破解Linux的用户密码:John [跨平台的密码解密工具] root@only:~# unshadow /etc/passwd /etc/shadow > ~/file_to_crack root@only:~# john --wordlist=/usr/share/john/password.lst ~/file_to_crack   rm -fr /root/.john/john.pot unshadow p s >file john --wordlist=zidian.txt file…

John Problem Description   Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. A…

函数:split() 功能:使用一个指定的分隔符把一个字符串分割存储到数组例子: var theString=”jpg|bmp|gif|ico|png”; var arr=theString.split(”|”); //arr是一个包含字符值”jpg”.”bmp”.”gif”.”ico”和”png”的数组 函数:John() 功能:使用您选择的分隔符将一个数组合并为一个字符串例子: var delimitedString=myArray.join(delimiter); var myList=n…

John the RipperJohn the Ripper(简称John)是一款著名的密码破解工具.它主要针对各种Hash加密的密文.它不同于Rainbow Table方式.它采用实时运算的方式和密文进行比较.为了提高运行效率,它提供各种辅助功能.例如,它提供辅助工具unshadow收集用户名等相关信息,内置了海量的常用密码,预编译了大量的Mangling Rule.针对用户需求,它提供了三种模式,应对不同的场景需求.例如,Single crack(单一破解)模式用于弱密码场景:Wordlis…

Description John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples…

Description 小约翰经常和他的哥哥玩一个非常有趣的游戏:桌子上有n堆石子,小约翰和他的哥哥轮流取石子,每个人取 的时候,可以随意选择一堆石子,在这堆石子中取走任意多的石子,但不能一粒石子也不取,我们规定取到最后一 粒石子的人算输.小约翰相当固执,他坚持认为先取的人有很大的优势,所以他总是先取石子,而他的哥哥就聪明 多了,他从来没有在游戏中犯过错误.小约翰一怒之前请你来做他的参谋.自然,你应该先写一个程序,预测一下 谁将获得游戏的胜利. Input 本题的输入由多组数据组成第一行包括一个…

Description Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Pleas…

对于任意一个 Anti-SG 游戏,如果我们规定当局面中所有的单一游戏的 SG 值为 0 时,游戏结束,则先手必胜当且仅当:  (1)游戏的 SG 函数不为 0 且游戏中某个单一游戏的 SG 函数大于 1:(2)游戏的 SG 函数为 0 且游戏中没有单一游戏的 SG 函数大于 1. John Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s):…

Priest John's Busiest Day Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1420    Accepted Submission(s): 415 Problem Description John is the only priest in his town. October 26th is the John's…

参见上一篇博客,里面有分析和结论. #include <cstdio> int main() { int T; scanf("%d", &T); while(T--) { , xorsum = ;//c为充裕堆的个数 scanf("%d", &n); ) c++; } ) || (xorsum && !c)) puts("Brother");//T2和S0状态必败 else puts("Joh…