题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4772 题面: Zhuge Liang's Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1404    Accepted Submission(s): 926 Problem Description In the anc…

Zhuge Liang's Password Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 59    Accepted Submission(s): 47 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous a…

这题前三段都是一堆吹爆赞助商的屁话,正式题目在图片下边,一个简单模拟题. 题目大意: 有n个男生,m个女生在进行舞会,其中一部分男生祥和比自己矮的女生跳舞,一部分男生想和比自己高的女生跳舞,一部分女生想和比自己高的男生跳舞,一部分女生想和比自己矮的男生跳舞: 所以,我们可以这样配对:想跟比自己矮的女生跳舞的男生——想跟比自己高的男生跳舞的女生:想跟比自己高的女生跳舞的男生——想跟比自己矮的男生跳舞的女生. 分组,排序,扫描配对,完事.(不过比赛中某人把代码里的n,m搞错了,从开头debug到封榜…

Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 236    Accepted Submission(s): 107 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous a…

Zhuge Liang's Mines Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4739 Description In the ancient three kingdom period, Zhuge Liang was the most famous and smartest military leader. His enemy was Shima Yi, who…

Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 831    Accepted Submission(s): 396 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous an…

Zhuge Liang's Stone Sentinel Maze Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 385    Accepted Submission(s): 106 Problem Description Zhuge Liang was a chancellor of the state of Shu Han dur…

Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 239    Accepted Submission(s): 110 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous an…

题目大意:给你一个完全二叉树,并且给他们编号,编号规则为左子树为2*k,右子树为2*k+1,每一个节点 上都有一个开关,初始时开关都处于关闭状态,小球碰到节点就会改变该点的开关的状态.然后给你I个小球, 要求输出第I个小球最终到达的叶子节点的编号. 思路分析:简单的模拟题,主要是要对二叉树的结构和编号原则清楚,有D层,则一共有(1<<D)-1个节点,然 后就可以开一个数组用来储存每一节点的状态,初识化为0,表示均处于关闭状态,然后就开始进行模拟,判断是否 越界通k>n?注意最后 要输出k…

e.... 虽然这是一道灰常简单的模拟题.但是米做的时候没有读懂第二个日历的计时方法.然后捏.敲完之后华丽的WA了进一个点.坑点就在一年的最后一天你是该输出本年的.e ...但是我好想并没有..看discuss里好想被坑的人还不少.总天数能直接整除260的时候.年数要减1. 不喜欢做模拟.....5555.... 附代码: #include<stdio.h>#include<string.h>#include<iostream>#include<string>…