安恒月赛之抗【疫】练习赛

安恒月赛之抗【疫】练习赛

WEB3 commix。

就是命令执行，不过cat和flag被过滤了，，，直接使用奇淫技巧加变量拼接可以进行绕过：
payload：

cmd=a=t;ca$a \ls`

MISC1 zip隐写

下载zip，发现zip中存在16进制的数字，504b开头，压缩包，保存下来到winhex，保存为压缩包
发现解压提示有损坏，查看文件，修改01为08，解压得到flag

得到：

MISC2 lemonEssence

这个就比较简单了
解压得到一张图片，发现是png，检查了一下发现有提示说不对
跑一下高度，，，

import struct
import binascii
import os
 
m = open("1.png","rb").read()
for i in range(0,65535):
    c = m[12:20] + struct.pack('>i', i) + m[24:29]
    crc = binascii.crc32(c) & 0xffffffff
    if crc == 0x986b9e93:
        print(hex(i))

修改高度得到flag：

CRYPTO1 古典密码

直接解密即可：
第一个凯撒加密：flag{36d9f2777
第二个摩斯密码：b92bac39aa2
第三个栅栏加密：ab206cd90d47}

CRYPTO2 RSA1

给了n,e1,e2,c1,c3，n一样，rsa的共模攻击，直接使用脚本就行：

from libnum import n2s, s2n
from gmpy2 import invert

# 扩展欧几里得算法
def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)


def main():
	n = 21550279102644053137401794357450944302610731390301294678793250727396089358072700658571260795910112265309568014296122288384516447895827201111531054386530016432904989927216701507587366446802666848322853781729905492728655474832512381505627940555854308364578108265962388044363133246414753768229564846275154311898383993892293297122428661960946207950994560898964054913194462187242818633295970027741085201122155726130759045957757833942616544066055081600792366411691979350744894938994915328874600229684477533220240489600171746943849179803693122081888324258987779131223150589953248929679931142134208151043000793272520874205933
	e1 = 65537
	e2 = 11187289
	c1 = 3398498381912395819190972489172462865619978412426461006637853132394421358554444085509204376417687407497725837275868696481008111895766215578504776574832032556271718345687763315140723387608016365200919607751172500433727679269003098314988424638473027123820847847826679169000817669427223462669128173658466684135284118199815059085013479646863344355311315928713888347485004116168388822942797985291207722712351376891776564431593839662958249777540851019964959285093222467104765037231393043482615879794268339523066822738215251088897330388858109680412562153811860413533184870172160079371279534423386236128033224501238509297353
	c2 = 3466733921305804638105947202761163747472618602445995245253771384553216569474005211746398256742813639292824489920799418551206486872148557599625985549276697777903434273072767901043963396047653458242735767809413051298636887840641872939342025101757793615068691040228073377366562557622977332819376942596081135968249279010542277871138668977160241877260538203101507006391433015105607006204397243716334344883925947719719479074061998068934050946968531874465924912747079003982022188875112147185558223515367430238618463189740762128953957802291125793882636020335117593003197811477506533564676975831899876919568948425610130348710
	s = egcd(e1, e2)
	s1 = s[1]
	s2 = s[2]
	# 求模反元素
	if s1 < 0:
		s1 = - s1
		c1 = invert(c1, n)
	elif s2 < 0:
		s2 = - s2
		c2 = invert(c2, n)

	m = pow(c1, s1, n) * pow(c2, s2, n) % n
	print(n2s(m))  # 二进制转string


if __name__ == '__main__':
	main()

得到：

RE1 easy-py

pyc文件反编译得到源码：

import base64
import string

def caser(flag):
    enc1 = ''
    for i in flag:
        enc1 += chr(ord(i) - 5)
    
    return enc1


def rail(flag):
    p1 = ''
    p2 = ''
    p3 = ''
    enc2 = ''
    for i in range(len(flag)):
        j = i % 3
        if j == 0:
            p1 += flag[i]
            continue
        if j == 1:
            p2 += flag[i]
            continue
        p3 += flag[i]
    
    enc2 = p1 + p2 + p3
    return enc2


def rep(flag):
    table1 = 'qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM'
    table2 = 'QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm'
    table = string.maketrans(table1, table2)
    return flag.translate(table, '=')

while True:
    flag = raw_input('please input flag to check:')
    if rep(base64.b64encode(rail(caser(flag)))) == 'ywjCytmRxI9CycWZngD2ncTDkZqYlJrGmhHCxISUnfWSlgfDlJi':
        print 'Success!you got it!'
        break
        continue
    print 'try a gain'

过程挺简单的，书写脚本即可，他是python2写的，所以我也是用的python2：

#!/usr/bin/python
# -*- coding: utf-8 -*-
import base64
import string

def rep(flag):
    table1 = 'qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM'
    table2 = 'QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm'
    table = string.maketrans(table1, table2)
    return flag.translate(table, '')
	
def rail(flag):
    s = ""
    s1 = flag[:13]
    s2 = flag[13:26]
    s3 = flag[26:] + "_"
    for i in range(13):
        s += s1[i]+s2[i]+s3[i]
    return s

def caser(flag):
    enc1 = ''
    for i in flag:
        enc1 += chr(ord(i) + 5)
    return enc1
    
s = "ywjCytmRxI9CycWZngD2ncTDkZqYlJrGmhHCxISUnfWSlgfDlJi="
print(caser(rail(base64.b64decode(rep(s)))))

RE2 maze1

画出地图就能走出来，生成地图：

得到地图：

1 1 1 1 1 1
1 0 1 0 0 1
1 0 1 0 0 1
1 0 1 0 0 1
1 0 1 0 0 1
1 0 1 0 0 2

走到2就行，输入的长度只能为10：


所以路径就是：dddddsssss
经过md5加密就是flag~~