2023 SHCTF-校外赛道 Crypto—Wp

🥇WEEK1

立正

wl hgrfhg 4gNUx4NgQgEUb4NC64NHxZLg636V6CDBiDNUHw8HkapH :jdoi vl vlkw  ~xrb wd nrrT Y:

凯撒解密，偏移量为3得到：

ti edoced 4dKRu4KdNdBRy4KZ64KEuWId636S6ZAYfAKREt8EhxmE :galf si siht  ~uoy ta kooQ V:

发现edoced并结合题目提示：注意要正，意识到要把它逆序:V Qook at you~ this is flag: EmxhE8tERKAfYAZ6S636dIWuEK46ZK4yRBdNdK4uRKd4 decode it

这个题还是有点脑洞的，观察中间的密文EmxhE8tERKAfYAZ6S636dIWuEK46ZK4yRBdNdK4uRKd4，对比一下flag{的base64密文：ZmxhZ3s

可知道，大写字母后移21位因为E–>Z，而小写字母不动，数字前移5位，因为Qook at you–>Look at you

exp:

import base64

c = "EmxhE8tERKAfYAZ6S636dIWuEK46ZK4yRBdNdK4uRKd4"
m = ""
for i in range(len(c)):
    if c[i].isupper():
        mm = 65 + (ord(c[i]) - 65 + 21) % 26 
        m += chr(mm)
    if c[i].isdigit():
        mm = ord('0') + ((ord(c[i]) - ord('0') -5) % 10)
        m += chr(mm)
    if c[i].islower():
        m += c[i]
flag = base64.b64decode(m.encode())
print(flag)
#flag{Y0U_MU57_5t4nd_uP_r1gHt_n0W}

Crypto_Checkin

QZZ|KQbjRRS8QZRQdCYwR4_DoQ7~jyO>0t4R4__aQZQ9|Rz+k_Q!r#mR90+NR4_4NR%>ipO>0s{R90|SQhHKhRz+k^S8Q5JS5|OUQZO}CQfp*dS8P&9R8>k?QZYthRz+k_O>0#>

Cyber一把梭: base85 –> base64 –> base32 –> base16

残缺的md5

苑晴在路边捡到了一张纸条，上面有一串字符串：KCLWG?K8M9O3?DE?84S9
问号是被污染的部分，纸条的背面写着被污染的地方为大写字母，还给了这串字符串的md5码值：F0AF????B1F463????F7AE???B2AC4E6
请提交完整的md5码值并用flag{}包裹提交

爆破

exp:

import string
import hashlib

s = "KCLWG?K8M9O3?DE?84S9"
table = string.ascii_letters + string.digits
for i in range(128):
    for j in range(128):
        for k in range(128):
            ss = "KCLWG" + chr(i) + "K8M9O3" + chr(j) + "DE" + chr(k) + "84S9"
            hash = hashlib.md5(ss.encode()).hexdigest().upper()
            if hash[:4] == "F0AF" and hash[-7:] == "B2AC4E6" and "B1F463" in hash:
                flag = "flag{" + hash + "}"
                flag1 = "flag{" + ss + "}"
                print(flag)
                print(flag1)
                
#flag{F0AF1443B1F463EAFFF7AEBB8B2AC4E6}
#flag{KCLWGXK8M9O3KDED84S9}

凯撒大帝

pvkq{mredsrkyxkx}

凯撒解码

key1 #10: flag{chutihaonan}

进制

好熟悉的进制，但不知道加密了几层
3636366336313637376236313638363636623661366336383662363136383764

okk

Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook! Ook? Ook! Ook! Ook. Ook? Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook? Ook. Ook? Ook! Ook. Ook? Ook. Ook. Ook. Ook. Ook! Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook. Ook? Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook! Ook? Ook! Ook! Ook. Ook? Ook! Ook! Ook! Ook!
Ook! Ook! Ook? Ook. Ook? Ook! Ook. Ook? Ook! Ook! Ook! Ook! Ook! Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook. Ook? Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook? Ook! Ook! Ook. Ook? Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook? Ook. Ook? Ook! Ook. Ook? Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook. Ook? Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook? Ook! Ook!
Ook. Ook? Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook!
Ook! Ook! Ook! Ook? Ook. Ook? Ook! Ook. Ook? Ook! Ook! Ook! Ook! Ook! Ook!
Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook!
Ook. Ook. Ook. Ook! Ook. Ook. Ook. Ook! Ook. Ook. Ook. Ook! Ook. Ook. Ook.
Ook! Ook. Ook. Ook. Ook! Ook. Ook. Ook. Ook! Ook. Ook. Ook. Ook! Ook. Ook.
Ook. Ook! Ook. Ook? Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook? Ook! Ook! Ook. Ook? Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook? Ook.
Ook? Ook! Ook. Ook? Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook. Ook?
Ook.

BrainFuck解码:  ★  flag{123456789}

熊斐特

熊斐特博士发现了一种新的密码。
uozt{zgyzhs xrksvi}

黑暗之歌

密文：

⠴⡰⡭⡳⠴⡰⡭⡰⡷⡲⡢⡩⡭⡡⠯⡩⡭⡡⡺⡩⡭⡡⠳⡩⡭⡡⡺⡩⡭⡡⡶⡩⡭⡡⡶⡩⡭⡡⡲⡩⡭⡡⡺⡩⡭⡡⠯⡩⡧⡊⡢⡩⡭⡡⠯⡩⡭⡡⡺⡃⡰⠫⡋⡚⡲⡍⡋⡮⠴⡰⡭⡶⡷⡲⡢⡩⡧⡊⡢⡃⡴⡵⡋⡁⡬⡵⡋⡁⡬⡵⡋⡁⡬⡳⡋⠲⠴⡯⡃⡗⠴⡰⡭⡴⠴⡰⡭⡶⡷⡲⡢⡩⡧⡊⡢⡩⡭⡡⡺⡩⡭⡡⡺⡩⡭⡡⠳⡩⡧⡊⡢⡩⡭⡡⠯⡩⡧⡊⡢⡃⡴⡵⡋⡚⡱⠫⡋⡚⡱⠫⡋⡚⡲⠵⠲⡺⠰⠽

一眼丁真，盲文解码

4pms4pmpwrbima/imazima3imazimavimavimarimazima/igJbima/imazCp+KZrMKn4pmvwrbigJbCtuKAluKAluKAlsK24oCW4pmt4pmvwrbigJbimazimazima3igJbima/igJbCtuKZq+KZq+KZrz0=

base64解码

♬♩¶♯♬♭♬♫♫♪♬♯‖♯♬§♬§♯¶‖¶‖‖‖¶‖♭♯¶‖♬♬♭‖♯‖¶♫♫♯=

音符解码

迷雾重重

题目描述：

morse？ASCII？


密文：

0010 0100 01 110 1111011 11 111 010 000 0 001101 00 000 001101 0001 0 010 1011 001101 0010 001 10 1111101

摩斯解码

flag{morse_is_very_fun}

难言的遗憾

题目描述：
我们本可以早些进入信息化时代的，但是清政府拒不采纳那份编码规则。 （注：flag为中文，使用flag{}包裹提交）



密文：

000111310008133175592422205314327609650071810649

中文电码查询

中文电码查询网址2

小兔子可爱捏

题目描述：宇宙的终极答案是什么？





U2FsdGVkX1/lKCKZm7Nw9xHLMrKHsbGQuFJU5QeUdASq3Ulcrcv9

你可能会需要一把钥匙，钥匙就是问题的答案。

宇宙的终极答案是42，作为密钥，Rabbit解码

flag{i_love_technology}

电信诈骗

050f000a7e407151537802540b747176075178027552756d0256726262627c
你好，我是秦始皇，我并没有死，我得到了长生不老药，但是药效有副作用，现在才醒。我刚花了一年在互联网上了解现在的时代，现在我要利用我地宫第四行第七列的无敌兵马俑军队卷土重来，但是我需要启动资金，vivo50作为启动资金，待我横扫天下，封你为大将军，赏你黄金万两！

先16进制转字节

c = "050f000a7e407151537802540b747176075178027552756d0256726262627c"
cc = []
for i in bytes.fromhex(c):
    cc.append(i)
print(cc)

得到这个数组[5, 15, 0, 10, 126, 64, 113, 81, 83, 120, 2, 84, 11, 116, 113, 118, 7, 81, 120, 2, 117, 82, 117, 109, 2, 86, 114, 98, 98, 98, 124]

提示给出47和50两个数字，拿部分明文flag{进行各种运算，最后发现每个明文字符和密文字符异或50的差值都是47，即47 = m[i] - (cc[i] ^ 50)

m = b"flag{"
for i in range(5):
    print(m[i] - (cc[i] ^ 50))

所以通过m[i] = cc[i] ^ 50，就能恢复明文

但是！这样操作的话，得到的是：flag{¡ry_hursdy_vv_o}，很明显有一些字符没有打出来

有一些明文的值是超出范围的，因为我们需要的值是在33~126中

所以要对超出126的值进行处理，把每个m的值打印出来

找一下规律发现，vv此处应该是vivo或者v1v0或者viv0或者v1vo，通过这样来找寻规律，最后通过v1v0找到了规律，要把超过126的值减去94

exp:

c = "050f000a7e407151537802540b747176075178027552756d0256726262627c"
cc = []
for i in bytes.fromhex(c):
    cc.append(i)
print(cc)

flag = ''
for i in range(len(cc)):
    mm = 47 + (cc[i] ^ 50)
    if mm > 126:
        mm -= 94
        flag += chr(mm)
    else:
        flag += chr(mm)
print(flag)
#flag{Cr42y_7hursd4y_v1v0_5o!!!}

what is m

from Crypto.Util.number import bytes_to_long
from secret import flag

m = bytes_to_long(flag)
print("m =",m)

# m = 7130439814059477365344643052269091933115903919817062018703864255164601096190338334430644132726759917910603893760972905213460376105525821771466384179562302119417654102079238248438835476382333

白给了

from Crypto.Util.number import *
m = 7130439814057451234816276827985624426378225444820448794672983850703174510423869748080084641434692304059989810437801952820137016591528446065914779069356815372089481476999952029232464963781245
print(long_to_bytes(m))
#flag{Th3r3_ar3_53vEraL_a1terNAtIvEs_7o_tHe_1ON9_T0_6YteS_1UNCti0N_dCGb796e3dSB}

really_ez_rsa

from Crypto.Util.number import getPrime, bytes_to_long

e = 65537
m = b''

p = getPrime(128)
q = getPrime(128)
n = p * q
m = bytes_to_long(m)
c = pow(m, e, n)

print("p =", p)
print("q =", q)
print("c =", c)
print("e =", e)

# p = 217873395548207236847876059475581824463
# q = 185617189161086060278518214521453878483
# c = 6170206647205994850964798055359827998224330552323068751708721001188295410644
# e = 65537

p,q都给了，又白给了

exp:

from Crypto.Util.number import *
import gmpy2

p = 217873395548207236847876059475581824463
q = 185617189161086060278518214521453878483
c = 6170206647205994850964798055359827998224330552323068751708721001188295410644
e = 65537

d = gmpy2.invert(e,(p-1)*(q-1))
m = pow(c,d,p*q)
print(long_to_bytes(m))
#flag{Y0ung_meiyou_xiaojj}

🥇WEEK2

XOR

n = 20810298530643139779725379335557687960281905096107101411585220918672653323875234344540342801651123667553812866458790076971583539529404583369246005781146655852295475940942005806084842620601383912513102861245275690036363402134681262533947475193408594967684453091957401932685922178406769578067946779033282889429596341714417295489842047781388337010440309434639274398589029236213499110100040841426995862849012466514170374143655264739023758914247116354182164550612494432327931655944868705959874670536031052370968354394583880324756639698871918124498442308334232127034553164826483441746719644515097123067550594588348951855987
c = 15294238831055894095745317706739204020319929545635634316996804750424242996533741450795483290384329104330090410419090776738963732127756947425265305276394058773237118310164375814515488333015347737716139073947021972607133348357843542310589577847859875065651579863803460777883480006078771792286205582765870786584904810922437581419555823588531402681156158991972023042592179567351862630979979989132957073962160946903567157184627177910380657091234027709595863061642453096671316307805667922247180282486325569430449985678954185611299166777141304330040782500340791721548519463552822293017606441987565074653579432972931432057376
e = 65537
p⊕q = 66138689143868607947630785415331461127626263390302506173955100963855136134289233949354345883327245336547595357625259526618623795152771487180400409991587378085305813144661971099363267511657121911410275002816755637490837422852032755234403225128695875574749525003296342076268760708900752562579555935703659615570

剪枝，通过p^q的值和n求得p,q

exp:

import sys
import gmpy2
sys.setrecursionlimit(3000)

gift = 66138689143868607947630785415331461127626263390302506173955100963855136134289233949354345883327245336547595357625259526618623795152771487180400409991587378085305813144661971099363267511657121911410275002816755637490837422852032755234403225128695875574749525003296342076268760708900752562579555935703659615570
N = 20810298530643139779725379335557687960281905096107101411585220918672653323875234344540342801651123667553812866458790076971583539529404583369246005781146655852295475940942005806084842620601383912513102861245275690036363402134681262533947475193408594967684453091957401932685922178406769578067946779033282889429596341714417295489842047781388337010440309434639274398589029236213499110100040841426995862849012466514170374143655264739023758914247116354182164550612494432327931655944868705959874670536031052370968354394583880324756639698871918124498442308334232127034553164826483441746719644515097123067550594588348951855987

def findp(p,q):
    if len(p) == 1024:
        pp = int(p,2)
        if N % pp == 0:
            print("p = ",pp)
            print("q = ",N // pp)
            
    else:
        l = len(p)
        pp = int(p,2)
        qq = int(q,2)
        if (pp ^ qq) % (2 ** l) == gift %(2**l) and pp * qq %(2 ** l) == N % (2**l):
            findp('1' + p,'1' + q)
            findp('1' + p,'0' + q)
            findp('0' + p,'1' + q)
            findp('0' + p,'0' + q)
            
findp('1','1')

p =  121707024037268877853347577606022680727723421754483145354598542291389059730952536897407296725631390435306824149329251450428520017032068370905491236371588783551854860933656246846861645834009102275363106544337746297119821339365287420703692873394398852690091625453178968493127766149991473384937200789954598163517
q =  170986832479534233007906048950464510414382588164533889416767650420928742690929190093999799507883047422413122991286355305384227808800633111611572663168246588357071419165779511128259447564377245832827901688451015954867004306626552500789867499455455629032408110167560346510245108938981288797349665759162752876911
c = 15294238831055894095745317706739204020319929545635634316996804750424242996533741450795483290384329104330090410419090776738963732127756947425265305276394058773237118310164375814515488333015347737716139073947021972607133348357843542310589577847859875065651579863803460777883480006078771792286205582765870786584904810922437581419555823588531402681156158991972023042592179567351862630979979989132957073962160946903567157184627177910380657091234027709595863061642453096671316307805667922247180282486325569430449985678954185611299166777141304330040782500340791721548519463552822293017606441987565074653579432972931432057376
e = 65537
phi = (p-1)*(q-1)

d = gmpy2.invert(e,phi)
m = pow(c,d,N)
print(bytes.fromhex(hex(m)[2:]))
#flag{7428fbd7-639b-11ee-b51b-64d69af3cb76}

easymath

from Crypto.Util.number import *
from random import *
p = getPrime(128)
seed = randint(2, p - 1)

class prng:
    n = p
    a,b = [randint(2, p - 1) for _ in range(2)]
    def __init__(self,seed):
        self.state = seed
    def next(self):
        self.state = (self.state * self.a + self.b) % self.n
        return self.state

    
def main():
    gen = prng(seed)
    s = [seed]
    s.append(gen.next())
    s.append(gen.next())
    s.append(gen.next())
    s.append(gen.next())
    s.append(gen.next())
    s.append(gen.next())
    f = open("output.txt",'w')
    json.dump(s,f)
    f.close()
    flag = "flag{"+str(gen.next())+"}"
    return flag
main()

LCG，已知seed,但是不知道a,b,n

参考：LCG

exp:

from Crypto.Util.number import *
import gmpy2

output = [288530505749272642500730917886204398531, 63547143998110685331032679758907988154, 15151206512028268617888756820805603406, 268092204209244869520724955865278855216, 261067075335188593563542448889694952077, 138067838531633886698552659065694918861, 201319433320428898153580935653793106657]
t = []
for i in range(1,len(output)):
    t.append(output[i]-output[i-1])

T = []
for i in range(1,len(t)-1):
    T.append(t[i+1]*t[i-1] - t[i]**2)

m = []
for i in range(len(T)-1):
    mm = gmpy2.gcd(T[i],T[i+1])
    if isPrime(mm):
        m.append(int(mm))
    else:
        for i in range(1,100):
            if isPrime(mm // i):
                mm = mm // i
                m.append(int(mm))
                break
print(m)

for i in m:
    if isPrime(i):
        a = gmpy2.invert(t[0],i) * t[1] % i
        b = output[1] - a*output[0] % i
        a_ = gmpy2.invert(a,i)

        seed = output[0]
        print("seed =",seed)
        for j in range(7):
            seed = (a * seed + b) % i
        flag = "flag{"+str(seed)+"}"
        print(flag)
#flag{302184756857257140159769321021979097116}

ez_rsa

# from flag import flag
from Crypto.Util.number import getPrime
from math import prod
import libnum

with open("flag.txt","rb") as f:
    flag = f.read().strip()

m = int.from_bytes(flag[:19],"big")
m1 = int.from_bytes(flag[19:],"big")

e = 65537

primes = [getPrime(64) for i in range(32)]
n = prod(primes)
c = pow(m1,e,n)
print("c =",c)
print("n =",n)

p = libnum.generate_prime(1024)
q = libnum.generate_prime(1024)
e1 = 13
e2 = 15
n1 = p * q
c1 = pow(m1, e1, n1)
c2 = pow(m1, e2, n1)
print("n1 =", n1)
print("e1 =", e1)
print("c1 =", c1)
print("n2 =", n1)
print("e2 =", e2)
print("c2 =", c2)

# c = 28535916699190273475273097091422420145718978597126134891571109006456944397344856577421369324831702083810238921719657496747722337086131545474384253288151783029981352196506749672783866527948391034258269669654392993063423671431837882584570973320095601407578443348352802850496429240170710269529489900871208384711844617081275862971410246759090936379744946527813691945129059991795202769186014306943707223831130752782380563227353615164053563120572722464543812139164048342504963081408349934180883607554389607335607410546630525512019818062185681153477671373000186961748278118124044645584490544698827467815360888525822167
# n = 114107341297408283801468814470303963122122556489590451040619457052827864984505912502462030175984161431709841571908269123131659496812467145870607611968843929870716066046232009282431653653484798819370087696248364531531706249180822839879862098012984590503284615395588919199545142177727328844260380842155437987767067800740569616584597507776426572206990858918111272636507821551592564540694671795374831548677720629329358177802890287837056940407030212276399942462042866947423728888561392653713356355778914658317507319575084393752755452971007289968044006561357799908892371839922838486713582082980752194204224263283004373
# n1 = 21235204662158833223664424963408105101885570855652885953922511758363954474947609854216589644512813634294435585894296340005122907229365513346971631594453999584706013889403572150499529308966742992668850443386284277210686717652643585324255759216699733045642544284406720854291604837774882256435503827543483289606177965628162259184958789025311291796067574924595051311298594432767265114154138693108465671184854794167878031822162731921299518989845784744659944947091213703810190708463199067553747177712259911724424547999547534441790125049383068377243727588278432796727885216967953646999183906479537750330738956233695342750567
# e1 = 13
# c1 = 5640630966585093229374938575158853304507369792931959909038819773057666482368490365383634362421839045569190487785222799103423460816096797210546343809620912249021763787314569982909943181390882015170344954037813745251119237402775124991005154299085147091159741067430623420349690886728161235034687649593258746455165172528681627568611599473627285223154284756417744280966157271904828156564067870877521824545300153084830020169048653830385763172792698591998191641849931039720453035065355411394516308865955772746815765864888631258825704788352584540380169938419618543124830541663995097651872542381
# n2 = 21235204662158833223664424963408105101885570855652885953922511758363954474947609854216589644512813634294435585894296340005122907229365513346971631594453999584706013889403572150499529308966742992668850443386284277210686717652643585324255759216699733045642544284406720854291604837774882256435503827543483289606177965628162259184958789025311291796067574924595051311298594432767265114154138693108465671184854794167878031822162731921299518989845784744659944947091213703810190708463199067553747177712259911724424547999547534441790125049383068377243727588278432796727885216967953646999183906479537750330738956233695342750567
# e2 = 15
# c2 = 5481001445755770090420425478456880914921441486935672376394423326451811448703288166341447356603281843336826624725965666634194700496514262129376916108926167953996689011980280761368893884042609095616407660087448963015169181749124738976578495911295096014725354350167650232970262765851074146687931181216305972147994236689422572940877763047930111954798962097847426932730342258169023809341164876019161104439561164839132092594444017039073155506935768658830659965630065643619399324102814118128802834719820426253836317043818687888302054465994498115387703382090351794495827905499417861507007863378916334790750453883661675063377

part1

yafu分解得到32各因子求φ（n）

part2

共模攻击

exp:

import gmpy2

n1 = 114107341297408283801468814470303963122122556489590451040619457052827864984505912502462030175984161431709841571908269123131659496812467145870607611968843929870716066046232009282431653653484798819370087696248364531531706249180822839879862098012984590503284615395588919199545142177727328844260380842155437987767067800740569616584597507776426572206990858918111272636507821551592564540694671795374831548677720629329358177802890287837056940407030212276399942462042866947423728888561392653713356355778914658317507319575084393752755452971007289968044006561357799908892371839922838486713582082980752194204224263283004373
c = 28535916699190273475273097091422420145718978597126134891571109006456944397344856577421369324831702083810238921719657496747722337086131545474384253288151783029981352196506749672783866527948391034258269669654392993063423671431837882584570973320095601407578443348352802850496429240170710269529489900871208384711844617081275862971410246759090936379744946527813691945129059991795202769186014306943707223831130752782380563227353615164053563120572722464543812139164048342504963081408349934180883607554389607335607410546630525512019818062185681153477671373000186961748278118124044645584490544698827467815360888525822167
primes = [10436802938040427139,12712357180113548549,11502613740816749197,10270880245559150279,13498192768855092449,17075632607344331131,10100522426677320149,10635881647150245973,15622487550947237203,11777892065426651999,10040612110882504553,13428970346605599557,9261040693807289549,16651625235320957803,10126802520926958821,14198042938738648387,16946507762934111301,9723861249937499279,15774106340553595249,16340211116882594287,13618885037077024279,9356350172425710359,11718181938374860349,16880270107514803247,9281508366366115669,15616762946597906161,10986943768724409089,17088379813205887661,11964584391817142269,14147604789494386003,10621161426185076191,10203735303764112277]
phi = 1
for i in primes:
    phi *= i-1 

d = gmpy2.invert(65537,phi)
m = pow(c,d,n1)
flag1 = bytes.fromhex(hex(m)[2:])

n2 = 21235204662158833223664424963408105101885570855652885953922511758363954474947609854216589644512813634294435585894296340005122907229365513346971631594453999584706013889403572150499529308966742992668850443386284277210686717652643585324255759216699733045642544284406720854291604837774882256435503827543483289606177965628162259184958789025311291796067574924595051311298594432767265114154138693108465671184854794167878031822162731921299518989845784744659944947091213703810190708463199067553747177712259911724424547999547534441790125049383068377243727588278432796727885216967953646999183906479537750330738956233695342750567
e1 = 13
c1 = 5640630966585093229374938575158853304507369792931959909038819773057666482368490365383634362421839045569190487785222799103423460816096797210546343809620912249021763787314569982909943181390882015170344954037813745251119237402775124991005154299085147091159741067430623420349690886728161235034687649593258746455165172528681627568611599473627285223154284756417744280966157271904828156564067870877521824545300153084830020169048653830385763172792698591998191641849931039720453035065355411394516308865955772746815765864888631258825704788352584540380169938419618543124830541663995097651872542381
e2 = 15
c2 = 5481001445755770090420425478456880914921441486935672376394423326451811448703288166341447356603281843336826624725965666634194700496514262129376916108926167953996689011980280761368893884042609095616407660087448963015169181749124738976578495911295096014725354350167650232970262765851074146687931181216305972147994236689422572940877763047930111954798962097847426932730342258169023809341164876019161104439561164839132092594444017039073155506935768658830659965630065643619399324102814118128802834719820426253836317043818687888302054465994498115387703382090351794495827905499417861507007863378916334790750453883661675063377
s,x,y = gmpy2.gcdext(e1,e2)
m = (pow(c1,x,n2)*pow(c2,y,n2))%n2
flag2 = bytes.fromhex((hex(m)[2:]))
flag = flag1 + flag2
print(flag)
#flag{05929ec9778ed739d94ee1a77b742714}

e？

p= 70724362259337647663584082414795381346569735601816096923682814277857463878289
q= 114427188167532721707398034034072867253267857672869034942206947096293901917007
e= 1314
c= 4308122681135507736058122041934864039713319497673888928736468819190185301630702240416683093700232966794026900978699666246019059398861283337865339404916304

e和φ(n)不互素

exp:

import gmpy2
p= 70724362259337647663584082414795381346569735601816096923682814277857463878289
q= 114427188167532721707398034034072867253267857672869034942206947096293901917007
e= 1314
c= 4308122681135507736058122041934864039713319497673888928736468819190185301630702240416683093700232966794026900978699666246019059398861283337865339404916304
n = p * q
phi = (p-1)*(q-1)
t = gmpy2.gcd(e,phi)
d = gmpy2.invert(e // t,phi)
M = pow(c,d,n)
m = gmpy2.iroot(M,t)[0]
print(bytes.fromhex((hex(m)[2:])))
#flag{This_e_is_real_or_not}

factorizing_n

n = 226515252384227990547287743140613580056836242860947832749754689048997071950972581790210817523352001702907675581567498443649554801433663166425134375454937126656357069687274036935331269594383360450823787099121079436459236734336130768046337169817940540921822023269188752420603975467384377614321048859304185067329741055517464271746238143742661897809442359331215501438861121047081117632626097939097519866099140569819965948998542652908170134545593659233229897003698175558888336706474178958535138595687148003367152624421106553412886263257022809480187410133186189435436294593588009551451899398811758511878324326255293307347560753524372663257044426744744426759970254203341706284024734042826158828749144322843934985927079504722440497388146240627249465363931951790326885478025237643
c = 52409805591744226507807531465616894934028463651864630447934395956954575834603756391651746535033902964658694070544877880970130028487381287088425209448038533705903737694267359561133766799228825599943891152463160326583722749586721691729062524310148743637505134465210906856660867852927837112666513674858029892207902196213784902541173835447263733760225682942461048573387925463479672527491229113710629340960375692432470493054415657845868577650170648157402682163577152288432313996310562452677399267755695644659367792066311336521698894993982901657735586844358679888210537898629281625526455444811591386493005341435516094660429968084363084301878446471676122069724608083578102382181382107225473535696274374370868301830807644939881080301668756603163431000745972823980427048672732291
e = 65537

n能分解成p的5次方

exp:

from Crypto.Util.number import *
import gmpy2

n = 226515252384227990547287743140613580056836242860947832749754689048997071950972581790210817523352001702907675581567498443649554801433663166425134375454937126656357069687274036935331269594383360450823787099121079436459236734336130768046337169817940540921822023269188752420603975467384377614321048859304185067329741055517464271746238143742661897809442359331215501438861121047081117632626097939097519866099140569819965948998542652908170134545593659233229897003698175558888336706474178958535138595687148003367152624421106553412886263257022809480187410133186189435436294593588009551451899398811758511878324326255293307347560753524372663257044426744744426759970254203341706284024734042826158828749144322843934985927079504722440497388146240627249465363931951790326885478025237643
c = 52409805591744226507807531465616894934028463651864630447934395956954575834603756391651746535033902964658694070544877880970130028487381287088425209448038533705903737694267359561133766799228825599943891152463160326583722749586721691729062524310148743637505134465210906856660867852927837112666513674858029892207902196213784902541173835447263733760225682942461048573387925463479672527491229113710629340960375692432470493054415657845868577650170648157402682163577152288432313996310562452677399267755695644659367792066311336521698894993982901657735586844358679888210537898629281625526455444811591386493005341435516094660429968084363084301878446471676122069724608083578102382181382107225473535696274374370868301830807644939881080301668756603163431000745972823980427048672732291
e = 65537

p = 11776588228599764849559519654482976956833367474471407292255776713760090338489966385328569279135095351660161277221351884258247731394014018172166064062551483
phi = p**4 * (p-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))
#flag{1f95f530f85b940db810fc917607ee22}

哈希猫

import hashlib
from secret import flag

assert flag[:5] == "flag{"
assert flag[-1:] == "}"
flag = flag[5:-1]
assert len(flag) == 43


print(hashlib.md5(flag[0:3].encode()).hexdigest())
print(hashlib.sha256(flag[3:5].encode()).hexdigest())
print(hashlib.sha512(flag[5:7].encode()).hexdigest())
print(hashlib.sha256(flag[7:9].encode()).hexdigest())
print(hashlib.sha256(flag[9:11].encode()).hexdigest())
print(hashlib.sha224(flag[11:13].encode()).hexdigest())
print(hashlib.sha224(flag[13:15].encode()).hexdigest())
print(hashlib.sha224(flag[15:17].encode()).hexdigest())
print(hashlib.sha1(flag[17:20].encode()).hexdigest())
print(hashlib.sha1(flag[20:23].encode()).hexdigest())
print(hashlib.sha512(flag[23:25].encode()).hexdigest())
print(hashlib.sha1(flag[25:28].encode()).hexdigest())
print(hashlib.sha256(flag[28:30].encode()).hexdigest())
print(hashlib.md5(flag[30:33].encode()).hexdigest())
print(hashlib.sha256(flag[33:35].encode()).hexdigest())
print(hashlib.sha1(flag[35:38].encode()).hexdigest())
print(hashlib.sha224(flag[38:40].encode()).hexdigest())
print(hashlib.md5(flag[40:43].encode()).hexdigest())


# bba78e7d3d6ea696bfef6e87a26a6c03
# 894461aba3e5e2724278cfc40fa15dc1510b0cd56430fec9fc8a90d6d8ebe5f7
# 092ca0da5f25c595eae6cc725dc12ae75889a639d205733223cb9cfaeaa702a521971e57f768079b03d53c8f092bb2c597cc3adb4a98224a144c5d62461beb52
# 122c597083bd438b7f6d72af75d025948899647711b806bdd2cd82fa69713db3
# 153812ae5fea0b73a011bf28bd7cea93644437c3fe3260b7b2d7e1e2f9f46bde
# 24296a0d1555ad424b04dd5f6268c5613dae32cde8a469ca8c9065b9
# 15d46f8eae648f5945c4605b15e0507836ee1251b74de2e95bf9394d
# 9afdf5d1027172492a32833975a69a1e501e6f824d3e5a7121ad18af
# 2b7655eb53f6f92086e58da465c6733bd8d8a1a3
# a44fce0412ae28173cf036700c3f1cc8c7a279c6
# 7356ea99a4fa12c6a3ed8ecc1e78a7bb28cdc1208962848646cf34e0d836f420cf594606cde74bec1dc2efe511dff0dd07b558bd0c7d01a526de2e9c25d47a6f
# cc5c3fe6e7356a26a134cff5633349f597c40a9d
# 757d22f85733770bae59cc857dcfc369db29031459421b808377cf06bc9881ad
# a2c7bdf97e3c761c431f8891dcc39fa9
# 0606c0e8f78de4d7a78522a5ae2f7b70e31e669e6e2a16e11c393e4443069813
# cb6c6c7cc8189868e1feaae773f947908233cdcc
# 2cb5169af85bd3122276f8b0a731ed2ea30c8bef366bfafe76b5e7db
# d1de3eac41792a7f498dae8f439bf16d

一堆hash

exp:

import hashlib
from itertools import product

c = ["bba78e7d3d6ea696bfef6e87a26a6c03",
"894461aba3e5e2724278cfc40fa15dc1510b0cd56430fec9fc8a90d6d8ebe5f7",
"092ca0da5f25c595eae6cc725dc12ae75889a639d205733223cb9cfaeaa702a521971e57f768079b03d53c8f092bb2c597cc3adb4a98224a144c5d62461beb52",
"122c597083bd438b7f6d72af75d025948899647711b806bdd2cd82fa69713db3",
"153812ae5fea0b73a011bf28bd7cea93644437c3fe3260b7b2d7e1e2f9f46bde",
"24296a0d1555ad424b04dd5f6268c5613dae32cde8a469ca8c9065b9",
"15d46f8eae648f5945c4605b15e0507836ee1251b74de2e95bf9394d",
"9afdf5d1027172492a32833975a69a1e501e6f824d3e5a7121ad18af",
"2b7655eb53f6f92086e58da465c6733bd8d8a1a3",
"a44fce0412ae28173cf036700c3f1cc8c7a279c6",
"7356ea99a4fa12c6a3ed8ecc1e78a7bb28cdc1208962848646cf34e0d836f420cf594606cde74bec1dc2efe511dff0dd07b558bd0c7d01a526de2e9c25d47a6f",
"cc5c3fe6e7356a26a134cff5633349f597c40a9d",
"757d22f85733770bae59cc857dcfc369db29031459421b808377cf06bc9881ad",
"a2c7bdf97e3c761c431f8891dcc39fa9",
"0606c0e8f78de4d7a78522a5ae2f7b70e31e669e6e2a16e11c393e4443069813",
"cb6c6c7cc8189868e1feaae773f947908233cdcc",
"2cb5169af85bd3122276f8b0a731ed2ea30c8bef366bfafe76b5e7db",
"d1de3eac41792a7f498dae8f439bf16d"]

flag = "flag{"
table = [chr(i) for i in range(32,128)]

for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.md5(m.encode()).hexdigest()
    if hash == c[0]:
        flag += m
print(flag)
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha256(m.encode()).hexdigest()
    if hash == c[1]:
        flag += m
print(flag)     
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha512(m.encode()).hexdigest()
    if hash == c[2]:
        flag += m
print(flag)    
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha256(m.encode()).hexdigest()
    if hash == c[3]:
        flag += m
print(flag)
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha256(m.encode()).hexdigest()
    if hash == c[4]:
        flag += m
print(flag)        
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha224(m.encode()).hexdigest()
    if hash == c[5]:
        flag += m
print(flag)
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha224(m.encode()).hexdigest()
    if hash == c[6]:
        flag += m
print(flag)
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha224(m.encode()).hexdigest()
    if hash == c[7]:
        flag += m
print(flag)     
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.sha1(m.encode()).hexdigest()
    if hash == c[8]:
        flag += m
print(flag)        
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.sha1(m.encode()).hexdigest()
    if hash == c[9]:
        flag += m
print(flag)      
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha512(m.encode()).hexdigest()
    if hash == c[10]:
        flag += m
print(flag)       
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.sha1(m.encode()).hexdigest()
    if hash == c[11]:
        flag += m
print(flag)       
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha256(m.encode()).hexdigest()
    if hash == c[12]:
        flag += m
print(flag)              
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.md5(m.encode()).hexdigest()
    if hash == c[13]:
        flag += m
print(flag)       
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha256(m.encode()).hexdigest()
    if hash == c[14]:
        flag += m
print(flag)
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.sha1(m.encode()).hexdigest()
    if hash == c[15]:
        flag += m
print(flag)      
for i in product(table,repeat=2):
    m = "".join(i)
    hash = hashlib.sha224(m.encode()).hexdigest()
    if hash == c[16]:
        flag += m
print(flag)       
for i in product(table,repeat=3):
    m = "".join(i)
    hash = hashlib.md5(m.encode()).hexdigest()
    if hash == c[17]:
        flag += m
flag += "}"
print(flag)
#flag{YOU'vE_c0m3_To_UNDerS7anD_Ha5H_GDIbgdF3EocT}

🥇WEEK3

Classical Master

import math
from secret import s
s = s.lower()
keyM = [?]
l = len(keyM)
assert(math.gcd(l,26)==1)
for i in range(len(s)):
    print(chr((ord(s[i])*l-97+(keyM[i % l]))%26+97),end="")

#

exp:

import collections
import string
max_lenth = 20
enc = ''
def group_text(ciphertext, key_length):
    # 根据密钥长度将密文分组
    groups = ['' for _ in range(key_length)]
    for i, char in enumerate(ciphertext):
        groups[i % key_length] += char
    return groups
def frequency_analysis(text):
    # 统计字母出现频率
    freq_counter = collections.Counter(text)
    total_chars = len(text)
    return {char: freq / total_chars for char, freq in freq_counter.items()}
def guess_key_length(ciphertext):
    max_key_length = min(len(ciphertext), max_lenth)  # 假设密钥最大长度为20
    best_key_length = 1
    best_index_of_coincidence = 0.0
    for key_length in range(1, max_key_length + 1):
        groups = group_text(ciphertext, key_length)
        index_of_coincidence_sum = 0.0
        for group in groups:
            freq_analysis = frequency_analysis(group)
            index_of_coincidence_sum += sum(freq ** 2 for freq in freq_analysis.values())
        average_index_of_coincidence = index_of_coincidence_sum / key_length
        if average_index_of_coincidence > best_index_of_coincidence:
            best_index_of_coincidence = average_index_of_coincidence
            best_key_length = key_length
    return best_key_length
key_length = guess_key_length(enc)
print(key_length)
# plain to cipher
ptoc = [["" for i in range(26)] for j in range(26)]
for i in range(26):
    for key in range(26):
        ptoc[i][key] = chr((ord(string.ascii_lowercase[i]) * key_length - 97 + (key)) % 26 + 97)
ctop = [["" for i in range(26)] for j in range(26)]
for i in range(26):
    for key in range(26):
        temp1 = ord(string.ascii_lowercase[i]) - 97
        while (1):
            if ((temp1 - key + 97) % key_length == 0):
                chr1 = (temp1 - key + 97) // key_length
                if (chr1 >= 97 and chr1 <= 122):
                    ctop[i][key] = chr(chr1)
                    break
            temp1 += 26
list_usual = [0.082, 0.015, 0.028, 0.043, 0.127, 0.022, 0.020, 0.061, 0.070, 0.002, 0.008, 0.040, 0.024, 0.067, 0.075, 0.019, 0.001, 0.060, 0.063, 0.091, 0.028, 0.010, 0.023, 0.001, 0.020, 0.001]
def guess_key(ciphertext, key_length, list_usual):
    groups = group_text(ciphertext, key_length)
    key = []
    for group in groups:
        freq_analysis = frequency_analysis(group)
        # 计算每个密钥字母的偏移量
        max_correlation = 0
        best_shift = 0
        for shift in range(26):
            correlation = sum(freq_analysis.get(letter, 0) * list_usual[string.ascii_lowercase.index(ctop[idx][shift])] for idx, letter in enumerate(string.ascii_lowercase))
            if correlation > max_correlation:
                max_correlation = correlation
                best_shift = shift
        key.append(best_shift)
    return key
key = guess_key(enc,key_length,list_usual)
print(key)
c = ''
for i in range(len(enc)):
    temp1 = ord(enc[i]) - 97
    tkey = key[i%key_length]
    while (1):
        if ((temp1 - tkey + 97) % key_length == 0):
            chr1 = (temp1 - tkey + 97) // key_length
            if (chr1 >= 97 and chr1 <= 122):
                c += chr(chr1)
                break
        temp1 += 26
print(c)
#flag{youaretherealmaster}

e=3

from Crypto.Util.number import *
import random
from secret import flag

M = 2**54
k = 6

def gen_prime(M, k):
    while True:
        prime = sum([random.getrandbits(16) * M**i for i in range(k)])
        if isPrime(prime) and (prime-1) % 3 == 0:
            return prime
        
p, q, r = [gen_prime(M, k) for i in range(3)]
N = p * q * r
e = 3
m = bytes_to_long(flag)
c = pow(m, e, N)
print(f'N = {N}')
print(f'c = {c}')

"""
N = 3298593732762513945346583663585189774036688951059270517149719979434109398447628726951796006700754759352430339647168415338320547665794785951232342902233013221132246450312038122695046634624323814318286314664160113738299465643128504110932989263063331290006313
c = 869489491924953293290699796392271834401780578884556874640489836779925847562085802848542382525324081900560761299059365684697233025590164192409062717942292142906458498707677300694595072310705415037345581289469698221468377159605973403471463296806900975548438
"""

p,q,r都可以表示为M进制，于是可以用多项式来分解N，因为e和phi不互素，所以用有限域开根，再用中国剩余定理

exp:

from Crypto.Util.number import *
import gmpy2

N = 3298593732762513945346583663585189774036688951059270517149719979434109398447628726951796006700754759352430339647168415338320547665794785951232342902233013221132246450312038122695046634624323814318286314664160113738299465643128504110932989263063331290006313
c = 869489491924953293290699796392271834401780578884556874640489836779925847562085802848542382525324081900560761299059365684697233025590164192409062717942292142906458498707677300694595072310705415037345581289469698221468377159605973403471463296806900975548438
M = 2^54
poly = sum(e * x^i for i,e in enumerate(Integer(N).digits(M)))
# poly = 483094776500*x^15 + 67737188732830*x^14 + 124085394790974*x^13 + 195683246513428*x^12 + 273742579935809*x^11 + 458380068952801*x^10 + 592623629284873*x^9 + 582132325688443*x^8 + 685453135920467*x^7 + 720971197330040*x^6 + 597860219518539*x^5 + 434712532854786*x^4 + 332870394840336*x^3 + 251268756615643*x^2 + 110730229142998*x + 17535693800233
(p,_),(q,_),(r,_) = poly.factor_list()

p, q ,r = p(x=M),q(x=M),r(x=M)
# print(p)
# print(q)
# print(r)
# 103574226729554375480512668967949133854292403117507474988278388756193462602107352821951
# 47963432552002818180880760250824590058982930733941748241661938238195705638187268342813
# 663998156522471100999941798165706402858681862228017448075268472245282758965006970051

e = 3
R.<x> = Zmod(p)[]
f = x^e-c
f = f.monic()
res1 = f.roots()

R.<x> = Zmod(q)[]
f = x^e-c
f = f.monic()
res2 = f.roots()

R.<x> = Zmod(r)[]
f = x^e-c
f = f.monic()
res3 = f.roots()

for i in res1:
    for j in res2:
        for k in res3:
            m = crt([int(i[0]),int(j[0]),int(k[0])],[int(p),int(q),int(r)])
            flag = long_to_bytes(int(m))
            if b"flag" in flag:
                print(flag)
                break
# flag{e1b7d2c2-e265-11eb-b693-98fa9b5bc5fe}

撤退！

from Crypto.Util.number import *

flag = *******
p = getPrime(1024)
q = getPrime(1024)
n = p * q

hb = len(flag)//2
hb1 = bytes_to_long(flag[:hb])
hb2 = bytes_to_long(flag[hb:])
D = 117
x = *******
y = *******
assert x**2 - D * y**2 == 1
enc1 = pow(334 * n ** 2 + 1, hb1, n ** 3)
enc2 = pow(y * n + 1, hb2, n ** 3)
print(n)
print(enc1)
print(enc2)

'''
22970461944771505344360312103272646796516672838005008112295760406393062653512719537671401409823031480497512491850701737384621917068068328814717390355072928714618936469722031401433712342846780800586803218279291870162605299119904016959036663767093191710796830156169925350938267584422752300171293262391805105435418210827517225439398971437884496416502510866914857269951072184669675339439115587325754431761172634305242650221404868035624879538862880516438147301289746375407945908866907822940285764276956194031840381838253923392794376568293056359058519233175242523219646628321609305890926063856400793680641992567798104042179
26380574883568223071748995929433720836641856899148821439556557592284999544802260386919172895274884666117488851000353221957579311943624258651646692068406462392980585841604755021251430357273933800209194484692955106014890051223465745443628784077844452303995642424661442978294757109040081050794398646640530714904683097650259060507908334791761124660725589404056356987726993518057050112725483482660202442987346646160168856264312557595890710521723067518303906942469282527855551751244126251698491010628369012024332666619702895796133780038584346428759785302542637171018926824843416176876077558936427399803328577151066597396550597352625005028261156114571696860700477410270949916316951150072218466374341394892405947793726872954497972795793421222424616005278493704125169150432275472846871295341469911428057621028515874978272004775903906188556908968810828510069826724631700523623584802605889173266453916347583720706846630531082266742377818663000322817114065116737931523412220137972079139507877669106470150742546914051556747087768279286696519700220233815812834114117581332234344024169109786527295900675653245014343393093832478814567179131966404207553408747774003319241150221488231674711614902743345516888975702483348011349617017294004761259419165663633915672647187482242462163420462987034240805524991
21190674872507845600786632640969893237129139877891071648594239906632201421611954626926407751780936578853046780585253060958265549804784845192757301417173404074965693840282568701968464564320290763073618132775799910356101999797720378313304899173154753858674284071499775857913937184713024788245068426198878834805943703426673512761178072458895973672088230653246356764681418231485563287856188079274727706554037799748595877069143254516390328019381867648697880975670688337068196993846986940286056873616919629721264139576692806770826129279380704466982862393203486037890448173834315360975464927583664991534571518159777852793416869350127023692816051992183670690315184731534611966603509867722931839839084915943647295195314171688904055674915382434841320612108023531722571519492067471405656160804893645713608592561788743509876384862097871840094582513721456962354498561006793609200187065931433827465455037397503619844768415369973322759940610358415184510344945559838007474725413347675347453443583610217539704055467297318282309867987435252614428856515259899385689971172417660178761139941056839133998928898528744331662995956041897599276732929020537698559927654297185422925737241274711904687894411308774527520523946951208805307060323875839353707549772052299847176824964552693112658495961070555882583739017417359463576705453026824255338859618053086622031941
'''

exp:

#sage
n = 
enc1 = 
enc2 = 

numTry = 1500
def solve_pell(N, numTry):
    cf = continued_fraction(sqrt(N))
    for i in range(numTry):
        denom = cf.denominator(i)
        numer = cf.numerator(i)
        if numer^2 - N * denom^2 == 1:
            return numer, denom
    return None, None

x,y = solve_pell(N,numTry)
# print(x,y)
# x = 649
# y = 60

var('m')
f = 334*m*n^2 + 1 == enc1
ans = solve(f,m)
# print(ans)
m1 = 149691910197777805350862530703771372803641869951585
flag1 = long_to_bytes(int(m1))

c1 = (enc2-1) // n
var('m')
f = m*y + (m*(m-1)/2) * y^2*n - c1 
ans = solve(f,m)
# print(ans)

m2 = 149371042625025154522769720206540986718252215526781
flag2 = long_to_bytes(int(m2))
print(flag1+flag2)
#flag{6c6eb27a-061b-baf4-4cae26-5a609588ce}

好好好！

***EGK*MAPZ**3TISLXYHW*B4*R*6CQV




e=65537
dp=89183181604123417010894108474901628410408206538085478807758137668201829058797702838603939730356798163745078443656032825128645105954284119126609502005130005399692420386460970318283171848176434285488698019425286328269756591931253074416895028845240978211030365697435579850343911269163064228581083838914477473793
n=17133884272385326910236146208723169235592379139078245324256146697759098524213354087333170410075813764497353656874360657828668202585141557095326829141561993608634568037533128091918704136052835609732443342167341276983343070200953604216445186924411131823487594273213380078485528148801722039459601896275130691200206027353715109606722659553700867073796386669768748305283547862565020499794358571741903375812063001390288166187510171105241363677243530996160649133253643422391688399573703498726489248479978887237752214015456924632092625018668632234215462091314384917176427670194819828555385014264912614752917792278216214856001
c=7297673446200396117470312266735704951424121735299327785232249350567349180167473433806232931862684106388722088953786183522191592452252650217579986150373463901393038386627370305688040315665037164819432754099421229466379901436696822022518438390977543864543590936753547325597766614648063328562516667604171990354928485383191174966274941678597887943784661684719053108281896697098991347034225406718530599672101743303723470910913422462764406680309933367328977341637394665138995676573466380198978810546689819954949832833954061771415463198737542769848298258925680570823701939997224167603657418270886620562332895947413332492672

dp泄露求解得到base32的密文7U25DUJJ7USYATEN5SREOFFG5NY57FPS77U5DFPY54JEG3NYKWSYA3YD5CXYTTNW53QS====，然后爆破base32的表完成求解

exp:

from Crypto.Util.number import long_to_bytes
import gmpy2

e=65537
dp=
n=
c=

def dp_leak(dp,c,n,e):
    for i in range(1,e):
        t = (dp * e - 1) % i        
        if t == 0:
            p = (dp * e - 1) // i + 1
            if n % p == 0:
                q = n // p
                d = gmpy2.invert(e,(p-1)*(q-1))
                print(long_to_bytes(pow(c,d,n)))
                
dp_leak(dp,c,n,e)
#7U25DUJJ7USYATEN5SREOFFG5NY57FPS77U5DFPY54JEG3NYKWSYA3YD5CXYTTNW53QS====

爆破base32的表

import base64

cipher = "7U25DUJJ7USYATEN5SREOFFG5NY57FPS77U5DFPY54JEG3NYKWSYA3YD5CXYTTNW53QS===="

table1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ234567"    #这是base32的表
table2 = "EGKMAPZ3TISLXYHWB4R6CQV"


list = []
for i in table1:
    if not (i in table2):
        list.append(i)

print(list)

for i in list:
    for j in list:
        if i != j:
            for k in list:
                if i != k and j != k:
                    for n in list:
                        if i != n and j != n and k != n:
                            for t in list:
                                if i != t and j != t and k != t and n != t:
                                    for tt in list:
                                        if i != tt and j != tt and k != tt and n != tt and t !=tt:
                                            for ttt in list:
                                                if i != ttt and j != ttt and k != ttt and n != ttt and t !=ttt and tt != ttt:
                                                    for tttt in list:
                                                        if i != tttt and j != tttt and k != tttt and n != tttt and t !=tttt and tt != tttt and ttt != tttt:
                                                            for ttttt in list:
                                                                if i != ttttt and j != ttttt and k != ttttt and n != ttttt and t !=ttttt and tt != ttttt and ttt != ttttt and tttt != ttttt:
                                                                    pad1 = '' + i + j + k
                                                                    table2 = pad1 + "EGK" + n + "MAPZ" + t + tt +"3TISLXYHW" + ttt + "B4" + tttt + "R" + ttttt + "6CQV"
                                                                    flag = base64.b32decode(cipher.translate(str.maketrans(table2,table1)))
                                                                    if b"flag" in flag:
                                                                        print(flag)

#flag{fa48a440-d0ff-0c2a-366243-a46b7e7853}

easyrsa

from Crypto.Util.number import *
from flag import flag

p1 = getPrime(512)
q1 = getPrime(512)
n1 = p1 * q1

e = 65537

p2 = getPrime(1024)
q2 = getPrime(1024)
n2 = p2 * q2

leak1 = (p2+q2) >> 400
leak2 = (p1 & ((1 << 350) - 1)) >> 5

enc = pow(leak2,e,n2)
c = pow(bytes_to_long(flag),e,n1)
f = open(f'output.txt','w')
f.write(f'n1 = {n1}\n')
f.write(f'n2 = {n2}\n')
f.write(f'leak1 = {leak1}\n')
f.write(f'enc = {enc}\n')
f.write(f'c = {c}')
f.close()

通过leak1来求leak2，leak2即是p1的低350位（但是被抹去了低5位，后面要爆破一下）

参考文章：2023 贵阳大数据安全精英赛 --- Crypto childrsa wp

联立leak1 = (p2 + q2) >> 400和n2 = p2 * q2求解的结果和p2的高位一样，只是低400位不一样

有了p2的高位后，利用coppersmith进行求解p2

exp:

n2 = 
leak1 = 
enc = 

RF = RealField(2048)
X = polygen(RF)
f = X * ((leak1 << 400)-X) - n2

# 求p2
sol = int(f.roots()[0][0])
p_high = (sol >> 430) << 430
PR.<x> = PolynomialRing(Zmod(n2))
f1 = x + p_high
x0 =f1.small_roots(X=2^430, beta=0.4)[0]
p2 = p_high + x0
print(f"p2 = {p2}")

# 求leak2
q2 =N // int(p2)
print(f"q2 = {q2}")
phi = (p2-1)*(q2-1)
d = gmpy2.invert(e,gmpy2.mpz(phi))
leak2 = pow(enc,d,n2)
print(f"leak2 = {leak2}")
# leak2 = 17360313144717510037537355688628643303848718125498770246433419674664931177802009718523022408554028650268575473276776652325042704063954902971874624580258429965610657615998701297143056706693836430020853673136823114239697815838085813271681130436537239288366997343978169099703086258100321964287523989425007035962983601510451873274120392974576079344343289173636686941721922540549894873583626722606223542204019760103801900148734981271526662512960432004374861666967746492254621730914184468921861371761515561318710135049915102041347748291394096585038509541586487535668372772139517571028910559303129731751631963650606704284688

再通过leak2求解flag

exp:

from Crypto.Util.number import *
import gmpy2

e = 65537
n = 
c = 
leak2 = 22334810767801800995021872014176778873829048161801414909315794486047873481911273730826962574216771288781
leak2 = leak2 << 5
for i in range(2^5):
    plow  = leak2 + i
    R.<x> = PolynomialRing(Zmod(n))
    f = x * 2^350 + plow
    f = f.monic()
    roots = f.small_roots(X = 2^162,beta=0.4)
    if roots:
        p = int(roots[0]) * 2^350 + plow
        print(p)
        q = n // p
        d = gmpy2.invert(e,(int(p)-1)*(int(q)-1))
        m = pow(c,d,n)
        print(long_to_bytes(int(m)))
        break
        # flag{9995eae8acaac286c7b72e50e5258dc3}

ECC

from Crypto.Util.number import bytes_to_long ,getPrime
from random import randint
from secret import flag , p, A, B
class LCG:
    def __init__(self, seed, multiplier, increment, modulus):
        self.state = seed
        self.multiplier = multiplier
        self.increment = increment
        self.modulus = modulus

    def round(self):
        self.state = (self.state * self.multiplier + self.increment) % self.modulus
        return self.state

LcG = LCG(p, A, B, getPrime(300))
hint = []
rounds = randint(9,999)
for i in range(rounds):
    hint.append(LcG.round())

print('N =', LcG.modulus)
print('hint =', hint[rounds-3:])
print('rounds =',rounds)

m = bytes_to_long(flag)
E = EllipticCurve(GF(p),[A,B])
P = E.random_point() 
Q = m*P

print ('P:',P)
print ('Q:',Q)


# N = 1394352669118552710431836954956199165049966552966771775234358569430031450944715218123897207 
# hint = [1355490957378506208208697701592950672672094387068804168531572728350762596829890135826003351, 112984469981193183570958290700584499305718437364882160259618096856045410765383317657942333, 116900480821095370484396815088833958436491607704440051600260019545590977856114612401304297]
# rounds = 39

# P:(25557735861606771704777704605284674181045459246221320438838413903928959660565 : 75510453815577804349033849040800841183099607056251418172827154938451453589656 : 1)
# Q:(56046292811652277371790197347740341815488809388747680953471201733291992185794 : 23661917912317435018072803232589214290695988219345248858537157579403272314 : 1)

先解LCG得到椭圆曲线的参数，然后发现椭圆曲线的阶和p一样，运用SmartAttack即可

exp:

from Crypto.Util.number import long_to_bytes
import gmpy2

n =  
output = 
rounds = 39

p1 = output[-1] - output[-2]
p2 = output[-2] - output[-3]

Ani = gmpy2.invert(p2,n)
a = p1 * Ani % n

b = (output[-1] - a*output[-2]) % n

ani = gmpy2.invert(a,n)
seed = output[-1]
for i in range(rounds):
    seed = ani * (seed-b) % n

p = seed
print("p =",p)
print("A =",a)
print("B =",b)
E = EllipticCurve(GF(p),[a,b])
print(E.order())

P = 
Q = 

def SmartAttack(P,Q,p):
    E = P.curve()
    Eqp = EllipticCurve(Qp(p, 2), [ ZZ(t) + randint(0,p)*p for t in E.a_invariants() ])

    P_Qps = Eqp.lift_x(ZZ(P.xy()[0]), all=True)
    for P_Qp in P_Qps:
        if GF(p)(P_Qp.xy()[1]) == P.xy()[1]:
            break

    Q_Qps = Eqp.lift_x(ZZ(Q.xy()[0]), all=True)
    for Q_Qp in Q_Qps:
        if GF(p)(Q_Qp.xy()[1]) == Q.xy()[1]:
            break

    p_times_P = p*P_Qp
    p_times_Q = p*Q_Qp

    x_P,y_P = p_times_P.xy()
    x_Q,y_Q = p_times_Q.xy()

    phi_P = -(x_P/y_P)
    phi_Q = -(x_Q/y_Q)
    k = phi_Q/phi_P
    return ZZ(k)

m = SmartAttack(P,Q,p)
print(long_to_bytes(int(m)))
# flag{7H4T5_e4sye5T_ECc_o9b8E3}

posted @ 2023-10-30 15:42 Kicky_Mu 阅读(742) 评论(0) 编辑收藏举报

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