题目链接:http://codeforces.com/problemset/problem/337/A 题意:有n个学生,m块puzzles,选出n块puzzles,但是需要满足这n块puzzles里的最大pieces(A)和最小pieces(B)之差最少,即the least possible difference. 这是一道贪心兼排序的题目,解决方法不难.首先对m块puzzles以非递减的顺序排序(可以保证每个n长度的区间difference最小),接着求出所有长度为n的区间中最大和最小的值…

QUESTION : What are the 10 algorithms one must know in order to solve most algorithm challenges/puzzles? ANSWER: Dynamic Programming (DP) appears to account for a plurality (some estimate up to a third) of contest problems. Of course, DP is also not…

题目:http://www.gowrikumar.com/c/ 参考:http://wangcong.org/blog/archives/291 http://www.cppblog.com/smagle/archive/2010/05/27/116211.html http://blog.chinaunix.net/uid-474889-id-2397033.html 博文索引: C puzzles详解[1-5题] C puzzles详解[6-8题] C puzzles详解[9-12题] C…

A. Puzzles Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/337/A Description The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to p…

Word Puzzles Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10782 Accepted: 4076 Special Judge Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using t…

这篇文章开始正式<algorithm puzzles>一书中的解谜之旅了! 狼羊菜过河: 谜题:一个人在河边,带着一匹狼.一只羊.一颗卷心菜.他需要用船将这三样东西运至对岸,然而,这艘船空间有限,只容得下他自己和另一样东西(狼.羊或卷心菜).若他不在场看管的话,狼就会吃掉羊,羊就会吃掉卷心菜.此人如何才能把这三个“乘客”没有损失的送到对岸? 提示:看到这个谜题的谜面非常小,利用穷举解决是绰绰有余的. 解谜:首先搬运的一定是羊,将其搬运到对岸,将其放下,随后驾空船回来,这次带狼或者卷心菜都可以,…

这个专题我们开始对<algorithm puzzles>一书的学习,这本书是一本谜题集,包括一些数学与计算机起源性的古典命题和一些比较新颖的谜题,序章的几句话非常好,在这里做简单的摘录. 手里拿着一把锤子,看什么都像钉子.我们这个年代最厉害的锤子就是算法.——William Poundstone.  解题是一种实用技能,怎么说呢,有点像游泳吧.我们学习任何使用技能的办法就是模仿和实践.——George Polya.  如果想使得上课不那么无聊,那么没有比加入带有创造力的主题更好的办法了,这些主…

Puzzles Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. Cities and roads of USC form a rooted tree (Barney's not sure why it is rooted). Root of the tree is the city num…

Problem Puzzles 题目大意 给一棵树,dfs时随机等概率选择走子树,求期望时间戳. Solution 一个非常简单的树形dp?期望dp.推导出来转移式就非常简单了. 在经过分析以后,我们发现期望时间戳其实只需要考虑自己父亲下来(步数加一)&从兄弟回来两种可能. 设size[i]为i节点子树大小(包括自身) 对于兄弟的情况,i节点的一个兄弟有1/2的可能已经被遍历完毕了,也就是步数加size该兄弟. 于是设ans[i]为到达i点的期望值,则 ans[i]=ans[Father i]+…

[cf contest697] D - Puzzles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 road…

闲下来后,需要讲最近涉及到的算法全部整理一下,有个indice,方便记忆宫殿的查找 MIT的算法课,地球上最好:https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-046j-design-and-analysis-of-algorithms-spring-2015/lecture-notes/ Why Puzzles? Solving puzzles will help you sharpen you…

概要 本章介绍<Java Puzzles>中关于异常的几个谜题.这一章都是以代码为例,相比上一章看起来更有意思.内容包括:谜题1: 优柔寡断谜题2: 极端不可思议谜题3: 不受欢迎的宾客谜题4: 您好,再见!谜题5: 不情愿的构造器谜题6: 域和流谜题7: 异常为循环而抛 转载请注明出处:http://www.cnblogs.com/skywang12345/p/3544353.html 谜题1: 优柔寡断 看看下面的程序,它到底打印什么? public class Indecisive {…

A. Puzzles Time Limit: 2 Sec  Memory Limit: 60 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373 Description The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. S…

/** 题目:pku1204 Word Puzzles 链接:http://poj.org/problem?id=1204 题意:给定一个L C(C <= 1000, L <= 1000)的字母矩阵, 再给定W(W <= 1000)个字符串,保证这些字符串都会在字母矩阵中出现(8种方向), 求它们的出现位置和方向. 思路:将单词构成ac自动机,然后对矩阵字符串从8个方向跑ac自动机, 向下方向:所有的(0,i) (0<=i<sm)为起点,一直跑到最下面. 其他方向类推: 注意…

B. Puzzles time limit per test  1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them.…

Word Puzzles Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 12090   Accepted: 4547   Special Judge Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started u…

Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client's percepti…

题目链接: B. Puzzles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between t…

Word Puzzles Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 10244   Accepted: 3864   Special Judge Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started u…

D. Puzzles time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. C…

Problem description The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her n students and give each of them a jigsaw puzzle (which, as wik…

Puzzles Time Limit: 1000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 337A64-bit integer IO format: %I64d      Java class name: (Any) The end of the school year is near and Ms. Manana, the teacher, will soon have t…

传送门:D - Puzzles 题意:在一个图中,从1开始dfs,求每一个点到达的期望: 思路:(下面是队长写的) 首先求的是到每一个点的步数的期望. 记fa( u ) = v, son( v )表示v的儿子的集合, z是son(v)中的点,其中 z != u ,  sum[z] 为 z 的子树的大小, p( z )表示z比u先访问到的概率; 那么可以发现对于u来说 ans[u] = ans[v] + 1 + x; 现在我要来算这个x, 如果 son(v).size == 1, 那么x为0; 否…

D. Puzzles Barney lives in country USC (United States of Charzeh). USC has n cities numbered from 1 through n and n - 1 roads between them. Cities and roads of USC form a rooted tree (Barney's not sure why it is rooted). Root of the tree is the city…

http://codeforces.com/problemset/problem/696/B 题目大意: 这是一颗有n个点的树,你从根开始游走,每当你第一次到达一个点时,把这个点的权记为(你已经到过不同的点的数量+1) 每一次只有当子树中所有的点都已经游走过了再会向父亲走,走到每个儿子上的概率是相同的 对于每个点,求他的权的期望 (1 ≤ n ≤ 10^5) 题解: 首先我们发现,所有子树中所有的点的编号都一定比父亲要大 而且子树中的大小关系和我们访问它的顺序有关 如果对于一个节点u它的儿子为v…

http://codeforces.com/problemset/problem/696/B (题目链接) 题意 给出一棵树,随机dfs遍历这棵树,求解每个节点的期望dfs序. Solution 考虑对于节点u,其某个儿子节点v的期望是多少. 首先,节点u的儿子的dfs的顺序是其儿子数son[x]的全排列.考虑在排列中有多少个节点在v的前面,不妨设x排在v的前面,那么满足的排列数为:${P_n^{n-2}}$,于是x对v的期望的贡献为:$${\frac{P_n^{n-2}×size[x]} {P…

期望计算的套路: 1.定义:算出所有测试值的和,除以测试次数. 2.定义:算出所有值出现的概率与其乘积之和. 3.用前一步的期望,加上两者的期望距离,递推出来. 题意: 一个树,dfs遍历子树的顺序是随机的.所对应的子树的dfs序也会不同.输出每个节点的dfs序的期望   思路: 分析一颗子树: 当前已知节点1的期望为1.0 ->anw[1]=1.0 需要通过节点1递推出节点2.4.5的期望值 1的儿子分别是2.4.5,那么dfs序所有可能的排列是6种: 1:1-2-4-5  (2.4.5节点的…

给一个L*C字符矩阵和W个字符串,问那些字符串出现在矩阵的位置,横竖斜八个向. 就是个多模式匹配的问题,直接AC自动机搞了,枚举字符矩阵八个方向的所有字符串构成主串,然后在W个模式串构造的AC自动机上跑. 另外,temp指针的那个找遗漏后缀的过程执行时标记一下,下一次再到这个结点就不需要再进行一次temp的过程,这样的时间复杂度就是O(W个模式串总长+LC). 一开始还想8个方向分别计算坐标= =写第二个方向懒得写了,然后就忽然想到可以一开始构造主串时就存坐标..最后代码很是挺长的.. #inc…

第五十一题 Write a C function which does the addition of two integers without using the '+' operator. You can use only the bitwise operators.(Remember the good old method of implementing the full-adder circuit using the or, and, xor gates....) 题目讲解: 参考:ht…

第四十六题 What does the following macro do? #define ROUNDUP(x,n) ((x+n-1)&(~(n-1))) 题目讲解: 参考:http://bbs.chinaunix.net/forum.php?mod=viewthread&tid=814501 用于内存对齐,n为2的幂. 第四十七题 Most of the C programming books, give the following example for the definitio…