NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解-Toy模板网

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一、brainfuck

附件信息

++++++++[>>++>++++>++++++>++++++++>++++++++++>++++++++++++>++++++++++++++>++++++++++++++++>++++++++++++++++++>++++++++++++++++++++>++++++++++++++++++++++>++++++++++++++++++++++++>++++++++++++++++++++++++++>++++++++++++++++++++++++++++>++++++++++++++++++++++++++++++<<<<<<<<<<<<<<<<-]>>>>>>>++++++.>----.<-----.>-----.>-----.<<<-.>>++..<.>.++++++.....------.<.>.<<<<<+++.>>>>+.<<<+++++++.>>>+.<<<-------.>>>-.<<<+.+++++++.--..>>>>---.-.<<<<-.+++.>>>>.<<<<-------.+.>>>>>++.

在线工具一把梭

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

二、Caesar's Secert

题目信息

kqfl{hf3x4w'x_h1umjw_n5_a4wd_3fed}

凯撒解码

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

三、Fence

题目信息

fa{ereigtepanet6680}lgrodrn_h_litx#8fc3

栏栅密码

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

四、Vigenère

题目信息

pqcq{qc_m1kt4_njn_5slp0b_lkyacx_gcdy1ud4_g3nv5x0}

维吉尼亚呀呀呀！！根据flag前缀通过偏移量手算key就行，是kfc呀嘿嘿

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

五、babyencoding

题目信息

part 1 of flag: ZmxhZ3tkYXp6bGluZ19lbmNvZGluZyM0ZTBhZDQ=
part 2 of flag: MYYGGYJQHBSDCZJRMQYGMMJQMMYGGN3BMZSTIMRSMZSWCNY=
part 3 of flag: =8S4U,3DR8SDY,C`S-F5F-C(S,S<R-C`Q9F8S87T`

flag分啦三部分，分别来解

part1：base64

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

part2：base32

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

part3：UUencode

NewStarCTF 2023 公开赛道 WEEK1|CRYPTO全解

六、babyrsa

题目信息

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

def gen_prime(n):
    res = 1

    for i in range(15):
        res *= getPrime(n)

    return res


if __name__ == '__main__':
    n = gen_prime(32)
    e = 65537
    m = bytes_to_long(flag)
    c = pow(m,e,n)
    print(n)
    print(c)
# 17290066070594979571009663381214201320459569851358502368651245514213538229969915658064992558167323586895088933922835353804055772638980251328261
# 14322038433761655404678393568158537849783589481463521075694802654611048898878605144663750410655734675423328256213114422929994037240752995363595

脚本：

import gmpy2
from Crypto.Util.number import long_to_bytes
n = 17290066070594979571009663381214201320459569851358502368651245514213538229969915658064992558167323586895088933922835353804055772638980251328261

e = 65537
c = 14322038433761655404678393568158537849783589481463521075694802654611048898878605144663750410655734675423328256213114422929994037240752995363595

p1=2217990919
p2=2338725373
p3=2370292207
p4=2463878387
p5=2706073949
p6=2794985117
p7=2804303069
p8=2923072267
p9=2970591037
p10=3207148519
p11=3654864131
p12=3831680819
p13=3939901243
p14=4093178561
p15=4278428893

phi = (p1 - 1) * (p2 - 1) * (p3 - 1) * (p4 - 1) * (p5 - 1) * (p6 - 1) * (p7 - 1) * (p8 - 1) * (p9 - 1) * (p10 - 1) * (p11 - 1) * (p12 - 1) * (p13 - 1) * (p14 - 1) * (p15 - 1)
d = gmpy2.invert(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m))
#flag{us4_s1ge_t0_cal_phI}

七、Small d

题目信息

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

p = getPrime(1024)
q = getPrime(1024)

d = getPrime(32)
e = inverse(d, (p-1)*(q-1))
n = p*q
m = bytes_to_long(flag)

c = pow(m,e,n)

print(c)
print(e)
print(n)

# c = 6755916696778185952300108824880341673727005249517850628424982499865744864158808968764135637141068930913626093598728925195859592078242679206690525678584698906782028671968557701271591419982370839581872779561897896707128815668722609285484978303216863236997021197576337940204757331749701872808443246927772977500576853559531421931943600185923610329322219591977644573509755483679059951426686170296018798771243136530651597181988040668586240449099412301454312937065604961224359235038190145852108473520413909014198600434679037524165523422401364208450631557380207996597981309168360160658308982745545442756884931141501387954248
# e = 8614531087131806536072176126608505396485998912193090420094510792595101158240453985055053653848556325011409922394711124558383619830290017950912353027270400567568622816245822324422993074690183971093882640779808546479195604743230137113293752897968332220989640710311998150108315298333817030634179487075421403617790823560886688860928133117536724977888683732478708628314857313700596522339509581915323452695136877802816003353853220986492007970183551041303875958750496892867954477510966708935358534322867404860267180294538231734184176727805289746004999969923736528783436876728104351783351879340959568183101515294393048651825
# n = 19873634983456087520110552277450497529248494581902299327237268030756398057752510103012336452522030173329321726779935832106030157682672262548076895370443461558851584951681093787821035488952691034250115440441807557595256984719995983158595843451037546929918777883675020571945533922321514120075488490479009468943286990002735169371404973284096869826357659027627815888558391520276866122370551115223282637855894202170474955274129276356625364663165723431215981184996513023372433862053624792195361271141451880123090158644095287045862204954829998614717677163841391272754122687961264723993880239407106030370047794145123292991433

识别：e特别大
在RSA中d也称为解密指数，当d比较小的时候，e也就显得特别大了，所以发现e特别大的时候可以考虑低解密指数攻击。

# Sage

def rational_to_contfrac(x, y):

    # Converts a rational x/y fraction into a list of partial quotients [a0, ..., an]

    a = x // y

    pquotients = [a]

    while a * y != x:

        x, y = y, x - a * y

        a = x // y

        pquotients.append(a)

    return pquotients

def convergents_from_contfrac(frac):

    # computes the list of convergents using the list of partial quotients

    convs = [];

    for i in range(len(frac)): convs.append(contfrac_to_rational(frac[0: i]))

    return convs

def contfrac_to_rational(frac):

    # Converts a finite continued fraction [a0, ..., an] to an x/y rational.

    if len(frac) == 0: return (0, 1)

    num = frac[-1]

    denom = 1

    for _ in range(-2, -len(frac) - 1, -1): num, denom = frac[_] * num + denom, num

    return (num, denom)

c = 6755916696778185952300108824880341673727005249517850628424982499865744864158808968764135637141068930913626093598728925195859592078242679206690525678584698906782028671968557701271591419982370839581872779561897896707128815668722609285484978303216863236997021197576337940204757331749701872808443246927772977500576853559531421931943600185923610329322219591977644573509755483679059951426686170296018798771243136530651597181988040668586240449099412301454312937065604961224359235038190145852108473520413909014198600434679037524165523422401364208450631557380207996597981309168360160658308982745545442756884931141501387954248
e = 8614531087131806536072176126608505396485998912193090420094510792595101158240453985055053653848556325011409922394711124558383619830290017950912353027270400567568622816245822324422993074690183971093882640779808546479195604743230137113293752897968332220989640710311998150108315298333817030634179487075421403617790823560886688860928133117536724977888683732478708628314857313700596522339509581915323452695136877802816003353853220986492007970183551041303875958750496892867954477510966708935358534322867404860267180294538231734184176727805289746004999969923736528783436876728104351783351879340959568183101515294393048651825
n = 19873634983456087520110552277450497529248494581902299327237268030756398057752510103012336452522030173329321726779935832106030157682672262548076895370443461558851584951681093787821035488952691034250115440441807557595256984719995983158595843451037546929918777883675020571945533922321514120075488490479009468943286990002735169371404973284096869826357659027627815888558391520276866122370551115223282637855894202170474955274129276356625364663165723431215981184996513023372433862053624792195361271141451880123090158644095287045862204954829998614717677163841391272754122687961264723993880239407106030370047794145123292991433

def egcd(a, b):

    if a == 0: return (b, 0, 1)

    g, x, y = egcd(b % a, a)

    return (g, y - (b // a) * x, x)

def mod_inv(a, m):

    g, x, _ = egcd(a, m)

    return (x + m) % m

def isqrt(n):

    x = n

    y = (x + 1) // 2

    while y < x:

        x = y

        y = (x + n // x) // 2

    return x

def crack_rsa(e, n):

    frac = rational_to_contfrac(e, n)

    convergents = convergents_from_contfrac(frac)

    for (k, d) in convergents:

        if k != 0 and (e * d - 1) % k == 0:

            phi = (e * d - 1) // k

            s = n - phi + 1

            # check if x*x - s*x + n = 0 has integer roots

            D = s * s - 4 * n

            if D >= 0:

                sq = isqrt(D)

                if sq * sq == D and (s + sq) % 2 == 0: return d

d = crack_rsa(e, n)

m = hex(pow(c, d, n))[2:]

print(bytes.fromhex(m))
#flag{learn_some_continued_fraction_technique#dc16885c}

八、babyxor

题目信息

from secret import *

ciphertext = []

for f in flag:
    ciphertext.append(f ^ key)

print(bytes(ciphertext).hex())
# e9e3eee8f4f7bffdd0bebad0fcf6e2e2bcfbfdf6d0eee1ebd0eabbf5f6aeaeaeaeaeaef2

简单的异或

脚本：

法一：爆破key

a = 'e9e3eee8f4f7bffdd0bebad0fcf6e2e2bcfbfdf6d0eee1ebd0eabbf5f6aeaeaeaeaeaef2'
c = bytes.fromhex(a)
for i in range(256):
    flag = []
    for j in c:
        flag.append(j ^ i)
    if b'flag' in bytes(flag):
        print(bytes(flag))
# flag{x0r_15_symm3try_and_e4zy!!!!!!}

法二：直接求key，key = ord(‘f’) ^ (密文第一个字节)

a = 'e9e3eee8f4f7bffdd0bebad0fcf6e2e2bcfbfdf6d0eee1ebd0eabbf5f6aeaeaeaeaeaef2'
c = bytes.fromhex(a)
key = ord('f') ^ c[0]
flag = []
for j in c:
    flag.append(j ^ key)
print(bytes(flag))
# flag{x0r_15_symm3try_and_e4zy!!!!!!}

九、Affine

题目信息

from flag import flag, key

modulus = 256

ciphertext = []

for f in flag:
    ciphertext.append((key[0]*f + key[1]) % modulus)

print(bytes(ciphertext).hex())

# dd4388ee428bdddd5865cc66aa5887ffcca966109c66edcca920667a88312064

题目分析：仿射密码

源码存在两个量 key[0] 和 key[1] 作为密钥a和b

其实仿射密码的本质就是解方程得到a和b

我们可以利用已知明文前缀flag{*****} 获得a和b的值

需要注意一点！！

使用f和l这两个已知字符是无法合适求解的所以一定要灵活整体往后取一位用l和a求解就好了

exp：

import gmpy2

#key = '****CENSORED***************'    #密钥 censored 被遮盖
flag = 'flag{*******CENSORED********}'    #部分明文

data = "dd4388ee428bdddd5865cc66aa5887ffcca966109c66edcca920667a88312064"   #密文待解密

encrypted = bytes.fromhex(data)
# encrypted = data
# print(encrypted)
plaindelta = ord(flag[2]) - ord(flag[1])
print(plaindelta)
cipherdalte = encrypted[2] - encrypted[1]
print(cipherdalte)

modulus = 256
a = gmpy2.invert(plaindelta, modulus) * cipherdalte % modulus
b = (encrypted[1] - a * ord(flag[1])) % modulus
print(a,b)
# a = 17
# b = 23
a_inv = gmpy2.invert(a, modulus)
result = ""
for c in encrypted:
    result += chr((c - b) * a_inv % modulus)
print(result)
#flag{4ff1ne_c1pher_i5_very_3azy}

十、babyaes

题目信息

from Crypto.Cipher import AES
import os
from flag import flag
from Crypto.Util.number import *


def pad(data):
    return data + b"".join([b'\x00' for _ in range(0, 16 - len(data))])


def main():
    flag_ = pad(flag)
    key = os.urandom(16) * 2
    iv = os.urandom(16)
    print(bytes_to_long(key) ^ bytes_to_long(iv) ^ 1)
    aes = AES.new(key, AES.MODE_CBC, iv)
    enc_flag = aes.encrypt(flag_)
    print(enc_flag)


if __name__ == "__main__":
    main()
# 3657491768215750635844958060963805125333761387746954618540958489914964573229
# b'>]\xc1\xe5\x82/\x02\x7ft\xf1B\x8d\n\xc1\x95i'

题目分析：
buu上有一道与这道题很类似
key等于字节a的前16位 * 2
iv = 字节a后16位 ^ key的前一半 ^ 1
key,iv都出来了，那么flag就好说了。。

exp:

from Crypto.Util.number import *
from Crypto.Cipher import AES

a = 3657491768215750635844958060963805125333761387746954618540958489914964573229
c = b'>]\xc1\xe5\x82/\x02\x7ft\xf1B\x8d\n\xc1\x95i'
key = long_to_bytes(a)[:16]
iv = bytes_to_long(key) ^ bytes_to_long(long_to_bytes(a)[16:]) ^ 1

aes = AES.new(key * 2,AES.MODE_CBC,long_to_bytes(iv))
flag = aes.decrypt(c)
print(flag)
# flag{firsT_cry_Aes}

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