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这是关于一个强大并且聪明的女人Rarapee的故事。她非常爱她的丈夫，但是在她怀孕期间，丈夫去世了，她决定成为一个单亲妈妈继续生活。在成长过程中，她的孩子痛恨爸爸的不负责任离开了他们。然而，Rarapee从不恨她的丈夫，相反，她坚信丈夫有一天会回来。很多年以后，她的愿望出乎意料地实现了。他的丈夫以一个年轻英俊的年仅24岁的男孩Thing的身份重生了，跟她的儿子年纪一样。当他们相遇时，不同年龄的神奇爱情开始了。那么Thing能够发现他的前世吗？他们会在爱情路上面临困境吗？因为父母的关系，离毕业还有半年的高中女生吉川菜绪（福原遥 饰）被迫开始了单身生活。她独自一人搬到新的公寓，谁知几乎同时有另一个男孩也搬了进来。对方名叫上原久志（白石隼也 饰），是菜绪所在目黑丘高中的同级生，以“御三家”后代而闻名。原来两人都掉入了房地产双重合同的陷阱，他们各执一词，互不相让。在谁也不退步的情况下，两人只好别别扭扭地展开了单身同居生活。在此期间，菜绪曾被男孩吻到，心里则有莫名的小鹿乱撞。看来青春烂漫的校园生活，注定不会以平淡结束……本片根据高须贺由枝的同名漫画改编。蕾卡是警察局办公室的秘书。她自告奋勇化装成大庄园主瓦拉蓬未曾见面的网上情人由美子，打入其内部，以调查瓦拉蓬农场贩毒的罪恶行径。蕾卡以甜美温柔的泰国和日本混血女子---由美子的形象出现在瓦拉蓬面前，成功进入瓦拉蓬家中。但瓦拉蓬的女儿诺菩坚决反对父亲再婚，不欢迎蕾卡的到来。瓦拉蓬的儿子信通对蕾卡的到来持怀疑态度，一再出难题刁难蕾卡。蕾卡机智地摆脱瓦拉蓬的反复纠缠，沉着应对信通的刁难，还要冒着暴露身份的危险千方百计去调查瓦拉蓬的罪行。蕾卡为了搜集资料，在瓦拉蓬的农场里拍了许多照片……Centos7, The service is no longer started using init-style scripts, Instead, the unit file is used, Therefore, commands like service iptables start can no longer be used in centos7, Therefore, service iptables save cannot be executed. At the same time, firewall is used to replace the original iptables service in centos7, but don't worry, we just need to install iptables and iptables-services through yum source (iptables are usually installed by default, but iptables-services are not installed by default in centos7). After installing iptables-services in centos7, the rules can be saved by the service iptables save command, as in centos6, and the rules are also saved in the/etc/sysconfig/iptables file.会友镖局在清同治年间极负盛名，官、商、贼三道无不留下面子三分。然而镖局其实外强中干，新任大掌柜尚智(马浚伟)决定来一场改革。智义子的身分和他大刀阔斧的改革却令他成镖局上下的箭靶，他最后更要跟兄弟尚忠(黎耀祥)、尚孝(黄贤智)和大伯尚正鹏(刘江)等分家。犹幸，局内的女镖师利祥凤(姚子羚)一直在旁支持。一波未平，一波又起。一方面，镖局被当了官的鹏逼得几乎要倒闭。另方面，清政府筹备兴建铁路，大货轮亦开进港口，镖局行业正日落西山。智留学回来的四弟尚义(黎诺懿)看清时势，鼓励智发展漕运。兄弟俩为镖局尽心尽力，终获朝廷颁发漕运的专利。昔日在陆路上随处可见的会友镖局旗帜，今日继续在海上迎风飘扬。In addition to solving the port problem of A, the serial number problem also needs to be paid attention to. The serial number corresponds to the sliding window. The fake TCP packet needs to be filled with the serial number. If the value of the serial number is not in the sliding window of B when it is sent to B before A, B will actively discard it. So we need to find the serial number that can fall into the sliding window between AB at that time. This can be solved violently, because the length of a sequence is 32 bits, and the value range is 0-4294967296. If the window size is like 65535 under Windows I caught in the above figure, I only need to divide, and I know that at most I only need to send 65537 (4294967296/65535=65537) packets to have a sequence number fall into the sliding window. The desired value can be calculated quickly by exhaustion on multiple control hosts.