Description  ``Accordian'' Patience  You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate n…

题目:52张扑克,从左到右在平面上排列,按着如下规则处理: 1.按照从左到右的顺序,如果一张牌和左边的第一张或者第三张匹配,就把它放到对应的牌上面. 2.如果可以移动到多个位置,移动到最左端的牌上面.(匹配:花色或者数值相同) 分析:数据结构.栈.模拟.对于每叠牌建立一个栈,进行模拟即可. 注意:每次只移动每叠牌的最顶上的牌. #include <iostream> #include <cstdlib> #include <cstdio> using namespace…

Description   Problem B - Generalized Matrioshkas   Problem B - Generalized Matrioshkas  Vladimir worked for years making matrioshkas, those nesting dolls that certainly represent truly Russian craft. A matrioshka is a doll that may be opened in two…

Description  ``Accordian'' Patience  You are to simulate the playing of games of ``Accordian'' patience, the rules for which are as follows: Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate n…

Description   Problem E: Expressions2007/2008 ACM International Collegiate Programming Contest University of Ulm Local Contest Problem E: Expressions Arithmetic expressions are usually written with the operators in between the two operands (which is…

Description   Matrix Chain Multiplication  Matrix Chain Multiplication  Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are per…

Description Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since every operator in an expression follows all of its operands. Bob is a student in…

以下以目前遇到题目开始记录,按发布时间排序 ACM之递推递归 ACM之数学题 拓扑排序 ACM之最短路径做题笔记与记录 STL学习笔记不(定期更新) 八皇后问题解题报告…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5512 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是给了初始的集合{a, b},然后取集合里的两个元素进行加或者减的操作,生成新的元素.问最后最多能生成多少个元素.问答案的奇偶性. 首先一开始有a, b.那么如果生成了b-a(b>a),自然原来的数同样可以由b-a, a生成(b != 2a). 于是如此反复下去,最后的数必然是可以由两个数p, 2p生成的. 于是所有的数肯定可以表示成xp+…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5521 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是一个人从1开始走,一个人从n开始走.让最后相遇的时间最短. 题目就是个最短路,不过唯一不同的是,题目图的描述方式比较特别. 从规模上来看,想把这张图描述成邻接矩阵或者邻接表是不可能的. 必然只能按照题目要求的意思来存. 于是第一步存图的方式,我采用了两个vector数组,(当然此处可以使用链式前向星),一个存了和点相关的集合有哪些in[]…