Babelfish DescriptionYou have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.InputInput consists of up to 100,000 diction…
水题,竟然花了那么多时间...主要是不知道为什么,明明在本机上编译过去了,但是用c++提交却编译错误...最后用g++提交AC 题意:给出n个学生的名字,然后给出m个测验. 每个测验给出n个学生的分数. 当给出第i次测验的成绩,求Li Ming在所有学生中,前i次成绩总和的排名(若分数相同,则Li Ming排在第一个) 开始没仔细看题,以为对于每次测验,只要给出Li Ming在此次测验中的排名. 后来才知道,原来题目中有这么一句话: In the i-th line, you should gi…
题目链接poj2503 Babelfish Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 38820 Accepted: 16578 Description You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunate…
2297: Carryon的字符串 Time Limit: C/C++ 1 s Java/Python 3 s Memory Limit: 128 MB Accepted: 11 Submit: 24 Submit My Status Problem Description Carryon最近喜欢上了一些奇奇怪怪的字符,字符都是英文小写字母,但奇怪的是a可能比b小,也可能比b大,好奇怪.与此同时,他拿到了好多的字符串,可是看着很不顺眼,因为他们很乱,所以他…
What Are You Talking About Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 28482 Accepted Submission(s): 9710 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…
487-3279 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 309257 Accepted: 55224 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable wor…
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789235.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 题意:给出n本书的id.名称.作者.多个关键词.出版社.出版年然后给出m个查询,每个查询包含查询的种类.对应的内容针对每个查询,让你输出所有符合的书的id,从小到大排序,没有的话则输出No Found 首先把每个book的信息都读取处理好,然后按照id排个序…
Chat Order Time Limit:3000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit Status Description Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a mes…
一丶匿名函数 语法: 函数名 = lambda参数:返回值 # 普通的正常的函数 def func(n): return n * n ret = func(9) print(ret) # 匿名函数 a = lambda n : n * n ret = a(9) print(ret) 说是匿名函数,可总该有个名字吧,我们可以用__name__来查看一下名字 b = lambda x: x+1 a = lambda n : n * n print(a.__name__) # __name__的值都是…