写了 j 行代码, 已经产生了 k 个 bugs . 根据题意,得知第 i 个程序员会写 r 行代码,那么相当于 dp[ i ][ j ][ k ] += dp[i - 1][j - r][k - ra[ i ]…

参考自:https://www.cnblogs.com/ECJTUACM-873284962/p/6423483.html B. Code For 1 time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Jon fought bravely to rescue the wildlings who were attacked by…