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…

题意:52张牌排一行,一旦出现任何一张牌与它左边的第一张或第三张"匹配",即花色或点数相同,则须立即将其移动到那张牌上面,将其覆盖.能执行以上移动的只有压在最上面的牌.直到最后没有牌能向左移动. 注意细则:如果同时有多张牌都可以移动,你应该采取的策略是移动最左边可移动的牌.当一张牌既可以移动到左边第一张,又可以移动到左边第三张时,应移动到左边第三张上面. 代码:(Accepted,0.100s) //UVa127 - "Accordian" Patience //A…

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 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   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[]…