VScode的live server插件问题!!!急

## **VScode的live server插件打开html,会显示 防火墙什么的也关了,弄了一下午了,各种方法都试过,网上能搜到的内容也找遍了............怎么肥四TT  **

MySQl子查询为null,导致主查询的固定值为空

此时from的查询结果此时查询全部null, 此时的t_date 也不会显示,有没有一个简单的办法,不管子查询是否为null,都会显示t_date 字段 ``` select peer_id,type,datausage,live_r, ADDDATE(curdate(),-1) t_date from( select peer_id,type, sum(total_speed) datausage,live_r from ds_transfer_data_session_202001 where start_time >= start_time >=1579305600 and start_time <= 1579392000 GROUP BY peer_id,type,live_r ) temp where datausage > 0 GROUP BY peer_id,type,live_r; ```

关于Mac安装和使用live-server的问题

本人小白一枚，问一下，关于mac上 live-server 的问题： live-server能否在mac上使用，如果能，怎么使用？目前安装好之后，使用的时候总是提示command not found，网上查询说mac使用npm需要配置.bash_profile文件，请问应该怎么配置？谢谢各位大神

Escape 逃脱问题

Problem Description 2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets. Input More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet. The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most.. 0 <= ai <= 100000 Output Determine whether all people can live up to these stars If you can output YES, otherwise output NO. Sample Input 1 1 1 1 2 2 1 0 1 0 1 1 Sample Output YES NO

java内存溢出问题，不确定哪里出的问题

项目运行的时候出现了内存溢出情况，复现方式无法确定，大概出现了五六次，每次复现的方式都不一样，下面是我用MAT工具分析的dump文件   除了hibernate里的那两个对象，其他的1000多万个对象里全是integer类型的对象 这是当时出现的时候产生的异常 ``` 2020-01-17 13:32:55 ERROR [DruidDataSource.java:2469] - create connection SQLException, url: jdbc:mysql://localhost:55060/mcs?useUnicode=true&characterEncoding=utf-8, errorCode 0, state 08S01 com.mysql.jdbc.exceptions.jdbc4.CommunicationsException: Communications link failure The last packet successfully received from the server was 1 milliseconds ago. The last packet sent successfully to the server was 0 milliseconds ago. at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method) at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:57) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45) at java.lang.reflect.Constructor.newInstance(Constructor.java:526) at com.mysql.jdbc.Util.handleNewInstance(Util.java:425) at com.mysql.jdbc.SQLError.createCommunicationsException(SQLError.java:990) at com.mysql.jdbc.MysqlIO.reuseAndReadPacket(MysqlIO.java:3517) at com.mysql.jdbc.MysqlIO.reuseAndReadPacket(MysqlIO.java:3417) at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3860) at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:864) at com.mysql.jdbc.MysqlIO.proceedHandshakeWithPluggableAuthentication(MysqlIO.java:1707) at com.mysql.jdbc.MysqlIO.doHandshake(MysqlIO.java:1217) at com.mysql.jdbc.ConnectionImpl.coreConnect(ConnectionImpl.java:2189) at com.mysql.jdbc.ConnectionImpl.connectOneTryOnly(ConnectionImpl.java:2220) at com.mysql.jdbc.ConnectionImpl.createNewIO(ConnectionImpl.java:2015) at com.mysql.jdbc.ConnectionImpl.<init>(ConnectionImpl.java:768) at com.mysql.jdbc.JDBC4Connection.<init>(JDBC4Connection.java:47) at sun.reflect.GeneratedConstructorAccessor118.newInstance(Unknown Source) at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45) at java.lang.reflect.Constructor.newInstance(Constructor.java:526) at com.mysql.jdbc.Util.handleNewInstance(Util.java:425) at com.mysql.jdbc.ConnectionImpl.getInstance(ConnectionImpl.java:385) at com.mysql.jdbc.NonRegisteringDriver.connect(NonRegisteringDriver.java:323) at com.alibaba.druid.pool.DruidAbstractDataSource.createPhysicalConnection(DruidAbstractDataSource.java:1513) at com.alibaba.druid.pool.DruidAbstractDataSource.createPhysicalConnection(DruidAbstractDataSource.java:1578) at com.alibaba.druid.pool.DruidDataSource$CreateConnectionThread.run(DruidDataSource.java:2466) Caused by: java.io.EOFException: Can not read response from server. Expected to read 4 bytes, read 0 bytes before connection was unexpectedly lost. at com.mysql.jdbc.MysqlIO.readFully(MysqlIO.java:2969) at com.mysql.jdbc.MysqlIO.reuseAndReadPacket(MysqlIO.java:3427) ... 19 more 2020-01-17 13:33:03 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_preset_rlat failed: java.lang.OutOfMemoryError: GC overhead limit exceeded at net.sf.ehcache.util.MemoryEfficientByteArrayOutputStream.getBytes(MemoryEfficientByteArrayOutputStream.java:65) at net.sf.ehcache.util.MemoryEfficientByteArrayOutputStream.serialize(MemoryEfficientByteArrayOutputStream.java:99) at net.sf.ehcache.store.disk.DiskStorageFactory.serializeElement(DiskStorageFactory.java:405) at net.sf.ehcache.store.disk.DiskStorageFactory.write(DiskStorageFactory.java:384) at net.sf.ehcache.store.disk.DiskStorageFactory$DiskWriteTask.call(DiskStorageFactory.java:485) at net.sf.ehcache.store.disk.DiskStorageFactory$PersistentDiskWriteTask.call(DiskStorageFactory.java:1088) at net.sf.ehcache.store.disk.DiskStorageFactory$PersistentDiskWriteTask.call(DiskStorageFactory.java:1072) at java.util.concurrent.FutureTask.run(FutureTask.java:262) at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$201(ScheduledThreadPoolExecutor.java:178) at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(ScheduledThreadPoolExecutor.java:292) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145) at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615) at java.lang.Thread.run(Thread.java:745) 2020-01-17 13:32:59 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:32:59 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:32:55 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:16 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:27 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:33:27 ERROR [SqlExceptionHelper.java:144] - wait millis 7401, active 16, maxActive 40, creating 1 2020-01-17 13:33:27 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-17 13:33:33 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:33 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:49 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:49 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:49 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:49 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:49 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:53 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:53 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:33:53 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:02 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:34:02 ERROR [SqlExceptionHelper.java:144] - wait millis 5503, active 18, maxActive 40, creating 1 2020-01-17 13:34:02 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-17 13:34:04 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:08 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:12 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:12 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:14 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:14 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:14 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:17 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:34:17 ERROR [SqlExceptionHelper.java:144] - wait millis 5485, active 19, maxActive 40, creating 1 2020-01-17 13:34:17 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.orm.hibernate3.HibernateJdbcException: JDBC exception on Hibernate data access: SQLException for SQL [n/a]; SQL state [null]; error code [0]; Could not open connection; nested exception is org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-17 13:34:17 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:27 INFO [DeviceCache.java:249] - 设备状态及报警维护过程中出现异常 2020-01-17 13:34:35 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:35 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:42 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:42 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:44 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:44 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:34:52 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:34:52 ERROR [SqlExceptionHelper.java:144] - wait millis 7301, active 22, maxActive 40, creating 1 2020-01-17 13:34:52 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-17 13:34:53 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:35:05 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_roll_preset failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-17 13:35:18 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:35:18 ERROR [SqlExceptionHelper.java:144] - Error 2020-01-17 13:35:50 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-17 13:35:50 ERROR [SqlExceptionHelper.java:144] - Error 2020-01-17 13:36:13 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-17 13:36:13 ERROR [SqlExceptionHelper.java:144] - Lock wait timeout exceeded; try restarting transaction 2020-01-17 13:36:13 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-17 13:36:24 ERROR [SqlExceptionHelper.java:144] - Lock wait timeout exceeded; try restarting transaction 2020-01-17 13:36:13 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-17 13:37:01 ERROR [SqlExceptionHelper.java:144] - Lock wait timeout exceeded; try restarting transaction 2020-01-17 13:37:43 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-17 13:36:37 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-17 13:36:33 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-17 13:36:22 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 ``` 现在就是无法确定到底哪里导致的内存溢出？ 这个是上次出现的异常和代码 ``` 2020-01-13 13:21:44 WARN [SqlExceptionHelper.java:143] - SQL Error: 1205, SQLState: 40001 2020-01-13 13:21:44 ERROR [SqlExceptionHelper.java:144] - Lock wait timeout exceeded; try restarting transaction 2020-01-13 13:23:55 INFO [MQTTProtocolHandler.java:283] - 接收到设备接入协议：{"topic":"info/deviceBaseInfo/BHIP118-S/00:00:01:A6:00:A2","reportTime":"256573751","decodeCapacity":{"totalLevel":2,"totalBlock":36,"totalPixel":16588800},"outputList":[{"status":"start","ratio":"UHD","pixel":"3840*2160","type":"UDP","url":"udp://231.0.100.80:7001"},{"status":"start","ratio":"HD","pixel":"960*540","type":"UDP","url":"udp://231.0.101.80:7001"},{"status":"start","ratio":"SD","pixel":"352*288","type":"UDP","url":"udp://231.0.102.80:7001"},{"status":"stop","ratio":"UHD","pixel":"3840*2160","type":"RTMP","url":"rtmp://192.168.15.80:1935/live/100"},{"status":"stop","ratio":"HD","pixel":"960*540","type":"RTMP","url":"rtmp://192.168.15.124:1935/live/101"},{"status":"stop","ratio":"SD","pixel":"352*288","type":"RTMP","url":"rtmp://192.168.15.80:1935/live/102"}],"baseInfo":{"deviceType":"BHIP118-S","code":"00:00:01:A6:00:A2","version":"v1.0.0.11","ip":"192.168.16.80"},"online":true} 2020-01-13 13:23:59 INFO [DeviceCache.java:127] - 接收到新增/更新设备信息:00:00:01:A6:00:A2 ：BaseInfo [type=0,name=BHIP118-S,ip=192.168.16.80,port=0,code=00:00:01:A6:00:A2,groupCode=null,groupIndex=0,online=true,multiCastTime=0,ver=v1.0.0.11,channel=0,reserve=0,lockStatus=null,workMode=null,videoPixerls=null,kvmMode=null,serialNumber=null,sdipPortInfo is null,hdIpPortInfo is null,audioIpPortInfo is null,outputList[SourceOutput [ratio=UHD, url=udp://231.0.100.80:7001, pixel=3840*2160, channel=0, SourceOutput [ratio=HD, url=udp://231.0.101.80:7001, pixel=960*540, channel=0, SourceOutput [ratio=SD, url=udp://231.0.102.80:7001, pixel=352*288, channel=0, SourceOutput [ratio=UHD, url=rtmp://192.168.15.80:1935/live/100, pixel=3840*2160, channel=0, SourceOutput [ratio=HD, url=rtmp://192.168.15.124:1935/live/101, pixel=960*540, channel=0, SourceOutput [ratio=SD, url=rtmp://192.168.15.80:1935/live/102, pixel=352*288, channel=0],callStatus=null] 2020-01-13 13:24:06 INFO [MultiCastDeviceInfoHandler.java:229] - 更新视频合成器设备信息: Device [id=ff8080816f694555016f6a4bf61e158c, code=00:00:01:A6:00:A2, name=192.168.16.80, ip=192.168.16.80, port=0, status=0, type=BHIP118, deviceType=BHIP118-S, stamp=2020-01-03 15:26:00, abilityInfo=null, netCardInfo=null] 2020-01-13 13:24:53 INFO [MultiCastDeviceInfoHandler.java:364] - 更新解码器信息:Decoder [IP=192.168.16.80, Port=0, Channel=0, totalBlocks=36, totalLevel=2, uRefWidth=1920, uRefHeight=1080, audioPort=0, totalPixel=16588800, deviceTypeName=BHIP118-S, pixelsWidth=null, pixelsHeight=null] 2020-01-13 13:25:17 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:17 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:17 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:19 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:28 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:42 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-13 13:25:42 ERROR [SqlExceptionHelper.java:144] - wait millis 14463, active 11, maxActive 40, creating 1 2020-01-13 13:25:46 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_playback_task failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:25:48 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-13 13:25:50 ERROR [SqlExceptionHelper.java:144] - wait millis 18056, active 11, maxActive 40, creating 1 2020-01-13 13:25:50 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_source_volume failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:25:51 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:51 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:51 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.orm.hibernate3.HibernateJdbcException: JDBC exception on Hibernate data access: SQLException for SQL [n/a]; SQL state [null]; error code [0]; Could not open connection; nested exception is org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-13 13:25:51 INFO [GetCWRCapHandler.java:114] - 解析录像服务器上报心跳协议过程中出现异常：org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.GenericJDBCException: Could not open connection 2020-01-13 13:25:58 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_kvm failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:25:56 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:25:56 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:03 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_video_terminal failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:05 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_pic_layer failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:05 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_template failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:05 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_multicast_address failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:08 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_mass_source failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:08 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_pic_encoder_block failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:08 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_ipc_preset_group failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:08 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:13 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_video_terminal failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:17 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_record_layer_block failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:24 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_video_meeting failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:28 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_audio_preset_rlat failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:28 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_preset failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:28 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_layer_block failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:28 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:32 WARN [SqlExceptionHelper.java:143] - SQL Error: 0, SQLState: null 2020-01-13 13:26:32 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:32 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:32 ERROR [DiskStorageFactory.java:495] - Disk Write of mcs_encoder failed: java.lang.OutOfMemoryError: GC overhead limit exceeded 2020-01-13 13:26:52 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:27:07 ERROR [SqlExceptionHelper.java:144] - wait millis 20150, active 12, maxActive 40, creating 1 2020-01-13 13:26:50 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:46 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:46 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:41 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:41 ERROR [SqlExceptionHelper.java:144] - wait millis 23244, active 12, maxActive 40, creating 1 2020-01-13 13:26:41 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:26:41 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:30:29 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:30:25 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:30:07 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:30:07 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 2020-01-13 13:30:07 WARN [DruidDataSource.java:1258] - get connection timeout retry : 1 ``` 这个是组播上报的设备信息，因为组播是一秒钟上报了好几次，是不是这里一直占用着连接 ``` public void run() { try { ms = new MulticastSocket(multiCastPort);//建立组播套接字 ms.setReceiveBufferSize(65535); // ms.setNetworkInterface(NetworkInterface.getByInetAddress(localAddress)); ms.joinGroup(InetAddress.getByName(multiCastIP));//加入组播组 LogHome.getLog().info("正常启动监听" + localAddress.getHostAddress()); byte[] buffer = null; DatagramPacket dp = null; while (true) { try { if (IsStop) { break; } buffer = new byte[1400]; dp = new DatagramPacket(buffer, buffer.length); LogHome.getLog().debug("等待接受组播信息："); ms.receive(dp); MultiCastInfoParser info = new MultiCastInfoParser(dp.getData()); DeviceCache.baseInfoOf1004Handler(info.parse()); } catch (Exception e) { if (IsStop) {// 如果换网卡可能会出问题 break; } else { if (ms.isClosed()) { ms = new MulticastSocket(multiCastPort); ms.setNetworkInterface(NetworkInterface.getByInetAddress(localAddress)); ms.joinGroup(InetAddress.getByName(multiCastIP)); } } LogHome.getLog().error("接受并设置组播信息失败",e); }finally{ Thread.sleep(1); } } } catch (Exception e) { e.printStackTrace(); try { if (listenerList.containsKey(localAddress.getHostAddress())) { listenerList.remove(localAddress.getHostAddress()); } } catch (Exception ex) { LogHome.getLog().error(ex); } } ```

ChaeYeon 如何实现呢

Problem Description Chae Yeon is a popular pop female singer who rose to fame with her amazing sexy dance style and the sounds of nature voice she has. She proved to be a great dancer, and she showed off her vocals in her live performances. If you had ever seen her dance, I bet you’d love it. I felt stage lighting interesting when I was enjoying Chae Yeon’s performance. We all know that stage lighting instruments are used for the concerts and other performances taking place in live performance venues. They are also used to light the stages. Actually this is a color mixing process. There are two types of color mixing: Additive and Subtractive. Most stages use the former and in this case there are three primary colors: red, green, and blue. In the absence of color, or when no colors are showing, the stage is black. If all three primary colors are showing, the result is white. When red and green combine together, the result is yellow. When red and blue combine together, the result is magenta. When blue and green combine together, the result is cyan. When two same color combine together, the result is still that color. We have got the coordinate and the primary color of the figure that each Stage Lighting Instrument sent out. For simplicity’s sake, we can just treat the figure as a circle. Of course we’ll know the radius of each colored circle. Some color may be changed based on the Color Mixed Theory we mentioned above. Can you find the area of each color? Input The first line consists of an integer T, indicating the number of test cases. The first line of each case consists of three integers R, G, B, indicating the number of red circles, green circles and blue circles. The next R + G + B lines, each line consists of three integer x, y, r, indicating the coordinate and the radius. The first R lines descript the red circles, the second G lines descript the green circles and the last B lines descript the blue circles. Output Output seven floating numbers, they are the area of red, green, blue, white, yellow, magenta and cyan. Please take each number with two factional digits. Constraints 0 < T <= 20 0 <= R, G, B <= 100 -100 <= x, y <= 100; 0 <= r <= 100 Sample Input 1 1 1 1 0 2 3 2 0 3 -2 0 3 Sample Output 9.28 15.04 15.04 4.92 7.04 7.04 1.28

Holiday's Accommodation 安排的算法问题

Problem Description Nowadays, people have many ways to save money on accommodation when they are on vacation. One of these ways is exchanging houses with other people. Here is a group of N people who want to travel around the world. They live in different cities, so they can travel to some other people's city and use someone's house temporary. Now they want to make a plan that choose a destination for each person. There are 2 rules should be satisfied: 1. All the people should go to one of the other people's city. 2. Two of them never go to the same city, because they are not willing to share a house. They want to maximize the sum of all people's travel distance. The travel distance of a person is the distance between the city he lives in and the city he travels to. These N cities have N - 1 highways connecting them. The travelers always choose the shortest path when traveling. Given the highways' information, it is your job to find the best plan, that maximum the total travel distance of all people. Input The first line of input contains one integer T(1 <= T <= 10), indicating the number of test cases. Each test case contains several lines. The first line contains an integer N(2 <= N <= 105), representing the number of cities. Then the followingN-1 lines each contains three integersX, Y,Z(1 <= X, Y <= N, 1 <= Z <= 106), means that there is a highway between city X and city Y , and length of that highway. You can assume all the cities are connected and the highways are bi-directional. Output For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y represents the largest total travel distance of all people. Sample Input 2 4 1 2 3 2 3 2 4 3 2 6 1 2 3 2 3 4 2 4 1 4 5 8 5 6 5 Sample Output Case #1: 18 Case #2: 62

oracle中一段sql查询优化的思考，第一段sql是原来的，第二段是优化的sql，想问下为什么第一段会那么慢？

oracle中一段sql查询优化的思考，第一段sql是原来的，第二段是优化的sql，想问下为什么第一段会那么慢？ INFO_TB_XXX_LIVE 数据量300万+，INFO_TB_XXX 数据量800万+ 第一段 ``` select b.id, (select count(l.id) from INFO_TB_XXX_LIVE l where l.build_id = b.id and l.del_flag = '0' and l.LIVE_EDATE IS NULL) as LIVE_COUNT from INFO_TB_XXX b ``` 第二段 ``` select b.id id, count(l.id) LIVE_COUNT from INFO_TB_XXX b join INFO_TB_XXXX_LIVE l on l.build_id = b.id and l.del_flag = '0' and l.LIVE_EDATE IS NULL group by b.id ``` 两段sql结果是一样的， 但是还不清楚第一段为啥那么慢？

Moles 的算法问题

Problem Description A mole is a strange mammal adapted to a subterranean lifestyle. It has invisible eyes and short, powerful limbs with large paws oriented for digging. Before the catastrophic earthquake, a group of moles have perceived the disaster. They move to a safe place and want to dig holes under the ground. There are n moles, numbered from 1 to n. They stand in a line and dig holes one by one. The mole line is not necessary sorted by the mole number. Every mole should live in a hole dug by itself and every mole just digs one hole. The first mole in the line digs the first hole with a channel to the ground. Then other moles go down through that channel and dig more holes and channels. A hole may have at most three neighbors connected by channels, one is on the upper level, and the other two holes are on the lower level lying on the left side and the right side. When a mole reaches a hole, if its number is smaller than the hole owner's, it will go to the left-lower hole(or digs a left-lower hole and stays there when there is no left-lower hole), otherwise it will go to the right-lower hole(or digs a right-lower hole and stays there when there is no right-lower hole). Due to the excellent ability and well-designed layout, these holes and channels will not cross with each other. Mouse Jerry is a friend of those moles. He comes to visit them and prepares gifts for every mole. There is a rule that the mole whose number is smaller must get the gift earlier than the mole whose number is larger. Jerry starts from the ground. He travels through the holes and channels to deliver his gifts. After giving out all his gifts, he comes back to the ground. In the mole world, it is interesting that the moles with odd numbers are males and others are females. When reaching a hole, Jerry takes a note about the gender of the hole owner(0 represents female and 1 represents male). When he gets back to the ground, he will get a 0-1 sequence. Now he wants to calculate the "harmony value". The harmony value represents the number of occurrences of a given "harmony string" in the above mentioned sequence. Occurrences may overlap. Please note that Jerry is very smart so his travel distance is as small as possible. Input The first line contains an integer t meaning that there are t test cases(t <= 10). For each test case : The first line is an integer n meaning that there are n moles. (n <= 600000). The second line contains n integers representing the mole line. Each integer is a mole's number and the first integer is the number of the first mole in the line. The third line is the harmony string. The string length is not large than 7000. Output For each test case, please output the case number and the harmony value. Sample Input 2 8 5 1 3 2 7 8 4 6 01 10 1 2 3 4 5 6 7 8 9 10 1010 Sample Output Case #1: 4 Case #2: 8

Evil Straw Warts Live 怎么解决

Description A palindrome is a string of symbols that is equal to itself when reversed. Given an input string, not necessarily a palindrome, compute the number of swaps necessary to transform the string into a palindrome. By swap we mean reversing the order of two adjacent symbols. For example, the string "mamad" may be transformed into the palindrome "madam" with 3 swaps: swap "ad" to yield "mamda" swap "md" to yield "madma" swap "ma" to yield "madam" Input The first line of input gives n, the number of test cases. For each test case, one line of input follows, containing a string of up to 8000 lowercase letters. Output Output consists of one line per test case. This line will contain the number of swaps, or "Impossible" if it is not possible to transform the input to a palindrome. Sample Input 3 mamad asflkj aabb Sample Output 3 Impossible 2

Building roads 道路修建问题

Problem Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he must build N * (N - 1) / 2 roads, which is so costly that cheapskate John will never do that, though that's the best choice for the cows. Clever John just had another good idea. He first builds two transferring point S1 and S2, and then builds a road connecting S1 and S2 and N roads connecting each barn with S1 or S2, namely every barn will connect with S1 or S2, but not both. So that every pair of barns will be connected by the roads. To make the cows don't spend too much time while dropping around, John wants to minimize the maximum of distances between every pair of barns. That's not the whole story because there is another troublesome problem. The cows of some barns hate each other, and John can't connect their barns to the same transferring point. The cows of some barns are friends with each other, and John must connect their barns to the same transferring point. What a headache! Now John turns to you for help. Your task is to find a feasible optimal road-building scheme to make the maximum of distances between every pair of barns as short as possible, which means that you must decide which transferring point each barn should connect to. We have known the coordinates of S1, S2 and the N barns, the pairs of barns in which the cows hate each other, and the pairs of barns in which the cows are friends with each other. Note that John always builds roads vertically and horizontally, so the length of road between two places is their Manhattan distance. For example, saying two points with coordinates (x1, y1) and (x2, y2), the Manhattan distance between them is |x1 - x2| + |y1 - y2|. Input The first line of input consists of 3 integers N, A and B (2 <= N <= 500, 0 <= A <= 1000, 0 <= B <= 1000), which are the number of barns, the number of pairs of barns in which the cows hate each other and the number of pairs of barns in which the cows are friends with each other. Next line contains 4 integer sx1, sy1, sx2, sy2, which are the coordinates of two different transferring point S1 and S2 respectively. Each of the following N line contains two integer x and y. They are coordinates of the barns from the first barn to the last one. Each of the following A lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows hate each other. The same pair of barns never appears more than once. Each of the following B lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows are friends with each other. The same pair of barns never appears more than once. You should note that all the coordinates are in the range [-1000000, 1000000]. Output You just need output a line containing a single integer, which represents the maximum of the distances between every pair of barns, if John selects the optimal road-building scheme. Note if there is no feasible solution, just output -1. Sample Input 4 1 1 12750 28546 15361 32055 6706 3887 10754 8166 12668 19380 15788 16059 3 4 2 3 Sample Output 53246

Roba number 是怎么写的

Problem Description As we know, in Roba's seraglio,there are many girls,according to this image. So,you will be curious.So many girls,Roba may don't have enough time. You start to doubt the authentic of this problem,assume Roba can meet 100 girls per day,and Roba can live 100years or so,then his dream will come true,but you don't know how many girs Roba will meet,so you have to guess. So,the bigger positive factor the number has,the more likely the number become the girls in Roba's seraglio!. But this is not the case,define Roba number,the least probably a number become the girls in Roba's seraglio,ie,the numbers have the least positive factors between 30000000 and 40000000,You are to output all the numbers Output all the Roba numbers between 3*10^7 and 4*10^7 Sample Output 30000001 30000023 30000037 ...... So many

Walk 旅行的问题

Problem Description Alice would like to visit Bob. However, they live in a hilly landscape, and Alice doesn’t like to walk in hills. She has a map of the area, showing the height curves. You have to calculate the total altitude climbed, and the total altitude descended, for the route which minimizes these numbers. It does not matter how far she has to walk to achieve this. Since you don’t know what the landscape looks like in between the height curves, you cannot know exactly how much climb and descent she will actually get in practice, but you should calculate the minimum possible under optimal conditions based on what you can deduce from the map. The map is represented as an xy grid. Alice lives in (0, 0), and Bob lives in (100 000, 0). The height curves are represented as polygons, where a polygon cannot intersect itself or another polygon. Furthermore, neither Alice nor Bob lives exactly on a height curve. Second test case from sample input (compressed). Input On the first line one positive number: the number of testcases, at most 100. After that per testcase: One line with 0 ≤ N ≤ 2 500, the number of height curves. One line for each height curve, with 1 ≤ Hi ≤ 1 000 being the height of the curve, 3 ≤ Pi ≤ 2 000 the number of vertices in the polygon, and the vertices x1, y1, …, xPi, yPi having integral values &#8722;300 000 ≤ xi, yi ≤ 300 000. There will be no more than 200 000 polygon vertices in total in all test cases. Output Per testcase: One line with two numbers: the total altitude climbed and the total altitude descended. Sample Input 2 2 20 3 10 10 0 -10 -10 10 25 3 20 20 0 -20 -20 20 3 100 4 -1 1 1 1 1 -1 -1 -1 300 8 -2 2 2 2 2 -2 5 -2 5 1 6 1 6 -3 -2 -3 50 8 3 3 100001 3 100001 -1 7 -1 7 2 4 2 4 -1 3 -1 Sample Output 5 0 200 250 Source

使用python实现rtsp客户端遇见的setup消息发送的问题

使用python实现rtsp客户端的setup消息发送时返回状态码为503；现在不知如何解决，遂请求帮助。有偿+wx：ww1194609610（10rmb） ``` import socket from urllib.parse import urlparse config_dict = { 'cseq': 2, 'user_agent': 'LibVLC/3.0.2 (LIVE555 Streaming Media v2016.11.28)', 'timeout': 3, 'recvbite': 4096, 'res_status': '200 OK', 'rtsp_status': 'flase' } clientports=[60784, 60785] def options_get(url): ''' options请求检测 url: rtsp流地址 return: options请求相应 ''' url = urlparse(url) host = url.netloc hostname = url.hostname path = url.path port = url.port str_options = 'OPTIONS rtsp://' + str(host) + \ path + ' RTSP/1.0\r\n' str_options += 'CSeq: ' + str(config_dict['cseq']) + '\r\n' str_options += 'User-Agent: ' + config_dict['user_agent'] + '\r\n' str_options += '\r\n' print(str_options) client = socket.socket(socket.AF_INET, socket.SOCK_STREAM) client.settimeout(config_dict['timeout']) client.connect((hostname, port)) client.send(str_options.encode()) d = client.recv(config_dict['recvbite']) return d def describe_get(url): ''' describe请求检测 url: rtsp流地址 return: describe请求相应 ''' url = urlparse(url) host = url.netloc hostname = url.hostname path = url.path port = url.port str_describe = 'DESCRIBE rtsp://' + str(host) + \ path + ' RTSP/1.0\r\n' str_describe += 'CSeq: ' + str(config_dict['cseq'] + 1) + '\r\n' str_describe += 'User-Agent: ' + config_dict['user_agent'] + '\r\n' str_describe += '\r\n' client = socket.socket(socket.AF_INET, socket.SOCK_STREAM) client.settimeout(config_dict['timeout']) client.connect((hostname, port)) client.send(str_describe.encode()) d = client.recv(config_dict['recvbite']) return d def setup_get(url): ''' setup请求检测 url: rtsp流地址 return: setup请求相应 ''' url = urlparse(url) host = url.netloc hostname = url.hostname path = url.path port = url.port str_setup = 'SETUP rtsp://' + str(host) + path + '/' + 'streamid=0' + ' RTSP/1.0\r\n' str_setup += 'CSeq: ' + str(config_dict['cseq'] + 2) + '\r\n' str_setup += 'User-Agent: ' + config_dict['user_agent'] + '\r\n' # config_dict['user_agent'] str_setup += 'Transport: RTP/AVP;unicast;client_port=61740-61741\r\n\r\n' str_setup += '\r\n' client = socket.socket(socket.AF_INET, socket.SOCK_STREAM) client.settimeout(config_dict['timeout']) client.connect((hostname, port)) client.send(str_setup.encode()) d = client.recv(config_dict['recvbite']) return d def teardown_get(url): ''' teardown请求检测 url: rtsp流地址 return: teardown请求相应 ''' url = urlparse(url) host = url.netloc hostname = url.hostname path = url.path port = url.port str_teardown = 'TEARDOWN rtsp://' + str(host) + path + ' RTSP/1.0\r\n' str_teardown += 'CSeq: ' + str(config_dict['cseq'] + 4) + '\r\n' str_teardown += 'User-Agent: ' + config_dict['user_agent'] + '\r\n' str_teardown += '\r\n' print(str_teardown) client = socket.socket(socket.AF_INET, socket.SOCK_STREAM) client.settimeout(config_dict['timeout']) client.connect((hostname, port)) client.send(str_teardown.encode()) d = client.recv(config_dict['recvbite']) return d def send_main(url): try: str_options = str(options_get(url)) print(str_options) if config_dict['res_status'] in str_options: str_des = str(describe_get(url)) print(str_des) if config_dict['res_status'] in str_des: str_setup = str(setup_get(url)) str_teardown = str(teardown_get(url)) print(str_setup) print(str_teardown) if config_dict['res_status'] in str_teardown: config_dict['rtsp_status'] = 'true' return True except Exception: return False else: return False print(send_main('rtsp://192.168.10.214:554/live/av0')) ``` 上方为我的代码。 b'RTSP/1.0 503 Service Unavailable\r\nCSeq: 4\r\nDate: Thu, 01 Jan 1970 07:11:16 GMT\r\n\r\n' 此段消息为setup请求的返回信息，我看过rtsp的相关文档，说describe请求返回的sdp信息，客户端再分析该SDP描述,并为会话中的每一个流发送一个RTSP建立命令( SETUP)。这个我没搞懂，也不知道如何写，这里进行请教一下。 下面是我的一次正常的rtsp交互信息。 ``` OPTIONS rtsp://192.168.10.214:554/live/av0 RTSP/1.0 CSeq: 2 User-Agent: LibVLC/3.0.5 (LIVE555 Streaming Media v2016.11.28) RTSP/1.0 200 OK CSeq: 2 Public: OPTIONS, DESCRIBE, SETUP, TEARDOWN, PLAY, PAUSE DESCRIBE rtsp://192.168.10.214:554/live/av0 RTSP/1.0 CSeq: 3 User-Agent: LibVLC/3.0.5 (LIVE555 Streaming Media v2016.11.28) Accept: application/sdp RTSP/1.0 200 OK CSeq: 3 Date: Thu, 01 Jan 1970 07:05:17 GMT Content-Base: rtsp://192.168.10.214:554/live/av0/ Content-Type: application/sdp Content-Length: 315 v=0 o=- 0 0 IN IP4 127.0.0.1 s=No Title c=IN IP4 0.0.0.0 t=0 0 a=tool:libavformat 55.12.100 m=video 0 RTP/AVP 96 a=rtpmap:96 H264/90000 a=fmtp:96 packetization-mode=1; sprop-parameter-sets=Z2QAKq2EAQwgCGEAQwgCGEAQwgCEK1A8ARPyzcBAQFAAAAMAEAAAAwPIQA==,aO48sA==; profile-level-id=64002A a=control:streamid=0 SETUP rtsp://192.168.10.214:554/live/av0/streamid=0 RTSP/1.0 CSeq: 4 User-Agent: LibVLC/3.0.5 (LIVE555 Streaming Media v2016.11.28) Transport: RTP/AVP;unicast;client_port=61740-61741 RTSP/1.0 200 OK CSeq: 4 Date: Thu, 01 Jan 1970 07:05:17 GMT Session: 093634014dba841b Transport: RTP/AVP/UDP;unicast;client_port=61740-61741;server_port=20006-20007 PLAY rtsp://192.168.10.214:554/live/av0/ RTSP/1.0 CSeq: 5 User-Agent: LibVLC/3.0.5 (LIVE555 Streaming Media v2016.11.28) Session: 093634014dba841b Range: npt=0.000- RTSP/1.0 200 OK CSeq: 5 Date: Thu, 01 Jan 1970 07:05:17 GMT Session: 093634014dba841b TEARDOWN rtsp://192.168.10.214:554/live/av0/ RTSP/1.0 CSeq: 6 User-Agent: LibVLC/3.0.5 (LIVE555 Streaming Media v2016.11.28) Session: 093634014dba841b RTSP/1.0 200 OK CSeq: 6 Date: Thu, 01 Jan 1970 07:05:21 GMT Session: 093634014dba841b ```

Dark Room 是怎么编写的呢

Problem Description One day Harry Potter came to a magic room, in which there are some shadow monsters sent by the black wizard waiting for him. Fortunately, shadow monsters can only live in areas where there are no lights shining, there are n*m lamps on the roof of the room, every little lamp can only illuminate the scope of the fixed area, and all the area of lights shining don't overlap, some lamps on, some lamps off, the areas below the closed lamps will appear the shadow, and in these areas shadow monsters will attack Harry Potter. Now Harry Potter comes to you and gives you the previous state of the room lights, Harry Potter can control the light switch in a remote place .The state of a light will change as follows: if a lamp state changed, the state of the lights which before it, after it, left to it or right to it will all change. Your task is to calculate how many times we need at least to turn on all the lights by changing the state of the lamps. Input in the first line there are two integers, n (1 < = n < = 100) and m (1 < = m < = 15), stands for there are n lines, each line have m data, 0 stands for the lights off, 1 stands for the lights on. Output for each group of input data, output an integer stands for the minimum number we need to change the state of the lights to put all the lights on, if we can't turn on all the lamps, output 'no solution'. Sample Input 3 3 1 0 1 0 0 0 1 0 1 Sample Output 1

Dark Room 程序的设计

Problem Description One day Harry Potter came to a magic room, in which there are some shadow monsters sent by the black wizard waiting for him. Fortunately, shadow monsters can only live in areas where there are no lights shining, there are n*m lamps on the roof of the room, every little lamp can only illuminate the scope of the fixed area, and all the area of lights shining don't overlap, some lamps on, some lamps off, the areas below the closed lamps will appear the shadow, and in these areas shadow monsters will attack Harry Potter. Now Harry Potter comes to you and gives you the previous state of the room lights, Harry Potter can control the light switch in a remote place .The state of a light will change as follows: if a lamp state changed, the state of the lights which before it, after it, left to it or right to it will all change. Your task is to calculate how many times we need at least to turn on all the lights by changing the state of the lamps. Input in the first line there are two integers, n (1 < = n < = 100) and m (1 < = m < = 15), stands for there are n lines, each line have m data, 0 stands for the lights off, 1 stands for the lights on. Output for each group of input data, output an integer stands for the minimum number we need to change the state of the lights to put all the lights on, if we can't turn on all the lamps, output 'no solution'. Sample Input 3 3 1 0 1 0 0 0 1 0 1 Sample Output 1

Groundhog Build Home 具体的编写

Problem Description Groundhogs are good at digging holes, their home is a hole, usually a group of groundhogs will find a more suitable area for their activities and build their home at this area .xiaomi has grown up, can no longer live with its parents.so it needs to build its own home.xiaomi like to visit other family so much, at each visit it always start from the point of his own home.Xiaomi will visit all of the groundhogs' home in this area(it will chose the linear distance between two homes).To save energy,xiaomi would like you to help it find where its home built,so that the longest distance between xiaomi's home and the other groundhog's home is minimum. Input The input consists of many test cases，ending of eof.Each test case begins with a line containing three integers X, Y, N separated by space.The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= N<= 1000. Groundhogs acivity at a rectangular area ,and X, Y is the two side of this rectangle, The number N stands for the number of holes.Then exactly N lines follow, each containing two integer numbers xi and yi (0 <= xi <= X, 0 <= yi <= Y) indicating the coordinates of one home. Output Print exactly two lines for each test case.The first line is the coordinate of xiaomi's home which we help to find. The second line is he longest distance between xiaomi's home and the other groundhog's home.The output round to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1). Sample Input 1000 50 1 10 10 1000 50 4 0 0 1 0 0 1 1 1 Sample Output (10.0,10.0). 0.0 (0.5,0.5). 0.7

Groundhog Build Home

Problem Description Groundhogs are good at digging holes, their home is a hole, usually a group of groundhogs will find a more suitable area for their activities and build their home at this area .xiaomi has grown up, can no longer live with its parents.so it needs to build its own home.xiaomi like to visit other family so much, at each visit it always start from the point of his own home.Xiaomi will visit all of the groundhogs' home in this area(it will chose the linear distance between two homes).To save energy,xiaomi would like you to help it find where its home built,so that the longest distance between xiaomi's home and the other groundhog's home is minimum. Input The input consists of many test cases，ending of eof.Each test case begins with a line containing three integers X, Y, N separated by space.The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= N<= 1000. Groundhogs acivity at a rectangular area ,and X, Y is the two side of this rectangle, The number N stands for the number of holes.Then exactly N lines follow, each containing two integer numbers xi and yi (0 <= xi <= X, 0 <= yi <= Y) indicating the coordinates of one home. Output Print exactly two lines for each test case.The first line is the coordinate of xiaomi's home which we help to find. The second line is he longest distance between xiaomi's home and the other groundhog's home.The output round to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1). Sample Input 1000 50 1 10 10 1000 50 4 0 0 1 0 0 1 1 1 Sample Output (10.0,10.0). 0.0 (0.5,0.5). 0.7

If You are my legend 求正确解答

Problem Description If you are my legend------If your are my legend, let it as long as the heaven and earth endure（Sky Long Earth Nine）. WhereIsHeroFrom loves a girl very much, we call her X Gu Niang. But today they can’t get together because they live in two different provinces and their distance is much farther. Now they want to build a castle for their future. The castle is so Complex, See the figure below: The castle surface is divided into N by M square cells. Some cubes are stacked one upon another over the cells, forming towers. For each cell the number of cubes stacked over it is given in the matrix A. Now, give A to you , please help them construct their castle, the unit cube represented as shown below: Input Input contains integers N , M, (1 <= N, M <= 30)followed by matrix A, row-by-row. The first row describes the cube tower furthest from the viewer, left to right, and the last row -- nearest to the viewer. Output Output must contain a string representation of the table view, with minimal number of lines required to show all cubes. Each line must contain a string of equal length, which is the minimal width required to show all cubes. Sample Input 2 4 2 2 3 2 1 1 2 3 Sample Output ************+---+**** ***********/ /|**** **********+---+---+** ****+---+-| / /|-+ ***/ / | +---+ |/| **+---+---+-| | + | **| | / | |/| + **| | +---+---+ |/| **+---+-| | | + | */ / | | |/| + +---+---+---+---+ |/* | | | | | +** | | | | |/*** +---+---+---+---+****

相见恨晚的超实用网站

相见恨晚的超实用网站 持续更新中。。。

Java学习的正确打开方式

在博主认为，对于入门级学习java的最佳学习方法莫过于视频+博客+书籍+总结，前三者博主将淋漓尽致地挥毫于这篇博客文章中，至于总结在于个人，实际上越到后面你会发现学习的最好方式就是阅读参考官方文档其次就是国内的书籍，博客次之，这又是一个层次了，这里暂时不提后面再谈。博主将为各位入门java保驾护航，各位只管冲鸭！！！上天是公平的，只要不辜负时间，时间自然不会辜负你。 何谓学习？博主所理解的学习，它是一个过程，是一个不断累积、不断沉淀、不断总结、善于传达自己的个人见解以及乐于分享的过程。

程序员必须掌握的核心算法有哪些？

由于我之前一直强调数据结构以及算法学习的重要性，所以就有一些读者经常问我，数据结构与算法应该要学习到哪个程度呢？，说实话，这个问题我不知道要怎么回答你，主要取决于你想学习到哪些程度，不过针对这个问题，我稍微总结一下我学过的算法知识点，以及我觉得值得学习的算法。这些算法与数据结构的学习大多数是零散的，并没有一本把他们全部覆盖的书籍。下面是我觉得值得学习的一些算法以及数据结构，当然，我也会整理一些看过...

有哪些让程序员受益终生的建议

从业五年多，辗转两个大厂，出过书，创过业，从技术小白成长为基层管理，联合几个业内大牛回答下这个问题，希望能帮到大家，记得帮我点赞哦。 敲黑板！！！读了这篇文章，你将知道如何才能进大厂，如何实现财务自由，如何在工作中游刃有余，这篇文章很长，但绝对是精品，记得帮我点赞哦！！！！ 一腔肺腑之言，能看进去多少，就看你自己了！！！ 目录： 在校生篇： 为什么要尽量进大厂？ 如何选择语言及方...

大学四年自学走来，这些私藏的实用工具/学习网站我贡献出来了

大学四年，看课本是不可能一直看课本的了，对于学习，特别是自学，善于搜索网上的一些资源来辅助，还是非常有必要的，下面我就把这几年私藏的各种资源，网站贡献出来给你们。主要有：电子书搜索、实用工具、在线视频学习网站、非视频学习网站、软件下载、面试/求职必备网站。 注意：文中提到的所有资源，文末我都给你整理好了，你们只管拿去，如果觉得不错，转发、分享就是最大的支持了。 一、电子书搜索 对于大部分程序员...

linux系列之常用运维命令整理笔录

本博客记录工作中需要的linux运维命令，大学时候开始接触linux，会一些基本操作，可是都没有整理起来，加上是做开发，不做运维，有些命令忘记了，所以现在整理成博客，当然vi，文件操作等就不介绍了，慢慢积累一些其它拓展的命令，博客不定时更新 free -m 其中：m表示兆，也可以用g，注意都要小写 Men：表示物理内存统计 total：表示物理内存总数(total=used+free) use...

比特币原理详解

一、什么是比特币 比特币是一种电子货币，是一种基于密码学的货币，在2008年11月1日由中本聪发表比特币白皮书，文中提出了一种去中心化的电子记账系统，我们平时的电子现金是银行来记账，因为银行的背后是国家信用。去中心化电子记账系统是参与者共同记账。比特币可以防止主权危机、信用风险。其好处不多做赘述，这一层面介绍的文章很多，本文主要从更深层的技术原理角度进行介绍。 二、问题引入 假设现有4个人...

程序员接私活怎样防止做完了不给钱？

首先跟大家说明一点，我们做 IT 类的外包开发，是非标品开发，所以很有可能在开发过程中会有这样那样的需求修改，而这种需求修改很容易造成扯皮，进而影响到费用支付，甚至出现做完了项目收不到钱的情况。 那么，怎么保证自己的薪酬安全呢？ 我们在开工前，一定要做好一些证据方面的准备（也就是“讨薪”的理论依据），这其中最重要的就是需求文档和验收标准。一定要让需求方提供这两个文档资料作为开发的基础。之后开发...

网页实现一个简单的音乐播放器（大佬别看。(⊙﹏⊙)）

今天闲着无事，就想写点东西。然后听了下歌，就打算写个播放器。 于是乎用h5 audio的加上js简单的播放器完工了。 演示地点演示 html代码如下` music 这个年纪 七月的风 音乐 ` 然后就是css`*{ margin: 0; padding: 0; text-decoration: none; list-...

Python十大装B语法

Python 是一种代表简单思想的语言，其语法相对简单，很容易上手。不过，如果就此小视 Python 语法的精妙和深邃，那就大错特错了。本文精心筛选了最能展现 Python 语法之精妙的十个知识点，并附上详细的实例代码。如能在实战中融会贯通、灵活使用，必将使代码更为精炼、高效，同时也会极大提升代码B格，使之看上去更老练，读起来更优雅。

数据库优化 - SQL优化

以实际SQL入手，带你一步一步走上SQL优化之路！

2019年11月中国大陆编程语言排行榜

2019年11月2日，我统计了某招聘网站，获得有效程序员招聘数据9万条。针对招聘信息，提取编程语言关键字，并统计如下： 编程语言比例 rank pl_ percentage 1 java 33.62% 2 cpp 16.42% 3 c_sharp 12.82% 4 javascript 12.31% 5 python 7.93% 6 go 7.25% 7 p...

通俗易懂地给女朋友讲：线程池的内部原理

餐盘在灯光的照耀下格外晶莹洁白，女朋友拿起红酒杯轻轻地抿了一小口，对我说：“经常听你说线程池，到底线程池到底是个什么原理？”

《奇巧淫技》系列-python！！每天早上八点自动发送天气预报邮件到QQ邮箱

将代码部署服务器，每日早上定时获取到天气数据，并发送到邮箱。 也可以说是一个小型人工智障。 知识可以运用在不同地方，不一定非是天气预报。

经典算法（5）杨辉三角

杨辉三角 是经典算法，这篇博客对它的算法思想进行了讲解，并有完整的代码实现。

英特尔不为人知的 B 面

从 PC 时代至今，众人只知在 CPU、GPU、XPU、制程、工艺等战场中，英特尔在与同行硬件芯片制造商们的竞争中杀出重围，且在不断的成长进化中，成为全球知名的半导体公司。殊不知，在「刚硬」的背后，英特尔「柔性」的软件早已经做到了全方位的支持与支撑，并持续发挥独特的生态价值，推动产业合作共赢。 而对于这一不知人知的 B 面，很多人将其称之为英特尔隐形的翅膀，虽低调，但是影响力却不容小觑。 那么，在...

腾讯算法面试题：64匹马8个跑道需要多少轮才能选出最快的四匹？

昨天，有网友私信我，说去阿里面试，彻底的被打击到了。问了为什么网上大量使用ThreadLocal的源码都会加上private static？他被难住了，因为他从来都没有考虑过这个问题。无独有偶，今天笔者又发现有网友吐槽了一道腾讯的面试题，我们一起来看看。 腾讯算法面试题：64匹马8个跑道需要多少轮才能选出最快的四匹？ 在互联网职场论坛，一名程序员发帖求助到。二面腾讯，其中一个算法题：64匹...

面试官：你连RESTful都不知道我怎么敢要你？

干货，2019 RESTful最贱实践

刷了几千道算法题，这些我私藏的刷题网站都在这里了！

遥想当年，机缘巧合入了 ACM 的坑，周边巨擘林立，从此过上了"天天被虐似死狗"的生活… 然而我是谁，我可是死狗中的战斗鸡，智力不够那刷题来凑，开始了夜以继日哼哧哼哧刷题的日子，从此"读题与提交齐飞， AC 与 WA 一色 "，我惊喜的发现被题虐既刺激又有快感，那一刻我泪流满面。这么好的事儿作为一个正直的人绝不能自己独享，经过激烈的颅内斗争，我决定把我私藏的十几个 T 的，阿不，十几个刷题网...

SQL-小白最佳入门sql查询一

不要偷偷的查询我的个人资料，即使你再喜欢我，也不要这样，真的不好；

JavaScript 为什么能活到现在？

作者 | 司徒正美 责编 |郭芮 出品 | CSDN（ID：CSDNnews） JavaScript能发展到现在的程度已经经历不少的坎坷，早产带来的某些缺陷是永久性的，因此浏览器才有禁用JavaScript的选项。甚至在jQuery时代有人问出这样的问题，jQuery与JavaScript哪个快？在Babel.js出来之前，发明一门全新的语言代码代替JavaScript...

项目中的if else太多了，该怎么重构？

介绍 最近跟着公司的大佬开发了一款IM系统，类似QQ和微信哈，就是聊天软件。我们有一部分业务逻辑是这样的 if (msgType = "文本") { // dosomething } else if(msgType = "图片") { // doshomething } else if(msgType = "视频") { // doshomething } else { // doshom...

致 Python 初学者

欢迎来到“Python进阶”专栏！来到这里的每一位同学，应该大致上学习了很多 Python 的基础知识，正在努力成长的过程中。在此期间，一定遇到了很多的困惑，对未来的学习方向感到迷茫。我非常理解你们所面临的处境。我从2007年开始接触 python 这门编程语言，从2009年开始单一使用 python 应对所有的开发工作，直至今天。回顾自己的学习过程，也曾经遇到过无数的困难，也曾经迷茫过、困惑过。开办这个专栏，正是为了帮助像我当年一样困惑的 Python 初学者走出困境、快速成长。希望我的经验能真正帮到你

Python 编程开发 实用经验和技巧

Python是一门很灵活的语言，也有很多实用的方法，有时候实现一个功能可以用多种方法实现，我这里总结了一些常用的方法和技巧，包括小数保留指定位小数、判断变量的数据类型、类方法@classmethod、制表符中文对齐、遍历字典、datetime.timedelta的使用等，会持续更新......

吐血推荐珍藏的Visual Studio Code插件

作为一名Java工程师，由于工作需要，最近一个月一直在写NodeJS，这种经历可以说是一部辛酸史了。好在有神器Visual Studio Code陪伴，让我的这段经历没有更加困难。眼看这段经历要告一段落了，今天就来给大家分享一下我常用的一些VSC的插件。 VSC的插件安装方法很简单，只需要点击左侧最下方的插件栏选项，然后就可以搜索你想要的插件了。 下面我们进入正题 Material Theme ...

实战：如何通过python requests库写一个抓取小网站图片的小爬虫

有点爱好的你，偶尔应该会看点图片文字，最近小网站经常崩溃消失，不如想一个办法本地化吧，把小照片珍藏起来！ 首先，准备一个珍藏的小网站，然后就可以开始啦！ 第一步 我们先写一个获取网站的url的链接，因为url常常是由page或者，其他元素构成，我们就把他分离出来，我找到的网站主页下有图片区 图片区内有标题页，一个标题里有10张照片大概 所以步骤是： 第一步：进入图片区的标题页 def getH...

“狗屁不通文章生成器”登顶GitHub热榜，分分钟写出万字形式主义大作

一、垃圾文字生成器介绍 最近在浏览GitHub的时候，发现了这样一个骨骼清奇的雷人项目，而且热度还特别高。 项目中文名：狗屁不通文章生成器 项目英文名：BullshitGenerator 根据作者的介绍，他是偶尔需要一些中文文字用于GUI开发时测试文本渲染，因此开发了这个废话生成器。但由于生成的废话实在是太过富于哲理，所以最近已经被小伙伴们给玩坏了。 他的文风可能是这样的： 你发现，...

程序员：我终于知道post和get的区别

是一个老生常谈的话题，然而随着不断的学习，对于以前的认识有很多误区，所以还是需要不断地总结的，学而时习之，不亦说乎

《程序人生》系列-这个程序员只用了20行代码就拿了冠军

你知道的越多，你不知道的越多 点赞再看，养成习惯GitHub上已经开源https://github.com/JavaFamily，有一线大厂面试点脑图，欢迎Star和完善 前言 这一期不算《吊打面试官》系列的，所有没前言我直接开始。 絮叨 本来应该是没有这期的，看过我上期的小伙伴应该是知道的嘛，双十一比较忙嘛，要值班又要去帮忙拍摄年会的视频素材，还得搞个程序员一天的Vlog，还要写BU...

加快推动区块链技术和产业创新发展，2019可信区块链峰会在京召开

11月8日，由中国信息通信研究院、中国通信标准化协会、中国互联网协会、可信区块链推进计划联合主办，科技行者协办的2019可信区块链峰会将在北京悠唐皇冠假日酒店开幕。 　　区块链技术被认为是继蒸汽机、电力、互联网之后，下一代颠覆性的核心技术。如果说蒸汽机释放了人类的生产力，电力解决了人类基本的生活需求，互联网彻底改变了信息传递的方式，区块链作为构造信任的技术有重要的价值。 　　1...

相关热词 c#中dns类 c#合并的excel c# implicit c#怎么保留3个小数点 c# 串口通信、 网络调试助手c# c# 泛型比较大小 c#解压分卷问题 c#启动居中 c# 逻辑或运算符