2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

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2024年11月6日14:00:34评论50 views字数 10842阅读36分8秒阅读模式

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前言：

本次比赛已经结束，用于赛后复现，欢迎大家交流学习！

文章同步CSDN，感谢大家观看，麻烦点个赞！

MISC-01

放入HxD，听别人说这题是 差分曼切斯特

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

#!/usr/bin/env python# -*- coding: utf-8 -*-import structimport math

def long_to_bytes(n):    s = b''    pack = struct.pack    while n > 0:        s = pack('>I', n & 0xffffffff) + s        n = n >> 32    for i in range(len(s)):        if s[i] != b'00'[0]:            break    else:        s = b'00'        i = 0    s = s[i:]    return s

# 字节逆序def byteinvert(str_bin):    ret = ''    for i in range(len(str_bin) // 8):        ret += str_bin[i * 8:i * 8 + 8][::-1]    return ret

# 标准曼彻斯特def MCST_stand(str_bin):    ret = ''    for i in range(len(str_bin) // 2):        x = str_bin[i * 2:i * 2 + 2]        if x == '01':            ret += '0'        elif x == '10':            ret += '1'        else:            return 'stand manchester decode wrong!'    return ret

# IEEE规范的曼彻斯特def MCST_IEEE(str_bin):    ret = ''    for i in range(math.ceil(len(str_bin) / 8)):        x = str_bin[i * 2:i * 2 + 2]        if x == '01':            ret += '1'        elif x == '10':            ret += '0'        else:            return 'stand manchester decode wrong!'    return ret

# 差分曼彻斯特def MCST_diff(str_bin):    ret = ''    for i in range(0, len(str_bin) // 2 - 1):        x1 = str_bin[i * 2:i * 2 + 2]        x2 = str_bin[i * 2 + 2:i * 2 + 4]        if x1 == x2:            ret += '0'        else:            ret += '1'    return ret

if __name__ == "__main__":    str_bin = open('data','r').read()
    m1 = MCST_IEEE(str_bin)    m2 = MCST_stand(str_bin)    m3 = MCST_diff(str_bin)    print('nIEEE曼彻斯特:')    # print(m1)    print(hex(int(m1, 2)))    print(long_to_bytes(int(m1, 2)))    print('n 标准曼彻斯特:')    # print(m2)    print(hex(int(m2, 2)))    print(long_to_bytes(int(m2, 2)))    print('n 差分曼彻斯特:')    # print(m3)    print(hex(int(m3, 2)))    print(long_to_bytes(int(m3, 2)))    print('n=============字节逆序=============')    m1 = byteinvert(m1)    m2 = byteinvert(m2)    m3 = byteinvert(m3)    print('nIEEE曼彻斯特:')    # print(m1)    print(hex(int(m1, 2)))    print(long_to_bytes(int(m1, 2)))    print('n 标准曼彻斯特:')    # print(m2)    print(hex(int(m2, 2)))    print(long_to_bytes(int(m2, 2)))    print('n 差分曼彻斯特:')    # print(m3)    print(hex(int(m3, 2)))    print(long_to_bytes(int(m3, 2)))

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

这里在差分曼彻斯特发现压缩包zip特征删除前面的然后进行提取压缩包，

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

这里进行Hex 转换发现确实是PK文件进行导出，发现压缩包损坏需要修复！！

f = open("flag.zip","rb")data = f.read().hex()
# print(data)   # 每 44 个字节分割一下zip_data = []zip_end = data[-10:]data = data[:-10]for i in range(0, len(data), 44):    # print(data[i:i+44])    # 将每 44 个字节后面的 6 个字节去掉, 除了最后 2 行    data1 = data[i:i+44]    data2 = data1[:-12]    zip_data.append(data2)    # print(data2)    # 在合并所有的字节，写入新的 zip 文件中zip_data.append(zip_end)# print(''.join(zip_data))with open("new_flag.zip", "ab") as f2:    f2.write(bytes.fromhex(''.join(zip_data)))

通过脚本修复得到正常压缩包需要密码

进行爆破得到密码：12345678

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

猜测所有带key的 png隐写最终发现是

PixelJihad 隐写得到flag。

https://sekao.net/pixeljihad/

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

MISC-02

一个 jar 包，先运行一下程序，进不去

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

这里反编译一下这个jar的包，发现有个AES加密程序

而且还给了密钥Y4SuperSecretKey

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

发现有很多Base64编码一个一个解最后得到flag

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

rCHPM64CJdSUR3ohx2EFokSG/hP11J/yor89iDiwUGh82uN1TSSvezEzZ4Ofj7NOvd4lZhJsMDkiOxf7MRgTTuK9u6swj92G/awWwJrbIIw=

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

MISC-03

给了一张图片，使用HxD发现文件有多余内容，

尝试分离 binwalk -e ，foremost进行分离。

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

得到两个压缩包文件，发现都需要密码，一个显示损坏

另一个发现大小是6 可以想到crc碰撞。

这里可以使用一个比较好用的工具bkcrack -L 可以进行查看

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

python3 crc32.py reverse 0x572a3c82python3 crc32.py reverse 0xe11d5266python3 crc32.py reverse 0x582cf195

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

最终拼接一下得到密码：This_WD_011cryptpW

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

然后根据题目提示可以看出来这里可以使用VeraCrypt 挂载输入密码挂载成功！

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

得到一个secret.key的rsa私钥文件

然后还有个压缩包是flag.txt 尝试解密！

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

from Crypto.Util.number import *from Crypto.PublicKey import RSAimport gmpy2import base64
privatekey='''-----BEGIN ENCRYPTED PRIVATE KEY-----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-----END ENCRYPTED PRIVATE KEY-----'''

# 解析私钥private_key = RSA.import_key(privatekey)# 提取dd = private_key.dp = private_key.pq = private_key.q
c = 'TtF75yk0Ucl888EaKvIyQIZD5UoWJRG/n9Y8QEK7RrFAjCgL9LVz14R0eLoT+vYrLavJHXLCNCkF8ngHjSAkEuBEiWP9r7+0ax6L04DtVqSwN/gSCimeHdpUfBg1FFz8EJ/Yf52ZI5wKeuW54vd3Sy8JDD+DuLJklcTvSb9V23o='# 解密m = pow(bytes_to_long(base64.b64decode(c)), d, p*q)# 计算m的模# m = gmpy2.iroot(m, 2)[0]# 输出明文print(long_to_bytes(m))

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

当然也可以使用在线网站：

https://www.lddgo.net/encrypt/rsa

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

Crypto-02

给出了p的高256位，爆破8位即可用copersmith

from Crypto.Util.number import *
from tqdm import trange
import gmpy2


n = 0x00b8cb1cca99b6ac41876c18845732a5cbfc875df346ee9002ce608508b5fcf6b60a5ac7722a2d64ef74e1443a338e70a73e63a303f3ac9adf198595699f6e9f30c009d219c7d98c4ec84203610834029c79567efc08f66b4bc3f564bfb571546a06b7e48fb35bb9ccea9a2cd44349f829242078dfa64d525927bfd55d099c024f
p_high = 0xe700568ff506bd5892af92592125e06cbe9bd45dfeafe931a333c13463023d4f
pbits = 512
e = 0x10001
c = bytes_to_long(open('flag.enc','rb').read())




for i in trange(2**8):
     p4 = p_high<<8      #这里需要先爆破8位，使得知道264位以后再恢复p
     p4 = p4 + i
     kbits = pbits - p4.nbits()
     p4 = p4 << kbits
     R.<x> = PolynomialRing(Zmod(n))
     f = x + p4
     x = f.small_roots(X=2^kbits, beta=0.4, epsilon=0.01)
     if x:
         p = p4 + int(x[0])
         q = n // p
         d = gmpy2.invert(e,(p-1)*(q-1))
         m = pow(c,d,n)
         print(long_to_bytes(int(m)))
         break

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

Crypto-03

2024年第四届“网鼎杯”网络安全比赛---朱雀组WriteUp

from Crypto.Util.number import *
import gmpy2
import sys


def HGCD(a, b):
    if 2 * b.degree() <= a.degree() or a.degree() == 1:
        return 1, 0, 0, 1
    m = a.degree() // 2
    a_top, a_bot = a.quo_rem(x^m)
    b_top, b_bot = b.quo_rem(x^m)
    R00, R01, R10, R11 = HGCD(a_top, b_top)
    c = R00 * a + R01 * b
    d = R10 * a + R11 * b
    q, e = c.quo_rem(d)
    d_top, d_bot = d.quo_rem(x^(m // 2))
    e_top, e_bot = e.quo_rem(x^(m // 2))
    S00, S01, S10, S11 = HGCD(d_top, e_top)
    RET00 = S01 * R00 + (S00 - q * S01) * R10
    RET01 = S01 * R01 + (S00 - q * S01) * R11
    RET10 = S11 * R00 + (S10 - q * S11) * R10
    RET11 = S11 * R01 + (S10 - q * S11) * R11
    return RET00, RET01, RET10, RET11
    
def GCD(a, b):
    print(a.degree(), b.degree())
    q, r = a.quo_rem(b)
    if r == 0:
        return b
    R00, R01, R10, R11 = HGCD(a, b)
    c = R00 * a + R01 * b
    d = R10 * a + R11 * b
    if d == 0:
        return c.monic()
    q, r = c.quo_rem(d)
    if r == 0:
        return d
    return GCD(d, r)


sys.setrecursionlimit(500000)


n1 = 46994335119752756509295563273644018207372437765472265663641969615870434649870172687711336119178476731707184071004952832066023502817858922050333696147278803794579361275375126259907918549399847213059844795524428354261714653121692582331027749211493858306453667935016549085360033613631067207061043880115121598962110616155562901846204247465460970460199073739292905462216078043392395114502994197819280075092818942982603177739751892161678322940701448492756420067768637842557026865307595566180909536549522253144031734977212534244571704403691638817139603787139183257554425149405808562098424603264387155486261738291201472340352391
n2 = 16976940907141191524915804499951951599097024510818489668879976514721679457799770728089555354636545350649925890442528319161801605921579588526654791259724057279137519961488033020252287291656224996559845743701416369311612554321667335443418374008955791237572115480942740883337659354946753897515362968288787775973679663087549902052834470568223797434895230309683907862913802778136943295386037545850573702404179634561691835378455616724949344536695840131493043451083253876120316451193093943057146831832699116608334542232364997054535258690377988973313999659509515527448391939826560776605356310573531843676266398565276101510029
n22 = 36411530853306134650264922004243230789578567724848633072235567209396268937793967740840652891507150988934473209257678404824253786278354088188390849648230052408289839466487412221995353670688546953660918964880243399052047334774328642066482644759267410757068380646549682108897462561270352289713894825761682839541110704279580406677692145731871923819604104525631319112589270885702852325198972148331330622015745319864858600319231335192994724027402115404237271512429501511126745777802735039211692592419147477615289600437180622597687832926895508311856277663498164572827860273972198929500903338268219335130382722268336214976042
n3 = 21226342903519534285704039882517125493795254854398682031383220268196433752075453643793049930458409496501461364842263780716599044573472994756306842377996628221505793163345085310959982378400333469066510661862233433012101801971093682929377066833668523917150410559059871941217104903416187874137499834716703973567800989192226779852732764629767211879833127201510149229683409885749849769992849238187983773625783302822977444881786880085459631032114161881530793073480211272526557216580382426240205299262699361842634152273785508127206291906598310431383694685205902446688152373743788426712423020706959735024415865092732865293579
n = 19200751218807086109753585680575754542082218941447307423137776192000278638223531900870979886440646382572944765791998502777800968087411516141316151136038574929174075944033279652394413631362165307728020601616420974300745352308691582297884104984804426129732048593928551212972068865439525515893127416831193018078920195294960658070386235303119382941397098688638535311813138494565312063872100234086554220918850216072480857287537545501163451784114917247368012496289240698205942331461841811192853876743048133216069555958862196216052716205918188818909294300568480346854813619378406414745549741993309728072933212639965842420209
c1 = 12141687753092765754669783940675830448713581449163645536701718573773069294641533345315774400524395912993282498354768039945837224261487080475299072168517385713156203158247568336654731834308671872786353160933613619005116377874975161570966140277643874697751373054295698713456796119408728017676683503913534440646385085389636575390015489314292617682540872972440031779129422252433290117697330366830572038270337066299696360385868714809628744897197112890179467453713145233723691881680439517219988389546696789593902726903494192576858861380642835270848700081419409960522215088312481748912596701475019010221975400877543191195027
c2 = 18671344975076988512476107719711622987157049217462407071825706570480527181436147001821738820485566989301383448587621862627034353591129052899016475842991199430712285197912470481262314752372552644754537903335502976906151622416039395240789105429194458158873730659899128044024578932624430327308839049226169398974549593196858893169326628555925601866140682257990896398722170890530721702632112280595631922141382527353635243846673957989765181865712175794172200610424145223744315723737355016529480510122584370575790398823526020995975020616450936696444211881773753140054718402336533120422812171276792369946574521223325184154908
e = 2999


# factor n1
t1 = gmpy2.iroot(n1 // 2024,2)[0]
for i in range(-100000,100000):
    p1 = t1 + i
    if n1 % p1 == 0:
        q1 = n1 // p1
        break
if p1 > q1:
    p1,q1 = q1,p1


# factor n2
p_add_q = gmpy2.iroot(n22 + 2*n2,2)[0]
p_sub_q = gmpy2.iroot(n22 - 2*n2,2)[0]
p2 = (p_add_q + p_sub_q) // 2
q2 = n2 // p2


if p2 > q2:
    p2,q2 = q2,p2
    
# factor n3
t3 = 2*gmpy2.iroot(n3,2)[0] + 2
p_add_q = t3
p_sub_q = gmpy2.iroot((p_add_q)**2 - 4*n3,2)[0]
p3 = (p_add_q + p_sub_q) // 2
q3 = (p_add_q - p_sub_q) // 2


if p3 > q3:
    p3,q3 = q3,p3


p1,q1 = int(p1),int(q1)
p2,q2 = int(p2),int(q2)
p3,q3 = int(p3),int(q3)


###### Franklin-Reiter
R.<x> = PolynomialRing(Zmod(n))
f = (p1 * x * x + p2 * x + p3)^e - c1
g = (q1 * x * x + q2 * x + q3)^e - c2


res = GCD(f,g)


m = -res.monic().coefficients()[0]
print(m)
flag = long_to_bytes(int(m))
print(flag)