2022年工业信息安全技能大赛“望岳杯”锦标赛 wp

 

0x01 恶意文件

动调动调

OD直接动调，更具xxx的流程图可以看到程序的整体结构，然后结合流程图使用OD动调

最终ip地址是存在了[ebp+ecx-0x14]的位置

将这一处跳转和两处的exit nop掉

跟踪到[ebp+ecx-0x14]在内存中的位置，一直F8就可以了

这里是端口，小端排序 1F 57，转换成十进制就是8023，前边在存储ip地址的时候就已经存到[ebp-0x1D0]的位置了，所以最终ip：1.15.228.170:8023

flag{1.15.228.170:8023}

 

0x02 攻击行为还原

FlaG.txt有密文

c=1566077545968868311749088444723100549024925747577903610033503

Id_rsa有

-----n=1606938044309278499168642398192229212629290234347717645487123-----

-----e=65537-----

在线分解n

#coding:utf-8
import binascii
from libnum import n2s,s2n
import base64
def gcd(a, b):
    if a < b:
        a, b = b, a
    while b != 0:
        temp = a % b
        a = b
        b = temp
    return a

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 modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

p = 1267650600235241044365044382487
q = 1267650600260888172240281085029
e = 65537
# tmp = base64.b64decode("qzogS7X8M3ZOpkUhJJcbukaRduLyqHAPblmabaYSm9iatuulrHcEpBmil7V40N7gbsQXwYx5EBH5r5V2HRcEIOXjgfk5vpGLjPVxBLyXh2DajHPX6KvbFpQ8jNpCQbUNq8Hst00yDSO/6ri9dk6bk7+uyuN0b2K1bNG5St6sCQ4qYEA3xJbsHFvMqtvUdhMiqO7tNCUVTKZdN7iFvSJqK2IHosIf7FqO24zkHZpHi31sYU7pcgYEaGkVaKs8pjq6nbnffr4URfoexZHeQtq5UAkr95zD6WgvGcxaTDKafFntboX9GR9VUZnHePiio7nJ3msfue5rkIbISjmGCAlj+w==")
#  = 
d = modinv(e, (p - 1) * (q - 1))
# c=s2n(tmp)
c = 1566077545968868311749088444723100549024925747577903610033503
#c = 225031483444634056931067907865853799650197225351377050632290334721073031287701730297815850654473721939907812470206115171738967740183098960272963323728747481560137205796840356532306950935686580268408289864109695494835661414073083573249882362332920722000099781994315336570711188934565379141406727420346806389405536474102730682155998263607095718543239272202402139286809779368710600842078606046563228470023546348908618719147790859257980882643030144242048154566691808688844513142261099020381730517293884263384819159874220288293023868919557980548807831273449743064237407705987056818011286315950476959812697067649075359373253
n = p*q
m=pow(c,d,n)
print n2s(m)

后半段直接google

没想到真对了，非预期。

flag{SSH_Brute_Force}

 

0x03 留痕取证

前半段

凯撒密码解密

后半段

flag{bufujiugana}

 

0x04 漏洞挖掘

Linux运行随便输个密码

Base16+栅栏解密得到flag

flag{d5000loginpassisDLBhh@0812}

 

0x05 密文破解

根据残缺的密文和MD5写脚本爆破

s = '0123456789!@#$%^&*`~()_+-=[]{};:,.<>?/\'\"\\|'
for i in s:
    for j in s:
        for k in s:
            for l in s:
                for m in s:
                    #print(1)
                    md5 = hashlib.md5()
                    c = str(i) + '1_1001_10_600_600_' + str(j) + str(k) + '00' + str(l) + str(m) + '7_202103301539'
                    #print(c)
                    md5.update(c.encode('utf-8'))
                    d = md5.hexdigest()
                    #print(c)
                    if d[:4] == 'b9fd' and d[6:9] =='9d5':
                        print(c)
                        print(d)

flag{1500.87}

 

0x06 文件格式分析

使用binwalk查看文件发现文件是一个zip压缩包，于是使用010editor修改文件头为504B，修改后缀为zip并解压文件，解压后打开文件

使用basecrack工具进行解码，得到flag

flag{W3lc0Me_t0_Xinx1}