eters x and y. use the euclidean algorithm to do this. return none if the gcd does not exist (i.e., if both parameters are 0). 2. (14 points) write a function mod inv(x, y) that returns the multiplicative inverse of x mod y (...
importsysfromCrypto.Util.numberimport*n=71451784354488078832557440841067139887532820867160946146462765529262021756492415597759437645000198746438846066445835108438656317936511838198860210224738728502558420706947533544863428802654736970469313030584334133519644746498781461927762736769115933249195917207059297145965502955615599481575507738939188415191defHGCD(a,b):if2*b....
该函数没有返回值,将docstring信息直接输出到屏幕或文件。 示例 下面是一个例子,调用write_docstringdict函数输出Python内置math模块的docstring信息,并以Markdown格式输出到屏幕上。 import math import turtle turtle.write_docstringdict(math.__dict__) 输出结果: ...
log_level = 'debug' class Gao: def __init__(self) -> None: # self.conn = process(['python3', 'another.py']) self.conn = remote('03.cr.yp.toc.tf', 31317) def gao_one(self): self.conn.recvuntil('Level ') s = self.conn.recvline().decode() a = int(s.split('=')[1...
“If you want to compute a new target…”,并在最后设置新的target。 重要提示:对数是以一个任意的(未知的)基数给出的。 如果你想根据一个特定的生成元g来定义它们,那么你将不得不计算g的对数,然后用所有的对数除以这个值。 见https://lists.gforge.inria.fr/pipermail/cado-nfs-discuss/2018-November/00...
from pwn import * - -debug = 0 -if debug: - p = process('./random') -else: - p = remote('49.4.66.242', 32150) - - -def refuse(a=1): - for x in xrange(a): - p.sendlineafter('note?(Y/N)', 'N') - - -def accept(): - p.sendlineafter('note?(Y/N)', 'Y') ...
An questions occured in my mind, how does this program check the input is valid or not? a white list? So we should dig deeper... I try to find some command string like: ls ps ps aux uname -a but failed, these command is not in the .data segment or other segments. these is a...
We have RSA ciphertext of two related messages, with small random paddings. This is standard case for coppersmith. I was able to find a script online and modify it to calculate the exact diff, then recover plaintext with gcddef short_pad_attack(c1, c2, e, n): PRxy.<x,y> = ...
find = find_empty(board) if not find: return True else: row, col = find for i in range(1, 10): if is_valid(board, i, (row, col)): board[row][col] = i if solve(board): return True board[row][col] = 0 return False ...
找个工具接收一下hashpython2 Responder.py -I eth0 -f 然后丢给john爆破一下就出来了 有密码了rdp或者net use都可以啦 Others: 在看wp的时候发现 这是因为前面做出来的队没有删ipc链接 其实我看到很多队伍传上来的工具都故意没有删,比如hydra,打包好的html.zip ...