前言

week2之后的内容，随缘更新，如果有，可能只会有Crypto的wp了

Week1

Misc

CyberChef’s Secret

basecrack一把梭

flag:

flag{Base_15_S0_Easy_^_^}

机密图片

lsb隐写，使用stegsolve查看

flag:

flag{W3lc0m3_t0_N3wSt4RCTF_2023_7cda3ece}

流量！鲨鱼！

追踪http流62，发现以base64的方式读取了flag

然后连续两次Base64解码

flag:

flag{Wri35h4rk_1s_u53ful_b72a609537e6}

压缩包们

先使用banzip修复压缩包，并且发现文件尾部有一串Base64字符串，解码得到hint：压缩包密码为6个数字
得到压缩包密码232311
解压得到flag

flag{y0u_ar3_the_m4ter_of_z1111ppp_606a4adc}

空白格

whitespace解密

flag:

flag{w3_h4v3_to0_m4ny_wh1t3_sp4ce_2a5b4e04}

隐秘的眼睛

一眼顶针，silentEye

flag:

flag{R0ck1ng_y0u_63b0dc13a591}

Web

泄露的秘密

慢扫描发现泄露www.zip,解压分别在index.php和robots.txt得到flag部分

flag:

flag{r0bots_1s_s0_us3ful_4nd_www.zip_1s_s0_d4ng3rous}

Begin of Upload

简简单单绕过前端JS上传，直接用bp抓包，然后修改后缀为php即可

获取flag

ErrorFlask

随便输入字符串触发报错

flag:

flag{Y0u_@re_3enset1ve_4bout_deb8g}

Begin of HTTP

按照题目提示分别在http各个字段添加对应值

POST /?ctf=1 HTTP/1.1
Host: node4.buuoj.cn:25964
Pragma: no-cache
Cache-Control: no-cache
Upgrade-Insecure-Requests: 1
User-Agent: NewStarCTF2023
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.7
Accept-Encoding: gzip, deflate
Accept-Language: zh-CN,zh;q=0.9,en;q=0.8
Cookie: td_cookie=4141505721; token=zZy1ZCBLubE5mDt0XNPVB6aFLV1FHsxu4XjmQdobEnIe%2B8ICxor4zaQUV4bUiOUS2s21RhndobJbf%2FXrREpqHQ%3D%3D; power=ctfer
Referer:newstarctf.com
X-real-ip:127.0.0.1
Connection: close
Content-Type: application/x-www-form-urlencoded
Content-Length: 28secret=n3wst4rCTF2023g00000d

Begin of PHP

按照每一步的要求进行绕过即可，都是常规知识点

http://c7a2b415-8554-4b1e-bb29-a5b243096721.node4.buuoj.cn:81/?key1=QNKCDZO&key2=aabg7XSs&key4[]=123&key5=2024akey3[]=1&flag5=("%14%12%15%05"|"%60%60%60%60")

R!C!E!

爆破一下hash值，然后bypass rce即可

password=114514&e[v.a.l=echo `tac /f*`;

EasyLogin

爆破admin的密码
得到密码000000
登陆查看响应包
flag在这.jpg，我真截图了（bushi

Crypto

brainfuck

brainfuck解密
flag:

flag{Oiiaioooooiai#b7c0b1866fe58e12}

Caesar’s Secert

遍历一下

flag:

flag{ca3s4r's_c1pher_i5_v4ry_3azy}

fence

flag:

flag{reordering_the_plaintext#686f8c03}

Vigenère

反推一下密钥即可，发现密钥为 key

flag:

flag{la_c1fr4_del_5ign0r_giovan_batt1st4_b3ll5s0}

babyrsa

因子较小，直接使用欧拉函数计算phi

#sage
n = 17290066070594979571009663381214201320459569851358502368651245514213538229969915658064992558167323586895088933922835353804055772638980251328261
c = 14322038433761655404678393568158537849783589481463521075694802654611048898878605144663750410655734675423328256213114422929994037240752995363595
e = 65537
phi = euler_phi(n)
d = inverse_mod(e,phi)
m = pow(c,d,n)
flag = bytes.fromhex(hex(m)[2:])
print(flag)
#flag{us4_s1ge_t0_cal_phI}

Small d

维纳攻击

from Crypto.Util.number import *
import gmpy2def continuedFra(x, y):"""计算连分数:param x: 分子:param y: 分母:return: 连分数列表"""cf = []while y:cf.append(x // y)x, y = y, x % yreturn cfdef gradualFra(cf):"""计算传入列表最后的渐进分数:param cf: 连分数列表:return: 该列表最后的渐近分数"""numerator = 0denominator = 1for x in cf[::-1]:# 这里的渐进分数分子分母要分开numerator, denominator = denominator, x * denominator + numeratorreturn numerator, denominatordef solve_pq(a, b, c):"""使用韦达定理解出pq，x^2−(p+q)∗x+pq=0:param a:x^2的系数:param b:x的系数:param c:pq:return:p，q"""par = gmpy2.isqrt(b * b - 4 * a * c)return (-b + par) // (2 * a), (-b - par) // (2 * a)def getGradualFra(cf):"""计算列表所有的渐近分数:param cf: 连分数列表:return: 该列表所有的渐近分数"""gf = []for i in range(1, len(cf) + 1):gf.append(gradualFra(cf[:i]))return gfdef wienerAttack(e, n):""":param e::param n::return: 私钥d"""cf = continuedFra(e, n)gf = getGradualFra(cf)for d, k in gf:if k == 0: continueif (e * d - 1) % k != 0:continuephi = (e * d - 1) // kp, q = solve_pq(1, n - phi + 1, n)if p * q == n:return dn = 19873634983456087520110552277450497529248494581902299327237268030756398057752510103012336452522030173329321726779935832106030157682672262548076895370443461558851584951681093787821035488952691034250115440441807557595256984719995983158595843451037546929918777883675020571945533922321514120075488490479009468943286990002735169371404973284096869826357659027627815888558391520276866122370551115223282637855894202170474955274129276356625364663165723431215981184996513023372433862053624792195361271141451880123090158644095287045862204954829998614717677163841391272754122687961264723993880239407106030370047794145123292991433
e = 8614531087131806536072176126608505396485998912193090420094510792595101158240453985055053653848556325011409922394711124558383619830290017950912353027270400567568622816245822324422993074690183971093882640779808546479195604743230137113293752897968332220989640710311998150108315298333817030634179487075421403617790823560886688860928133117536724977888683732478708628314857313700596522339509581915323452695136877802816003353853220986492007970183551041303875958750496892867954477510966708935358534322867404860267180294538231734184176727805289746004999969923736528783436876728104351783351879340959568183101515294393048651825
c = 6755916696778185952300108824880341673727005249517850628424982499865744864158808968764135637141068930913626093598728925195859592078242679206690525678584698906782028671968557701271591419982370839581872779561897896707128815668722609285484978303216863236997021197576337940204757331749701872808443246927772977500576853559531421931943600185923610329322219591977644573509755483679059951426686170296018798771243136530651597181988040668586240449099412301454312937065604961224359235038190145852108473520413909014198600434679037524165523422401364208450631557380207996597981309168360160658308982745545442756884931141501387954248
d = wienerAttack(e, n)
m = pow(c, d, n)
flag = long_to_bytes(m)
print(flag)
#flag{learn_some_continued_fraction_technique#dc16885c}

babyxor

异或第一个明文得到key为143,后续直接密文循环异或143即可

from binascii import *enc = 'e9e3eee8f4f7bffdd0bebad0fcf6e2e2bcfbfdf6d0eee1ebd0eabbf5f6aeaeaeaeaeaef2'
c = unhexlify(enc)
flag = ''
for i in c:flag += chr(i^143)
print(flag)
#flag{x0r_15_symm3try_and_e4zy!!!!!!}

babyencoding

part1,base64

flag{dazzling_encoding#4e0ad4

part2,base32

f0ca08d1e1d0f10c0c7afe422fea7

part3,uuencode

c55192c992036ef623372601ff3a}

flag:

flag{dazzling_encoding#4e0ad4f0ca08d1e1d0f10c0c7afe422fea7c55192c992036ef623372601ff3a}

Affine

先利用已知的flag格式flag{爆破符合条件的a和b,再去遍历a,b解密密文,最后把包含flag{的明文打印出来

from binascii import *
import gmpy2def find_keys(c,part_flag):keys = []for k in range(5):for i in range(1,101):for j in range(1, 101):a = ib = jtry:plain = (c[k]-j)*gmpy2.invert(a,256) % 256if chr(plain) == part_flag[k]:keys.append([i,j])except:passreturn keysdef get_flag(c,keys):for key in keys:flag = ''a = key[0]b = key[1]for i in c:plain = (i-b)*gmpy2.invert(a,256) % 256flag += chr(plain)if 'flag{' in flag:print(flag)breakif __name__ == '__main__':enc = 'dd4388ee428bdddd5865cc66aa5887ffcca966109c66edcca920667a88312064'c = unhexlify(enc)part_flag = 'flag{'keys = find_keys(c,part_flag)get_flag(c,keys)#flag{4ff1ne_c1pher_i5_very_3azy}

babyaes

先异或1得到xor_key_iv,然后取其前16字节得到key,再异或xor_key_iv得到iv

from Crypto.Util.number import *
from Crypto.Cipher import AESc = b'>]\xc1\xe5\x82/\x02\x7ft\xf1B\x8d\n\xc1\x95i'
xor_data = 3657491768215750635844958060963805125333761387746954618540958489914964573229
xor_key_iv = xor_data^1
key = long_to_bytes(int(xor_key_iv))[:16]*2
iv = long_to_bytes(xor_key_iv^bytes_to_long(key))
aes = AES.new(key, AES.MODE_CBC, iv)
flag = aes.decrypt(c).split(b'\00')[0]
print(b'flag{'+flag+b'}')
#flag{firsT_cry_Aes}

【活着得，怎么就轻松惬意了，无需愧疚。】