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1269. Obscene Words Filter Time limit: 0.5 secondMemory limit: 8 MB There is a problem to check messages of web-board visitors for the obscene words. Your elder colleagues commit this problem to you. You are to write a program, which check if there i…
Rope in the Labyrinth Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice URAL 1145 Description A labyrinth with rectangular form and size m × n is divided into square cells with sides' length 1 by lines…
凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:packet reader failure 2. 使用lsnrctl status检查监听,一直没有响应,这个是极少见的情况. 3. 检查数据库状态为OPEN,使用nmon检查系统资源.如下一张截图所示,CPU利用率不高,但是CPU Wait%非常高.这意味着I/O不正常.可能出现了IO等待和争用(IO…
题意描述:给定K支队伍,每队三个队员,不同队伍之间队员可能部分重复,输出这些队员同时能够组成多少完整的队伍: DFS,利用DFS深度优先搜索,如果该队所有队员都没有被访问过,那么将该队计入结果,再去选择下一队~ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <stdlib.h> #include <map>…
题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串中心位置,RMQ询问LCP = min (height[rank[l]+1] to height[rank[r]]),注意的是RMQ传入参数最好是后缀的位置,因为它们在树上的顺序未知,且左边还要+1. #include <cstdio> #include <algorithm> #in…
2071. Juice Cocktails Time limit: 1.0 secondMemory limit: 64 MB Once n Denchiks come to the bar and each orders a juice cocktail. It could be from 1 to 3 different juices in each cocktail. There are three juices in the bar: apple, banana and pineappl…
2073. Log Files Time limit: 1.0 secondMemory limit: 64 MB Nikolay has decided to become the best programmer in the world! Now he regularly takes part in various programming contests, attentively listens to problems analysis and upsolves problems. But…
2070. Interesting Numbers Time limit: 2.0 secondMemory limit: 64 MB Nikolay and Asya investigate integers together in their spare time. Nikolay thinks an integer is interesting if it is a prime number. However, Asya thinks an integer is interesting i…
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