Description Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a broken Caps Lock key, so Edward never presses the broken Caps Lock key. Luckily, all the other keys on the QWERTY keyboard work well.…

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…

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5502  The 12th Zhejiang Provincial Collegiate Programming Contest - J Convert QWERTY to Dvorak Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, a poor copy typist, is a user o…

题目传送门 /* 模拟:手敲map一一映射,累! 除了忘记读入字符串不能用gets用getline外还是很顺利的AC了:) */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set>…

Convert QWERTY to Dvorak Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a broken Caps Lock key, so Edward never presses the broken Caps Lock key…

链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5526 Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a broken Caps Lock key, so Ed…

  Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3878 Description Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a broken Caps Lock key, so…

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[]…