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…