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combinations组合密码the | 邮电通讯系统 2022-07-14 60 0star收藏 版权: . 保留作者信息 . 禁止商业使用 . 禁止修改作品
来历:常思一二,不思八九PJ简单解说:nbsp<imgsrc="static/image/smiley/default/titter.gif"smilieid="9"border="0"alt=""/>IN码分前4和后4,先破前4只要最多一万个组合,破后4中的前3只要一千个组合,一共就是一万一千个密码组合。10的4次方+10的3次方=11000个密码组合。当reaver断定前4位PIN密码后,其命令进展数值将直接跳跃至90.9%以上,也就是说只剩下一千个密码组合了。一共一万一千个密码!

(Origin: often thinking about one or two, not thinking about eight or nine PJ simple explanation: nbsp<imgsrc="static/image/smiley/default/titter.gif"smilieid="9"border="0"alt=""/>IN code Dividing the top 4 and the bottom 4, the top 4 only needs a maximum of 10,000 combinations, and the top 3 of the latter 4 only need a thousand combinations, which is a total of 11,000 password combinations. 10 to the 4th power + 10 to the 3rd power = 11,000 password combinations. When reaver determines the first 4-digit PIN password, its command progress value will jump directly to more than 90.9?which means that there are only a thousand password combinations left. Eleven thousand passwords in total!)

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