CTF之crpto练习三

DES弱加密之easy_BlockCipher

下载附件得到2个文件：
https://adworld.xctf.org.cn/media/task/attachments/5b8bcb28546b4423b481b13149abc99f.zip

分析题目，题目中给出了加密时的代码。

des-ofb.py:

from Crypto.Cipher import DES

f = open('key.txt', 'r')
key_hex = f.readline()[:-1] # discard newline
f.close()
KEY = key_hex.decode("hex")
IV = '13245678'
a = DES.new(KEY, DES.MODE_OFB, IV)

f = open('plaintext', 'r')
plaintext = f.read()
f.close()

ciphertext = a.encrypt(plaintext)
f = open('ciphertext', 'w')
f.write(ciphertext)
f.close()

可知加密时采用了DES算法，并且在OFB模式下对明文进行加密。

因此在已知 IV = ‘12345678’ 的情况下，只需要知道Key，即可对密文进行破解。

根据已知信息，仅有IV以及未知的Key，所以想到DES加密种存在弱密钥。在 DES 的计算中，56bit 的密钥最终会被处理为 16 个轮密钥，每一个轮密钥用于 16 轮计算中的一轮，DES 弱密钥会使这 16 个轮密钥完全一致，所以称为弱密钥。

其中四个弱密钥为：

0x0000000000000000
0xFFFFFFFFFFFFFFFF
0xE1E1E1E1F0F0F0F0
0x1E1E1E1E0F0F0F0F

利用四组若密钥尝试对密文进行破解。

from Crypto.Cipher import DES

f = open('ciphertext', 'r')
ciphertext = f.read()
f.close()
IV = '13245678'
KEY=b'\x00\x00\x00\x00\x00\x00\x00\x00'
a = DES.new(KEY, DES.MODE_OFB, IV)
plaintext = a.decrypt(ciphertext)
print plaintext

KEY=b'\x1E\x1E\x1E\x1E\x0F\x0F\x0F\x0F'
a = DES.new(KEY, DES.MODE_OFB, IV)
plaintext = a.decrypt(ciphertext)
print plaintext

KEY="\xE1\xE1\xE1\xE1\xF0\xF0\xF0\xF0"
a = DES.new(KEY, DES.MODE_OFB, IV)
plaintext = a.decrypt(ciphertext)
print plaintext

KEY="\xFF\xFF\xFF\xFF\xFF\xFF\xFF\xFF"
a = DES.new(KEY, DES.MODE_OFB, IV)
plaintext = a.decrypt(ciphertext)
print plaintext

从得到的结果中得到明文为莎士比亚的一首诗。

或者脚本：
#coding:utf-8
from Crypto.Cipher import DES
import libnum
ct=open('ciphertext','rb').read()
KEY=libnum.n2s(0xe0e0e0e0f1f1f1f1)
IV='13245678'
a=DES.new(KEY,DES.MODE_OFB,IV)
print a.decrypt(ct)
最终得到flag:
flag{_poor_single_dog_has_found_an_echo_from_it}


RSA算法之special-rsa
题目描述：在学习RSA算法时，我发现了一种和RSA具有同等安全性的算法。 对msg.txt加密得到了msg.enc。 $ python special_rsa.py enc msg.txt msg.enc 你能从flag.enc中恢复flag.txt么？
下载附件，里面包含了4个文件，如下： 
https://adworld.xctf.org.cn/media/task/attachments/7a407f44a073442c91fd395b20594f01.zip
flag.enc
special_rsa.py
msg.enc
msg.txt  

题目的思路是用隐藏的key解密flag.enc文件，阅读special_rsa.py文件加密和解密过程后，我做了简单的公式来找到隐藏的key。
v4
伪代码如下：
flag.sage:
N = 23927411014020695772934916764953661641310148480977056645255098192491740356525240675906285700516357578929940114553700976167969964364149615226568689224228028461686617293534115788779955597877965044570493457567420874741357186596425753667455266870402154552439899664446413632716747644854897551940777512522044907132864905644212655387223302410896871080751768224091760934209917984213585513510597619708797688705876805464880105797829380326559399723048092175492203894468752718008631464599810632513162129223356467602508095356584405555329096159917957389834381018137378015593755767450675441331998683799788355179363368220408879117131L

c1 = 14548997380897265239778884825381301109965518989661808090688952232381091726761464959572943383024428028270717629953894592890859128818839328499002950828491521254480795364789013196240119403187073307558598496713832435709741997056117831860370227155633169019665564392649528306986826960829410120348913586592199732730933259880469229724149887380005627321752843489564984358708013300524640545437703771424168108213045567568595093421366224818609501318783680497763353618110184078118456368631056649526433730408976988014678391205055298782061128568056163894010397245301425676232126267874656710256838457728944370612289985071385621160886
c2 = 12793942795110038319724531875568693507469327176085954164034728727511164833335101755153514030256152878364664079056565385331901196541015393609751624971554016671160730478932343949538202167508319292084519621768851878526657022981883304260886841513342396524869530063372782511380879783246034751883691295368172069170967975561364277514063320691930900258017293871754252209727301719207692321798229276732198521711602080244950295889575423383308099786298184477668302842952215665734671829249323604032320696267130330613134368640401070775927197554082071807605399448960911234829590548855031180158567578928333030631307816223152118126597

m1 = 8246074182642091125578311828374843698994233243811347691229334829218700728624047916518503687366611595562099039411430662968666847086659721231623198995017758424796091810259884653332576136128144958751327844746991264667007359518181363522934430676655236880489550093852524801304612322373542296281962196795304499711006801211783005857297362930338978872451934860435597545642219213551685973208209873623909629278321181485010964460652298690058747090298312365230671723790850998541956664376820820570709272500330966205578898690396706695024001970727864091436518202414166919020415892764617055978488996164642229582717493375419993187360
m2 = 15575051453858521753108462063723750986386093067763948316612157946190835527332641201837062951012227815568418309166473080588354562426066694924364886916408150576082667797274000661726279871971377438362829402529682825471299861814829463510659258586020732228351258291527965822977048954720558973840956731377322516168809373640494227129998871167089589689796024458501705704779109152762373660542684880052489213039920383757930855300338529058000330103359636123251274293258

r1 = 12900676191620430360427117641859547516838813596331616166760756921115466932766990479475373384324634210232168544745677888398849094363202992662466063289599443
r2 = 7718975159402389617924543100113967512280131630286624078102368166185443466262861344357647019797762407935675150925250503475336639811981984126529557679881059

_, a, b = xgcd(r1, r2)
k = pow((c1/m1 % N), a, N) * pow((c2/m2 % N), b, N)
print (k)

得到key：
175971776542095822590595405274258668271271366360140578776612582276966567082080372980811310146217399585938214712928761559525614866113821551467842221588432676885027725038849513527080849158072296957428701767142294778752742980766436072183367444762212399986777124093501619273513421803177347181063254421492621011961  

得到key，解密flag.enc，得到答案：
port msgpack

def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def modinv(a, m):
    g, x, y = egcd(a, m)
    assert g == 1
    return x % m

def pad_even(x):
    return ('', '0')[len(x)%2] + x

def decrypt(c, k):
    out = ''
    for r_s, c_s in msgpack.unpackb(c):
        r = int(r_s.encode('hex'), 16)
        c = int(c_s.encode('hex'), 16)
        k_inv = modinv(k, N)
        out += pad_even(format(pow(k_inv, r, N) * c % N, 'x')).decode('hex')
    return out

N = 23927411014020695772934916764953661641310148480977056645255098192491740356525240675906285700516357578929940114553700976167969964364149615226568689224228028461686617293534115788779955597877965044570493457567420874741357186596425753667455266870402154552439899664446413632716747644854897551940777512522044907132864905644212655387223302410896871080751768224091760934209917984213585513510597619708797688705876805464880105797829380326559399723048092175492203894468752718008631464599810632513162129223356467602508095356584405555329096159917957389834381018137378015593755767450675441331998683799788355179363368220408879117131
k = 175971776542095822590595405274258668271271366360140578776612582276966567082080372980811310146217399585938214712928761559525614866113821551467842221588432676885027725038849513527080849158072296957428701767142294778752742980766436072183367444762212399986777124093501619273513421803177347181063254421492621011961
print decrypt(open("flag.enc").read(), k)
最终得到flag:
Flag: BCTF{q0000000000b3333333333-ju57-w0n-pwn20wn!!!!!!!!!!!!}



题目描述：It seems easy, right?Tip: openssl rsautl -encrypt -in FLAG -inkey public.pem -pubin -out flag.enc

题目给了一个flag.enc，还有一个public.pem，附件下载地址：
https://adworld.xctf.org.cn/media/task/attachments/9244cc370caa43f491636f8c4670fe7d.zip
安装openssl可以读取到n和e，因为n不大，可以在yafu或者factordb.com上分解得到n = p * q * r
根据flag.enc，可以得到密文m
根据中国剩余定理，我们要求得m在p，q，r下的余数，不妨设为pmod，qmod，rmod
然后根据模三次剩余，即：proot ^ 3 ≡ pmod ( mod p），求得：proot，同理求得qroot，rroot
利用网页工具可以直接计算得到：
http://www.wolframalpha.com/input/?i=x%5E3+%3D+19342563376936634263836075415482+(mod+27038194053540661979045656526063)
我们从openssl命令行得到一个似乎是通过RSA加密的密文。我们还可以访问公钥，因此我们像使用标准RSA密码一样，通过恢复参数来执行以下操作：
e = 3
n = 23292710978670380403641273270002884747060006568046290011918413375473934024039715180540887338067
使用YAFU，我们将模量分为：
p = 26440615366395242196516853423447
q = 27038194053540661979045656526063
r = 32581479300404876772405716877547
我们得到三个质数。这仍然是好的，这可能只是多质RSA。这一点也不奇怪，总的来说很简单(p-1)(q-1)(r-1)，其余的计算照常进行。但它不存在，因为我们发现模乘逆不存在。原因很明显：gcd(e，to客户端)=3，应该是1。这不是我们第一次遇到类似的情况（如https://github.com/p4-team/ctf/tree/master/2015-10-18-hitcon/crypto 314-u rsabin-35；eng版本），所以我们对如何处理这个问题有了一些想法。
在应用RSA解码之前，我们需要去掉这3个。这意味着加密是：
ciphertext = plaintext^e mod n = (plaintext^e')^3 mod n
所以如果我们能形成方程两边的模三次根(mod n)，我们就可以用e'=e/3进行RSA译码。因为e=3，所以e'=e/3=1，所以这里并不容易，这意味着我们的加密是简单的：
ciphertext = plaintext^3 mod n
所以整个解密过程需要密文中的模立方根(mod n)。
一些关于模根的阅读使我们得出结论，这是可能的，但只有在有限的领域。所以它不能对n做，这是一个复合数，我们知道它是，因为它是pqr。
这个问题让我们想起了中文提醒定理（ https://en.wikipedia.org/wiki/Chinese_remainder_theorem ），我们考虑了一会儿，我们想到了一个想法，如果我们能从密文（mod prime）中计算出3个素数的三次模根，我们就能计算出合并根。我们可以用高斯算法（http://www.di-mgt.com.au/crt.html#gaussalg）来实现这一点。
所以我们继续计算：
pt^3 mod p = ciperhtext mod p = 20827907988103030784078915883129
pt^3 mod q = ciperhtext mod q = 19342563376936634263836075415482
pt^3 mod r = ciperhtext mod r = 10525283947807760227880406671000
然后我们花了一段时间来解决pt的这个方程，最后我们发现wolframalpha实现了这个功能，例如：
http://www.wolframalpha.com/input/?i=x^3+%3D+20827907988103030784078915883129+%28mod+26440615366395242196516853423447%29
这给了我们一套可能的解决方案：
roots0 = [5686385026105901867473638678946, 7379361747422713811654086477766, 13374868592866626517389128266735]
roots1 = [19616973567618515464515107624812]
roots2 = [6149264605288583791069539134541, 13028011585706956936052628027629, 13404203109409336045283549715377]
我们对这些根应用高斯算法：
# -*- coding:utf-8 -*-
# 0CTF 2016: RSA? (Crypto 2pt)

'''
# openssl rsa -in public.pem -pubin -text -modulus
Public-Key: (314 bit)
Modulus:
    02:ca:a9:c0:9d:c1:06:1e:50:7e:5b:7f:39:dd:e3:
    45:5f:cf:e1:27:a2:c6:9b:62:1c:83:fd:9d:3d:3e:
    aa:3a:ac:42:14:7c:d7:18:8c:53
Exponent: 3 (0x3)
Modulus=2CAA9C09DC1061E507E5B7F39DDE3455FCFE127A2C69B621C83FD9D3D3EAA3AAC42147CD7188C53
writing RSA key
-----BEGIN PUBLIC KEY-----
MEEwDQYJKoZIhvcNAQEBBQADMAAwLQIoAsqpwJ3BBh5Qflt/Od3jRV/P4Seixpti
HIP9nT0+qjqsQhR81xiMUwIBAw==
-----END PUBLIC KEY-----
'''

# Chinese Remainder Theorem
def chinese_remainder(n, a):
   sum = 0
   prod = reduce(lambda a, b: a*b, n)
   for n_i, a_i in zip(n, a):
      p = prod / n_i
      sum += a_i * mul_inv(p, n_i) * p
   return sum % prod

def mul_inv(a, b):
    b0 = b
    x0, x1 = 0, 1
    if b == 1: return 1
    while a > 1:
        q = a / b
        a, b = b, a%b
        x0, x1 = x1 - q * x0, x0
    if x1 < 0: x1 += b0
    return x1


e = 3
n = 0x2CAA9C09DC1061E507E5B7F39DDE3455FCFE127A2C69B621C83FD9D3D3EAA3AAC42147CD7188C53

# Factorize n using factordb.com
p = 26440615366395242196516853423447
q = 27038194053540661979045656526063
r = 32581479300404876772405716877547

assert p * q * r == n

ct = int(open("flag.enc", "rb").read().encode("hex"), 16)
# ct = 2485360255306619684345131431867350432205477625621366642887752720125176463993839766742234027524

# pt^3 mod p = ct mod p = 20827907988103030784078915883129
# pt^3 mod q = ct mod q = 19342563376936634263836075415482
# pt^3 mod r = ct mod r = 10525283947807760227880406671000

# Calculate possible cube-root
# using wolflam alpha (or using modified Tonelli-Shanks algorithm)
# like this: "x^3 = 20827907988103030784078915883129 (mod 26440615366395242196516853423447)"
p_root = [5686385026105901867473638678946, 7379361747422713811654086477766, 13374868592866626517389128266735]
q_root = [19616973567618515464515107624812]
r_root = [6149264605288583791069539134541, 13028011585706956936052628027629, 13404203109409336045283549715377]

# For every compination of roots, mix them using Chinese Remainder Theorem
for x in p_root:
   for y in q_root:
      for z in r_root:
         m = chinese_remainder([p, q, r], [x, y, z])
         pt = hex(m)[2:-1]
         if len(pt) % 2 != 0:
            pt = "0"+pt
         pt = pt.decode("hex")
         if pt.find("0ctf{") >= 0:
            # Get the flag
            print pt.strip()
            break
最终flag:
0ctf{HahA!Thi5_1s_n0T_rSa~}

RSA解密之babyRSA1

1.下载附件，发现有3个文件：

https://adworld.xctf.org.cn/media/task/attachments/5a456b6a66c04c02bf754d540e5b531d.zip

在挑战中，我们获得代码，公钥和结果。

加密相当简单：

R.<a> = GF(2^2049)

def encrypt(m):
    global n
    assert len(m) <= 256
    m_int = Integer(m.encode('hex'), 16)
    m_poly = P(R.fetch_int(m_int))
    c_poly = pow(m_poly, e, n)
    c_int = R(c_poly).integer_representation()
    c = format(c_int, '0256x').decode('hex')
    return c

看来n这是GF（2）上的PolynomialRing的多项式。加密基本上将消息更改为GF（2 ^ 2049）的元素，然后将其表示为环P的元素，然后将该消息的多项式表示形式提高为emod多项式的幂n。

所以，它真的可以归结为经典的RSA，与小搓那m和n现在是多项式，一切都在PolynomialRing在GF（2）calcualted。

我们发现了一篇很好的论文，描述了这个想法：www.diva-portal.se/smash/get/diva2:823505/FULLTEXT01.pdf（Izabela Beatrice Gafitoiu的理学学士学位论文）。

如果遵循此论点，我们会发现在这种情况下，d解密指数可以计算为emod的模乘逆s。特殊值s是等效的phi，从经典RSA，并且基本上是(2^p_d-1)(2^q_d-1)其中p_d和q_d正度多项式的p和q，使得p*q == n。

所以这个想法很简单：

1、因子多项式n成p和q 2、计算 s 3、计算 d 4、解密标志

def decrypt(m, d):
    m_int = Integer(m.encode('hex'), 16)
    m_poly = P(R.fetch_int(m_int))
    c_poly = pow(m_poly, d, n)
    c_int = R(c_poly).integer_representation()
    c = format(c_int, '0256x').decode('hex')
    return c

if __name__ == '__main__':
    p,q = n.factor()
    p,q = p[0],q[0]
    s = (2^p.degree()-1)*(2^q.degree()-1)
    d = inverse_mod(e,s)
    with open('flag.enc', 'rb') as f:
        ctext = f.read()
        print(decrypt(ctext,d))

最终得到flag:flag{P1ea5e_k33p_N_as_A_inTegeR~~~~~~}


AES解密之in-plain-sight

题目描述：这次的挑战并不难：你只需要对隐藏的密文进行解密。为了让解密更加简单，我会给你除了HiddenCiphertext以外你需要的所有东西，你要做的就是自己将密文找出来！ 你需要： 算法：AES-256-CBC 密钥：c086e08ad8ee0ebe7c2320099cfec9eea9a346a108570a4f6494cfe7c2a30ee1 IV：0a0e176722a95a623f47fa17f02cc16a



通过网站，https://morsecode.world/international/decoder/audio-decoder-adaptive.html（音频解密器）得到C1和C2的值

C1:4314251881242803343641258350847424240197348270934376293792054938860756265727535163218661012756264314717591117355736219880127534927494986120542485721347351
C2:485162209351525800948941613977942416744737316759516157292410960531475083863663017229882430859161458909478412418639172249660818299099618143918080867132349
利用openssl对两个公钥进行解密，得到n1n2和e1e2：

import gmpy2
import libnum

n1 = gmpy2.mpz(10285341668836655607404515118077620322010982612318568968318582049362470680277495816958090140659605052252686941748392508264340665515203620965012407552377979)
n2 = gmpy2.mpz(8559553750267902714590519131072264773684562647813990967245740601834411107597211544789303614222336972768348959206728010238189976768204432286391096419456339)
e1 = 41221
e2 = 41221
p = gmpy2.gcd(n1,n2)
q1 = n1 / p
q2 = n2 / p
c1 = 4314251881242803343641258350847424240197348270934376293792054938860756265727535163218661012756264314717591117355736219880127534927494986120542485721347351
c2 = 485162209351525800948941613977942416744737316759516157292410960531475083863663017229882430859161458909478412418639172249660818299099618143918080867132349
phin1 = (p - 1)*(q1 - 1)
phin2 = (p - 1)*(q2 - 1)
d1 = gmpy2.invert(e1,phin1)
d2 = gmpy2.invert(e2,phin2)

print(libnum.n2s(pow(c1,d1,n1))+libnum.n2s(pow(c2,d2,n2)))
最终得到flag:
UNCTF{ac01dff95336aa470e3b55d3fe43e9f6}


题目描述：安全分析人员截获间谍发出的秘密邮件，该邮件只有一个mp3文件，安全人员怀疑间谍通过某种private的方式将信息传递出去，尝试分析该文件，获取藏在文件中的数据。 flag形式为 flag{}

题目附件连接：
https://adworld.xctf.org.cn/media/task/attachments/206c0533300340b19c3a18d82d806a98.mp3
解题步骤：
题目提示文件使用了private加密信息，在010Editor中打开mp3文件，发现存在private bit，因此，只需要提取每一个mf组中的该字节，组合起来，就是答案。可以从图中看到 ms 开始位为1 C1B8H，即第115128字节，如图所示：，如图所示：
uint32 frame_sync : 12
    uint32 mpeg_id : 1
    uint32 layer_id : 2
    uint32 protection_bit : 1
    uint32 bitrate_index : 4
    uint32 frequency_index : 2
    uint32 padding_bit : 1
    uint32 private_bit : 1
    uint32 channel_mode : 2
    uint32 mode_extension : 2
    uint32 copyright : 1
    uint32 original : 1
    uint32 emphasis : 2

12+1+2+1+4+2+1+1+2+2+1+1+2=32，即总共4字节，private_bit 为24，所在的字节为第3个字节因此要从前一个，即第二个字节开始提取内容，该字节对应的地址为 115130观察每一个mf组，大小都为414h，即1044字节，因此可以得到以下脚本：
# coding:utf-8
import re
import binascii
n = 115130
result = ''
fina = ''
file = open('flag-woody.mp3','rb')
while n < 2222222 :
    file.seek(n,0)
    n += 1044
    file_read_result = file.read(1)
    read_content = bin(ord(file_read_result))[-1]
    result = result + read_content
textArr = re.findall('.{'+str(8)+'}', result)
textArr.append(result[(len(textArr)*8):])
for i in textArr:
    fina = fina + hex(int(i,2))[2:].strip('\n')
fina = fina#.decode('hex')
print (fina)

将得到的字符串
464c41477b707231763474335f6269377d25a1cedc3e69888894dac4dd3a87c5e1c5276fa6d626832148d39288a0c596c95abaac3f09f9f524647595ae4894f9b82b3f4c1b47537c365d8d69d84a353c1a93ae436761d430e666e4111752d479746d1828f9c07c27ab1c3eaf1948f8a9e839b280a4342f321e89eb73b237a2b55d5310b77811c0975cfc1365e146f6c9212e244751398f73c17ee1a6664b4fd712d4b0a297275fa471fb65e440bc7bdc12fb0a39d81a1d374f2d55b8faabf9bf2c342f1046fbab7e66ac7896ffac672d277b89f8606759a8ac21a58fbb4b9b51d45f126a7f67c1a297e1fcb638356ec739b89555568816
转换对应的ASCII码，得到Flag：
最终 flag:
flag{pr1v4t3_bi7}

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原文地址：https://www.cnblogs.com/backlion/p/15693608.html