Passwords are widely used in our lives: for ATMs, online forum logins, mobile device unlock and door access. Everyone cares about password security. However, attackers always find ways to steal our passwords. Here is one possible situation: Assume th…

https://code.google.com/codejam/contest/4214486/dashboard#s=p0 排列组合.DP递推式,如下代码.dp[m][n]表示长度为n的字符串里有m个字符,那么可以先用m-1个字符拼一个长度为n-1的字符串,然后再C(n,1)里面挑一个放最后一个字符;这是最后一种字符是一个的情况,后面还有两个三个等等.所以代码如下: 要注意的是,可以先计算组合数combination[n][m],用C(n,m)=C(n−1,m)+C(n−1,m−1)来算. /…

Password Attacker 题意就是给 M 个关键字,组合成 N 字符长度的结果,每一个关键字都必须在 N 位的字符中出现,有多少种可能结果. 范围 1 ≤ M ≤ N ≤ 100. 举例假设 M = 3(key = 3, 7, 5)  N = 4 位字符长度 结果可以为3577, 3557, 7353, 5735. 下面的结果是不合法的 1357 // 1 没有在key中,不是合法关键字 3355 // 7 是关键字,但是结果中没有出现 357 // 结果必须是一个4位长度 思路1:…

hdu3625 题意: 酒店发生一起谋杀案.作为镇上最好的侦探,您应该立即检查酒店的所有N个房间.但是,房间的所有门都是锁着的,钥匙刚锁在房间里,真是个陷阱!您知道每个房间里只有一把钥匙,并且所有可能的分配可能性均等.例如,如果N = 3,则有6种可能的分布,每种分布的概率为1/6.为方便起见,我们将房间编号从1到N,房间1的键编号为键1,房间2的键编号为2,依此类推.要检查所有房间,您必须用力摧毁一些门.但是您不想破坏太多,因此您采取以下策略:首先,您手中没有钥匙,因此您会随机破坏一扇锁着的门…

A complex 16-Level XSS Challenge, held in summer 2014 (+1 Hidden Level) Index Level 0 Level 1 Level 2 Level 3 Level 4 Level 5 Level 6 Level 7 Level 8 Level 9 Level 10 Level 11 Level 12 Level 13 Level 14 Level 15 Hidden Level Rules Call prompt(1) to w…

实验环境 https://github.com/TouwaErioH/security/tree/master/web1 Windows10 Oracle VM VirtualBox Ubuntu16.04 i386 安装Ruby和rails,http://gorails.com/setup/ubuntu/16.04 下载实验提供的project 2源码 重定位到/bitbar目录下,执行bundle install 开启服务器 (rails server) 可以在http://localhos…

Spring Security’s PasswordEncoder interface is used to support the use of passwords which are encoded in some way in persistent storage. You should never store passwords in plain text. Always use a one-way password hashing algorithm such as bcrypt wh…

  What I don't fully understand is the use of ClientId and Secret vs Username and Password. The code I pasted generates a token by username and password and I can work with that token (until it expires), but when I try to get a refresh token, I must…

if we know some user's email, the we will can reset the user's email by host header attack. The attack vector is that the user binded email and we know the user's email. The following POST request is the normal request. as the picture shows,the reque…

p { margin-bottom: 0.1in; line-height: 120% } 一.一开始我是按照网友所说的 : rm -f ~/.gnome2/keyrings/login.keyring 删除秘钥文件, 按照网友的解决办法,这步之后,会打开新的程序会提示输入新的密码.但是重新打开谷歌浏览器,还是提示unlocking, 然后我开始排除问题所在,首先  cd ~/.gnome2/ 里面是空的了,有两种可能一种是我删掉了,一种就是本来就是空的,但keyrings文件夹应该还在才对.猜…