2022-01-22week1&2&3&4

老年人复健，但依旧在舒适圈晃悠（……

crypto-Easy rsa

# 已知p,q,e,c, 求m
# m = pow(c,d,n), n=p*q
import gmpy2

def countD(e,p,q):     
    s = (p-1)*(q-1)
    d = gmpy2.invert(e,s)
    return d

a=[(12433, 149, 197, 104), (8147, 131, 167, 6633), (10687, 211, 197, 35594), (19681, 131, 211, 15710), (33577, 251, 211, 38798), (30241, 157, 251, 35973), (293, 211, 157, 31548), (26459, 179, 149, 4778), (27479, 149, 223, 32728), (9029, 223, 137, 20696), (4649, 149, 151, 13418), (11783, 223, 251, 14239), (13537, 179, 137, 11702), (3835, 167, 139, 20051), (30983, 149, 227, 23928), (17581, 157, 131, 5855), (35381, 223, 179, 37774), (2357, 151, 223, 1849), (22649, 211, 229, 7348), (1151, 179, 223, 17982), (8431, 251, 163, 30226), (38501, 193, 211, 30559), (14549, 211, 151, 21143), (24781, 239, 241, 45604), (8051, 179, 131, 7994), (863, 181, 131, 11493), (1117, 239, 157, 12579), (7561, 149, 199, 8960), (19813, 239, 229, 53463), (4943, 131, 157, 14606), (29077, 191, 181, 33446), (18583, 211, 163, 31800), (30643, 173, 191, 27293), (11617, 223, 251, 13448), (19051, 191, 151, 21676), (18367, 179, 157, 14139), (18861, 149, 191, 5139), (9581, 211, 193, 25595)]
l=''
for i in a:
    d=countD(i[0],i[1],i[2])
    m=pow(i[3],d,i[1]*i[2])
    l+=chr(m)
print(l)

crypto-Matryoshka

将盲文替换为摩斯密码

..-/--.../--..--/-----/--.../--..--/---../...--/--..--/.-../....-/--..--/..---/....-/--..--/..---/--.../--..--/.-.-/-..../--..--/...-/....-/--..--/....-/--.../--..--/---../...../--..--/---../--.../--..--/...-/-..../--..--/---../...--/--..--/.----/--.../--..--/--.../...--/--..--/...-/....-/--..--/-./....-/--..--/----./--.../--..--/--.../...../--..--/..-/-..../--..--/.-../....-/--..--/.----/--.../--..--/./-..../--..--/----./...../--..--/---../...--/--..--/.----/--.../--..--/..-/....-/--..--/...-/....-/--..--/.----/...--/--..--/.-.-/....-/--..--/--.../...../--..--/-..../....-/--..--/--.../....-/--..--/-----/--.../--..--/./-..../--..--/.----/....-/--..--/-..../--.../--..--/-----/...--/--..--/---../--.../--..--/..---/...../--..--/....-/--.../--..--/.-.-/-..../--..--/-..../...--/--..--/.----/...../--..--/.----/....-/--..--/--.../...--/--..--/--.../--.../--..--/-..../-..../--..--/....-/--.../--..--/--.../...--/--..--/.-.-/-..../--..--/..-/....-/--..--/./....-/--..--/....-/-..../--..--/...../-..../--..--/....-/....-/--..--/...../...--/--..--/---../-..../--..--/....-/....-/--..--/....-/-....

将结果逆序，解摩斯密码，hex2chr

FfBufEFnmLs6D3siYtL6X2p4iN0cdSlykm9rQN9oMS1jks9rK2R6kL8hor0=

维吉尼亚（hgame），base64，凯撒（21）

h0gralmde_{0Wfe_lccr0ympet_0tg0r_atphhey_!w}

a='h0gralmde_{0Wfe_lccr0ympet_0tg0r_atphhey_!w}'
a=a[::2]+a[1::2]
# hgame{Welc0me_t0_the_w0rld_0f_crypt0graphy!}

Re - easy asm

seg1=['91','61','1','C1','41','A0','60','41','D1','21','14','C1','41','E2','50','E1','E2','54','20','C1','E2','60','14','30','D1','51','C0','17']
seg1=[int(i,16) for i in seg1]
# flag='hgame'
# for i in range(5):
#     tmp=((ord(flag[i])<<4)+(ord(flag[i])>>4))&int('11111111',2)
#     tmp=tmp^23
#     if tmp==seg1[i]:
#         print('111111')
l=''        
for i in seg1:
    tmp=str(bin(i^23))[2:].zfill(8)
    l+=chr(int(tmp[4:]+tmp[:4],2))
print(l)

Re - flag checker

import base64
from Crypto.Cipher import ARC4 as rc4cipher
s2 = base64.b64decode('mg6CITV6GEaFDTYnObFmENOAVjKcQmGncF90WhqvCFyhhsyqq1s=')
enc = rc4cipher.new('carol'.encode("utf-8"))
res = enc.decrypt(s2)
print(res)

Web - 蛛蛛...嘿嘿♥我的蛛蛛

import requests
import re
header={'Host': 'hgame-spider.vidar.club',
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:96.0) Gecko/20100101 Firefox/96.0',
'Accept': 'text/html,application/xhtml+xml,application/xml',
'Accept-Language': 'zh-CN,zh',
'Accept-Encoding': 'gzip, deflate',
'Connection': 'close'}

url='https://hgame-spider.vidar.club/7724d28954'
key=''
while True:
    a=requests.get(url+key,headers=header)
    try:
        key=re.findall('<a href="?.*%3D">点我试试</a>',a.text)[0][9:-10]
    except:
        print(a.headers)
        break

Web - Tetris plus

f12找JS文件

开控制台解密弹flag

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Misc - 这个压缩包有点麻烦

典型的AZPR使用教学题（乐）
爆破，字典爆破，明文攻击，binwalk，伪加密
加密位：

Misc - 好康的流量

追踪TCP流 ->将base64编码的图像另存 -> Green plane 2发现条码，解密得到flag前半部分 -> LSB发现flag后半部分
条形码在线解码：https://online-barcode-reader.inliteresearch.com/

import base64
with open('img.txt')as f:
    data=base64.b64decode(f.read())
with open('res.jpg','wb')as f2:
    f2.write(data)

Misc-群青

文件尾有提示使用SilentEye

Audacity频谱图可见

解密可得SSTV.wav，下载

安装VirtualAudioCable，将虚拟声卡设置为输入输出设备

安装RX-SSTV，选择robot36，播放音频，扫描二维码得到flag

Week1 - Web - easy auth - 复现

想到要从token下手，但去爆了密钥（……）
https://jwt.io/

Week1 - Fujiwara Tofu Shop - 复现

学到了新知识（痴呆）

这里正确的做法应该是设置 X- Real-IP 为 127.0.0.1 。IP 伪造和代理服务器有关，相关的请求头有 X-Forworded-For , X-Real- IP, X-Client-IP等，至于那个请求头能成功伪造 IP，得参考具体的网络环境，编程语言，服务端框架和服务端配置。

Week1 - Re - crack me - 复现

现在tea都成密码题标配了么（痛苦）
抄了（抄了）

enced = [0x48D93488, 0x030C144C, 0x52EB78C2, 0xED9CE5ED,
          0xAE1FEDE6, 0xBA5A126D, 0xCF9284AA, 0x65E0F2E3]
messg = 0
mask = 0xffffffff 
offset = 0x12345678 
v10_0 = 0x44434241 
v10_1 = 0x48474645 
v10_2 = 0x4C4B4A49 
v10_3 = 0x504F4E4D 
flag=''
for i in range(4): 
    v3 = (offset * 32) & mask 
    v5 = enced[2 * i] 
    v6 = enced[2 * i + 1] 
    for _ in range(32): 
        v6 -= v3 ^ (v3 + v5) ^ (v10_0 + (16 * v5)&mask) ^ (v10_1 + (v5 >> 5)) 
        v6 &= mask 
        v5 -= v3 ^ (v3 + v6) ^ (v10_2 + (16 * v6)&mask) ^ (v10_3 + (v6 >> 5)) 
        v5 &= mask 
        v3 -= offset 
        v3 &= mask 
        # print(hex(v5)) 
        # print(hex(v6)) 
    v5t = v5 
    v6t = v6 
    for _ in range(4): 
        flag+=chr(v5t&0xff)
        v5t >>= 8 
    for _ in range(4): 
        flag+=chr(v6t&0xff)
        v6t >>= 8

Week2 - Crypto - RSA Attack

factordb即可得到q、p

Week2 - Crypto - RSA Attack 2

import re
from math import ceil
from Crypto.Util.number import getPrime
from libnum import s2n,n2s
##from secret import flag

##flag_parts = list(map(s2n, re.findall(rf".{{,{ceil(len(flag) / 3)}}}", flag)))

##print("# task1")
##m = flag_parts[0]
##e = 65537
##p = getPrime(1024)
##q = getPrime(1024)
##r = getPrime(1024)
##n1 = p * q
##c1 = pow(m, e, n1)
##n2 = r * q
##c2 = pow(m, e, n2)
##print("e =", e)
##print("n1 =", n1)
##print("c1 =", c1)
##print("n2 =", n2)
##print("c2 =", c2)

import gmpy2

e = 65537
n1 = 14611545605107950827581005165327694782823188603151768169731431418361306231114985037775917461433925308054396970809690804073985835376464629860609710292181368600618626590498491850404503443414241455487304448344892337877422465715709154238653505141605904184985311873763495761345722155289457889686019746663293720106874227323699288277794292208957172446523420596391114891559537811029473150123641624108103676516754449492805126642552751278309634846777636042114135990516245907517377320190091400729277307636724890592155256437996566160995456743018225013851937593886086129131351582958811003596445806061492952513851932238563627194553
c1 = 965075803554932988664271816439183802328812013694203741320763105376036912584995031647672348468111310423680858101990670067065306237596121664884353679987689532305437801346923070145524106271337770666947677115752724993307387122132705797012726237073550669419110046308257408484535063515678066777681017211510981429273346928022971149411064556225001287399141306136081722471075032423079692908380267160214143720516748000734987068685104675254411687005690312116824966036851568223828884335112144637268090397158532937141122654075952730052331573980701136378212002956719295192733955673315234274064519957670199895100508623561838510479
n2 = 20937478725109983803079185450449616567464596961348727453817249035110047585580142823551289577145958127121586792878509386085178452171112455890429474457797219202827030884262273061334752493496797935346631509806685589179618367453992749753318273834113016237120686880514110415113673431170488958730203963489455418967544128619234394915820392908422974075932751838012185542968842691824203206517795693893863945100661940988455695923511777306566419373394091907349431686646485516325575494902682337518438042711296437513221448397034813099279203955535025939120139680604495486980765910892438284945450733375156933863150808369796830892363
c2 = 11536506945313747180442473461658912307154460869003392732178457643224057969838224601059836860883718459986003106970375778443725748607085620938787714081321315817144414115589952237492448483438910378865359239575169326116668030463275817609827626048962304593324479546453471881099976644410889657248346038986836461779780183411686260756776711720577053319504691373550107525296560936467435283812493396486678178020292433365898032597027338876045182743492831814175673834198345337514065596396477709839868387265840430322983945906464646824470437783271607499089791869398590557314713094674208261761299894705772513440948139429011425948090

def gcd(a, b):
   if a < b:
     a, b = b, a
   while b != 0:
     temp = a % b
     a = b
     b = temp
   return a

p=gcd(n1,n2)
q=n1//p
d = gmpy2.invert(e,(p-1)*(q-1))
m=pow(c1,d,n1)
f1=n2s(int(m))

##print("# task2")
##m = flag_parts[1]
##e = 7
##p = getPrime(1024)
##q = getPrime(1024)
##n = p * q
##c = pow(m, e, n)
##print("e =", e)
##print("n =", n)
##print("c =", c)
##

e = 7
n = 14157878492255346300993349653813018105991884577529909522555551468374307942096214964604172734381913051273745228293930832314483466922529240958994897697475939867025561348042725919663546949015024693952641936481841552751484604123097148071800416608762258562797116583678332832015617217745966495992049762530373531163821979627361200921544223578170718741348242012164115593777700903954409103110092921578821048933346893212805071682235575813724113978341592885957767377587492202740185970828629767501662195356276862585025913615910839679860669917255271734413865211340126544199760628445054131661484184876679626946360753009512634349537
c = 10262871020519116406312674685238364023536657841034751572844570983750295909492149101500869806418603732181350082576447594766587572350246675445508931577670158295558641219582729345581697448231116318080456112516700717984731655900726388185866905989088504004805024490513718243036445638662260558477697146032055765285263446084259814560197549018044099935158351931885157616527235283229066145390964094929007056946332051364474528453970904251050605631514869007890625

k = 0
while 1:
    res = gmpy2.iroot(c+k*n,e)
    if(res[1] == True):
        f2=n2s(int(res[0]))
        break
    k+=1

##print("# task3")
##m = flag_parts[2]
##p = getPrime(1024)
##q = getPrime(1024)
##n = p * q
##e1 = getPrime(32)
##e2 = getPrime(32)
##c1 = pow(m, e1, n)
##c2 = pow(m, e2, n)
##print("n =", n)
##print("e1 =", e1)
##print("c1 =", c1)
##print("e2 =", e2)
##print("c2 =", c2)


def common_modulus(n, c1, c2, e1, e2):
    _, s1, s2 = gmpy2.gcdext(e1, e2)#扩展欧几里得算法
    #若s1<0，则c1^s1==(c1^-1)^(-s1)，其中c1^-1为c1模n的逆元。
    if s1 < 0:
        s1 = -s1
        c1 = gmpy2.invert(c1, n)
    if s2 < 0:
        s2 = -s2
        c2 = gmpy2.invert(c2, n)
    return pow(c1, s1, n)*pow(c2, s2, n)%n


n = 18819509188106230363444813350468162056164434642729404632983082518225388069544777374544142317612858448345344137372222988033366528086236635213756227816610865045924357232188768913642158448603346330462535696121739622702200540344105464126695432011739181531217582949804939555720700457350512898322376591813135311921904580338340203569582681889243452495363849558955947124975293736509426400460083981078846138740050634906824438689712748324336878791622676974341814691041262280604277357889892211717124319329666052810029131172229930723477981468761369516771720250571713027972064974999802168017946274736383148001865929719248159075729
e1 = 2519901323
c1 = 3230779726225544872531441169009307072073754578761888387983403206364548451496736513905460381907928107310030086346589351105809028599650303539607581407627819797944337398601400510560992462455048451326593993595089800150342999021874734748066692962362650540036002073748766509347649818139304363914083879918929873577706323599628031618641793074018304521243460487551364823299685052518852685706687800209505277426869140051056996242882132616256695188870782634310362973153766698286258946896866396670872451803114280846709572779780558482223393759475999103607704510618332253710503857561025613632592682931552228150171423846203875344870
e2 = 3676335737
c2 = 940818595622279161439836719641707846790294650888799822335007385854166736459283129434769062995122371073636785371800857633841379139761091890426137981113087519934854663776695944489430385663011713917022574342380155718317794204988626116362865144125136624722782309455452257758808172415884403909840651554485364309237853885251876941477098008690389600544398998669635962495989736021020715396415375890720335697504837045188626103142204474942751410819466379437091569610294575687793060945525108986660851277475079994466474859114092643797418927645726430175928247476884879817034346652560116597965191204061051401916282814886688467861

res=common_modulus(n, c1, c2, e1, e2)
f3=n2s(int(res))

print (f1+f2+f3)

# just in case you want less code
# ...
# p=gmpy2.gcd(n1,n2)
# q=n1//p
# f1=n2s(int(pow(c1,gmpy2.invert(e,(p-1)*(q-1)),n1)))
# ...
# f2=n2s(int(gmpy2.iroot(c, e)[0]))
# ...

Week3 - Crypto - Multi Prime RSA

总觉得以前做过差不多的题，在哪儿来着

##from Crypto.Util.number import getPrime
##from libnum import s2n
##from secret import flag
##
##p = getPrime(256)
##q = getPrime(256)
##r = getPrime(256)
##s = getPrime(256)
##n = p ** 2 * q ** 3 * r ** 5 * s ** 7
##e = 65537
##c = pow(s2n(flag), e, n)
##print(f"p = {p}")
##print(f"q = {q}")
##print(f"r = {r}")
##print(f"s = {s}")
##print(f"n = {n}")
##print(f"e = {e}")
##print(f"c = {c}")

import gmpy2
from libnum import n2s
p = 61789932148719477384027458333380568978056286136137829092952317307711908353477
q = 91207969353355763685633284378833506319794714507027332929290701748727534193861
r = 105471299607375388622347272479207944509670502835651250945203397530010861809367
s = 83153238748903772448138307505579799277162652151244477391465130504267171881437
n = 1039344372165087100001063920598151812324151064684841845250974758525265148567706103784958424873181721352440209284812493753972556519482026327282644619091466886523804841248277210353173383407944598453848113815866908595335619458549486958764490103808475329598085842184963065068499489886467911087295087163762599284622055185456905774507245781667293199205317692029829495961487347944813874415423771980660778986211145841712412631156369129146470119135136378158203459576596246169191419488560832734046076107673091995860021863239882608638458149930255944184863801278386551031980146460231515747754411678651752698881001464973981424240781413084941947261875289725538959720572496329348499870580057997540844488309111059240745081048324762866572948371222839278718034435739827677190025500802453626872356208612718417249649474571197167076916403582394186357812640566250930361276229969553128128312736245440129556020108188835966131425956431796417720436474093381770796431629523054378258497546013222494974549262140415585158985940966415459478150722832119691308697510189026447359189994055885090735411738332296254011208547676914004864732327863884217733456287369771087094514708468685641820375220835485053482570852619363091173324203334503461823983610886849930944250553928855506012684504211525542998575275626784129736345142772399109273619522445919
e = 65537
c = 844677395496466411520394190869787261209960246734415406217975986418865760680024542119231873259131861208878522030009923057991526761346423130242121884493257732067700857897379859545356609151834223804262174935191718271211809221730601602827122249238086030580971376104724987801049500689134122609834321586609223761140538079460830213824674361601046367637227094018381901291488659642720549583856812747877519600804325570421770575999289389175021646347371879234023647657507178519047236746071420327155188213839293382288787853777540226192644761028822256165706787395891134765908229036044468473519166141610604791485071702808854944672418124203289328124793348198048601338476086482318248264508789781967910205393740835345086784345145351367491197717933757414967811594913692588314161669333147733048171044386546892346475181197482702164468542430187885074163177843285948999943328049159021873821254267471067523609151007885131921896462161216356454116929796355815756642621369974260365378070336290542971599886325232821981080341858950609157813769416455337935096696635623426418166316737131174435618543058086342714723330814586496030805366321181723292731710369013923285787724941830672247377301048663929453294620044701627159066468762709113137517559435822623284148112827473010030736329596829357275518641576798298066541516764673029908084962144713
phi = (p ** 1 * (p-1)) * (q ** 2 *(q-1)) * (r ** 4 *(r-1)) * (s ** 6 * (s-1))
print (n2s(pow(c, int(gmpy2.invert(e, phi)), n)))

Week3 - Crypto - RSA Attack 3

搜来一个wienerAttack的脚本，抄了

##from Crypto.Util.number import getPrime
##from gmpy2 import invert
##from libnum import s2n
##from secret import flag
##
##p = getPrime(2048)
##q = getPrime(2048)
##n = p * q
##d = getPrime(64)
##e = invert(d, (p - 1) * (q - 1))
##c = pow(s2n(flag), e, n)
##print(f"n = {n}")
##print(f"e = {e}")
##print(f"c = {c}")
##


import gmpy2
import libnum
def transform(x,y):       #使用辗转相处将分数 x/y 转为连分数的形式
    res=[]
    while y:
        res.append(x//y)
        x,y=y,x%y
    return res
    
def continued_fraction(sub_res):
    numerator,denominator=1,0
    for i in sub_res[::-1]:      #从sublist的后面往前循环
        denominator,numerator=numerator,i*numerator+denominator
    return denominator,numerator   #得到渐进分数的分母和分子，并返回

    
#求解每个渐进分数
def sub_fraction(x,y):
    res=transform(x,y)
    res=list(map(continued_fraction,(res[0:i] for i in range(1,len(res)))))  #将连分数的结果逐一截取以求渐进分数
    return res

def get_pq(a,b,c):      #由p+q和pq的值通过维达定理来求解p和q
    par=gmpy2.isqrt(b*b-4*a*c)   #由上述可得，开根号一定是整数，因为有解
    x1,x2=(-b+par)//(2*a),(-b-par)//(2*a)
    return x1,x2

def wienerAttack(e,n):
    for (d,k) in sub_fraction(e,n):  #用一个for循环来注意试探e/n的连续函数的渐进分数，直到找到一个满足条件的渐进分数
        if k==0:                     #可能会出现连分数的第一个为0的情况，排除
            continue
        if (e*d-1)%k!=0:             #ed=1 (mod φ(n)) 因此如果找到了d的话，(ed-1)会整除φ(n),也就是存在k使得(e*d-1)//k=φ(n)
            continue
        
        phi=(e*d-1)//k               #这个结果就是 φ(n)
        px,qy=get_pq(1,n-phi+1,n)
        if px*qy==n:
            p,q=abs(int(px)),abs(int(qy))     #可能会得到两个负数，负负得正未尝不会出现
            d=gmpy2.invert(e,(p-1)*(q-1))     #求ed=1 (mod  φ(n))的结果，也就是e关于 φ(n)的乘法逆元d
            return d
    print("该方法不适用")

if __name__ == "__main__":
    n = 507419170088344932990702256911694788408493968749527614421614568612944144764889717229444020813658893362983714454159980719026366361318789415279417172858536381938870379267670180128174798344744371725609827872339512302232610590888649555446972990419313445687852636305518801236132032618350847705234643521557851434711389664130274468354405273873218264222293858509477860634889001898462547712800153111774564939279190835857445378261920532206352364005840238252284065587291779196975457288580812526597185332036342330147250312262816994625317482869849388424397437470502449815132000588425028055964432298176942124697105509057090546600330760364385753313923003549670107599757996810939165300581847068233156887269181096893089415302163770884312255957584660964506028002922164767453287973102961910781312351686488047510932997937700597992705557881172640175117476017503918294534205898046483981707558521558992058512940087192655700351675718815723840568640509355338482631416345193176708501897458649841539192993142790402734898948352382350766125000186026261167277014748183012844440603384989647664190074853086693408529737767147592432979469020671772152652865219092597717869942730499507426269170189547020660681363276871874469322437194397171763927907099922324375991793759
    e = 77310199867448677782081572109343472783781135641712597643597122591443011229091533516758925238949755491395489408922437493670252550920826641442189683907973926843505436730014899918587477913032286153545247063493885982941194996251799882984145155733050069564485120660716110828110738784644223519725613280140006783618393995138076030616463398284819550627612102010214315235269945251741407899692274978642663650687157736417831290404871181902463904311095448368498432147292938825418930527188720696497596867575843476810225152659244529481480993843168383016583068747733118703000287423374094051895724494193455175131120243097065270804457787026492578916584536863548445813916819417857064037664101684455000184987531252344582899589746272173970083733130106407810619258077266603898529285634495710846838011858287024329514491058790557305041389614650730267774482954666726949886313386881066593946789460028399523245777171320319444673551268379126203862576627540177888290265714418064334752499940587750374552330008143708562065940245637685833371348603338834447212248648869514585047871442060412622164276894766238383894693759347590977926306581080390685360615407766600573527565016914830132066428454738135380178959590692145577418811677639050929791996313180297924833690095
    c = 165251729917394529793163344300848992394021337429474789711805041655116845722480301677817165053253655027459227404782607373107477419083333844871948673626672704233977397989843349633720167495862807995411682262559392496273163155214888276398332204954185252030616473235814999366132031184631541209554169938146205402400412307638567132128690379079483633171535375278689326189057930259534983374296873110199636558962144635514392282351103900375366360933088605794654279480277782805401749872568584335215630740265944133347038070337891035560658434763924576508969938866566235926587685108811154229747423410476421860059769485356567301897413767088823807510568561254627099309752215808220067495561412081320541540679503218232020279947159175547517811501280846596226165148013762293861131544331444165070186672186027410082671602892508739473724143698396105392623164025712124329254933353509384748403154342322725203183050328143736631333990445537119855865348221215277608372952942702104088940952142851523651639574409075484106857403651453121036577767672430612728022444370874223001778580387635197325043524719396707713385963432915855227152371800527536048555551237729690663544828830627192867570345853910196397851763591543484023134551876591248557980182981967782409054277224
    d=wienerAttack(e,n)
    m = pow(c, d, n)
    print(libnum.n2s(int(m)))

Week3 - Crypto - Block Cipher - 复现

我：什么东西，block c…区块链，跑了跑了不看这题
还是我：大意了，是块加密

import operator
##import random
##import re
from functools import reduce
##from secret import flag

##def pad(s):
##    padding_length = (8 - len(s)) % 8
##    return s + chr(padding_length) * padding_length

def xor(a, b):
    assert len(a) == len(b)
    return bytes(map(operator.xor, a, b))

##
##def encrypt(s):
##    iv = bytes(random.randint(0, 255) for _ in range(8))
##    key = bytes(random.randint(0, 255) for _ in range(8))
##    parts = list(map(str.encode, map(pad, re.findall(r'.{1,8}', s))))
##    results = []
##    for index, part in enumerate(parts):
##        results.append(reduce(xor, [part, iv if index == 0 else results[-1], key]))
##    return iv, key, results
##
##iv, key, parts = encrypt(flag)
##print(f"iv = {iv}")
##print(f"key = {key}")
##print(f"parts = {parts}")

def decrypt(iv, key, parts):
    result = []
    for index, part in enumerate(parts):
        result.append(reduce(xor, [part, iv if index == 0 else parts[index - 1], key]))
    return b''.join(result)
iv = b'Up\x14\x98r\x14%\xb9'
key = b'\r\xe8\xb86\x9c33^'
parts = [b'0\xff\xcd\xc3\x8b\\T\x8b', b'RT\x1e\x89t&\x17\xbd', b'\x1a\xee\x8d\xd6\x9b>w\x8c', b'9CT\xb3^pF\xd0']
print(decrypt(iv, key, parts))

Week4 - Misc - 摆烂 - 复现

大意了，没有闪，它一个零宽字节隐写打过来ry
在线decode：
https://330k.github.io/misc_tools/unicode_steganography.html
binwalk分离压缩包和图片，stegsolve打开图片可见横向条纹，推测考点为盲水印

010editor打开

存在acTL和fcTL数据块，搜索可知为apng文件。使用apngdis分离，得到两张一样的图片，stegsolve打开，xor两张图片，可见明显横向条纹，确定为盲水印。decode得到水印图，可得压缩包解压密码：

解压后拼好二维码，扫码即可得到一段含有零宽字节隐写的废话生成器文章（……）使用在线网站decode可得flag

Week4 - Misc - TOP SERECT 2 - 复现

是我不会用hashcat，学到了.jpg

.\hashcat.exe -a 3 -m 1800 --force '$6$6ldh0bSoluytiIS2$vs6caYlhCSX1rmWaPrdDYI55hIg3Ls75PnbJmEBfYOsZRaknhS4cTdmyIbHPWz/2dTAdQUEM7IEmOc.GPo5UO.' --custom-charset1 'abcdefgnopqrs' --custom-charset2 'hijklmtuvwxyz' co?2b?2nd_7?up?dm6Q_?1nd_4br?d9Pc

Week4 - crypto - ECC - 复现

以前遇到sage题懒得复现终究是天道好轮回了.jpg
在线运行sage：https://sagecell.sagemath.org/

p = 74997021559434065975272431626618720725838473091721936616560359000648651891507
a = 61739043730332859978236469007948666997510544212362386629062032094925353519657
b = 87821782818477817609882526316479721490919815013668096771992360002467657827319
k = 93653874272176107584459982058527081604083871182797816204772644509623271061231
E = EllipticCurve(GF(p), [a, b]) 
c1 = E(14455613666211899576018835165132438102011988264607146511938249744871964946084 : 25506582570581289714612640493258299813803157561796247330693768146763035791942 : 1)
c2 = E(37554871162619456709183509122673929636457622251880199235054734523782483869931 : 71392055540616736539267960989304287083629288530398474590782366384873814477806 : 1)
m = c1 - k * c2 
cipher_left = 68208062402162616009217039034331142786282678107650228761709584478779998734710
cipher_right = 27453988545002384546706933590432585006240439443312571008791835203660152890619
print (int(cipher_left//m[0])) 
print (int(cipher_right//m[1]))

结果n2s一下即可得到flag

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