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It is divided into two groups AB. Group A has three signatures A1, A2 and A3. If A3 is drawn, it will be empty and will advance directly to the next match. If A1A2 is drawn for the match, the winner will compete with A3 for a total of 2 matches, and the first and second place in Group A will be obtained.　　厨艺满天下的孤女方一勺自小就孤苦无依，但是却有一手好厨艺，因为替人烧菜却遭到暗算，不得不代嫁给知府家的恶霸少爷沈勇。生性乐观、汤圆脸、大眼睛却耐看的方一勺对诗词歌赋却一窍不通，生气起来哪怕是一根擀面杖也能舞得虎虎生风。她厨艺精湛并略懂武艺，被骗嫁入沈家，颇得公公婆婆喜爱。真可说是一物降一物，沈勇不情不愿的娶了方一勺，却渐渐喜欢上这个不懂诗书但却能陪自己打架喝酒划拳的冒牌才女，更何况这个小丫头还固执的认定他沈勇是个好人，方一勺也华丽上演了训夫记。精诚所至金石为开，浪子终是要回头的，在方一勺的陪伴下，沈勇终有所成……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.　　弗兰德饰演詹姆斯·怀特豪斯，是一位雄心勃勃，富有魅力的初级部长，一直以来分享着妻子的好运。米勒饰演他的妻子苏菲，牛津大学毕业，育有两个孩子，看似完美的生活即将被引爆。道克瑞饰演主攻性犯罪的刑事律师＆王室法律顾问凯特·伍德克罗夫特，她在自己的领域迅速崛起，刚刚接手了一个百年难遇的案子。　　小姨子顾芸结婚，开哥去岳母家送亲途中遇到追债人，慌不择路的他误入顾芸的婚纱之内，丢尽颜面。谷小燕的父亲谷八抬担任婚庆司仪，却因找不到电动车钥匙而错过了时辰，婆家人无奈之下邀请能说会道的开哥前去救场。由于紧张，开哥误将“二拜高堂”说成了“二拜灵堂”，惹怒婆家人。同时，他的救场行为也被谷八抬误认为是故意抢生意。