1: De beaux souvenirs
Ouvrez le fichier de code source et lisez que nous avons constaté que l'algorithme de cryptage est CBC, le mode de cryptage est le mode CBC et le cryptage est de 8 bits, comme indiqué dans la figure ci-dessous
Nous savons que le texte en clair de début est "passe un bon moment" et le texte chiffré chiffré, c'est-à-dire le texte en clair1, le texte en clair2 et tous les textes chiffrés sont connus. À ce stade, nous pouvons obtenir le texte en clair [i] = texte chiffré [i-1] ^ clé chiffrée [i] ^, afin que tout le texte en clair puisse être résolu. Le code source est le suivant:
def xor(data,key):
return bytes([x ^ key[i%len(key)] for i, x in enumerate(data)])
with open('flag.txt.encrypted', 'rb') as f:
s=f.read()
s1="have a good time".encode()
c1=s[:8]
c2=s[8:16]
c3=s[16:24]
key=xor(xor(c1,s1[8:]),c2)
print(key)
m1=xor(xor(c2,key),c3)
sssssss=""
for i in range(8,len(s)-8,8):
m1=xor(xor(s[i:i+8],key),s[i+8:i+16])
sssssss=sssssss+m1.decode('utf-8')
print(sssssss)2: Fin triste
Le même que le principe de cryptage de la question précédente, mais nous obtenons le texte brut de fin "éloignez-vous de xiaocui!", Mais nous avons un remplissage dans le processus de cryptage, nous ne connaissons pas l'avant-dernier texte brut dans le cryptage, l'avant-dernier paragraphe Quel est le texte brut, nous pouvons deviner que le remplissage est de 0 à 7 (selon le principe de chiffrement), essayez de deviner le remplissage de différentes longueurs pour obtenir la clé, puis déchiffrez et vérifiez, obtenez la vraie clé, déchiffrez
def xor(data,key):
return bytes([x ^ key[i%len(key)] for i, x in enumerate(data)])
with open('flag.txt.encrypted', 'rb') as f:
s=f.read()
BLOCK_SIZE=8
s1="keep away from xiaocui!".encode()
c1=s[-16:-8]
c2=s[-24:-16]
c3=s[-32:-24]
keylist=[]
for i in range(8):
m1=s1[i+8:i+16]
keylist.append(xor(xor(c1,m1),c2))
key=b""
for i in range(8):
mm=xor(xor(c2,keylist[i]),c3)
if mm==s1[i:i+8]:
key=keylist[i]
print(key)
sssssss=""
for i in range(8,len(s)-8,8):
m1=xor(xor(s[i:i+8],key),s[i+8:i+16])
sssssss=sssssss+m1.decode('utf-8')
print(sssssss)3. Le secret perdu
Nous avons obtenu la clé publique et la clé privée RSA incomplètes, c'est-à-dire que les paramètres n, p, q et autres dans le fichier sont incomplets. Recherche et nous trouvons https://www.40huo.cn/blog/rsa-private-key-recovery Les questions dans -and-oaep.html sont similaires à nos questions actuelles, mais notre n n'est pas complet et l'ordre inférieur n'existe pas. Mais les bits bas de p et q existent, donc nous pouvons utiliser les bits bas de p et q pour calculer les bits bas de n, c'est-à-dire 0x8b * 0x05 = 0x2b7. Nous pouvons mettre 'b7' à la fin de n pour compléter n, puis nous utilisons le script ci-dessus Peut résoudre p, q et clé privée
#!/usr/bin/python
#-*- coding:utf-8 -*-
import re
import pickle
from itertools import product
from libnum import invmod, gcd
def solve_linear(a, b, mod):
if a & 1 == 0 or b & 1 == 0:
return None
return (b * invmod(a, mod)) & (mod - 1) # hack for mod = power of 2
def to_n(s):
s = re.sub(r"[^0-9a-f]", "", s)
return int(s, 16)
def msk(s):
cleaned = "".join(map(lambda x: x[-2:], s.split(":")))
return msk_ranges(cleaned), msk_mask(cleaned), msk_val(cleaned)
def msk_ranges(s):
return [range(16) if c == " " else [int(c, 16)] for c in s]
def msk_mask(s):
return int("".join("0" if c == " " else "f" for c in s), 16)
def msk_val(s):
return int("".join("0" if c == " " else c for c in s), 16)
E = 65537
N = to_n("""00:c4:9d:36:a4:77:76:12:12:85:24:6c:74:1d:7d:
b3:ce:f4:c3:a4:69:cd:0b:2e:8f:d6:75:e3:80:b8:
e8:1c:ce:e8:60:90:45:56:73:ab:32:32:00:7f:6a:
76:3e:b6:10:d3:a2:74:da:f9:4e:a5:7e:ae:ef:f4:
da:82:57:6d:68:82:50:d8:b1:fc:92:b1:5c:7d:54:
f5:7e:d0:06:8a:60:ff:82:70:72:20:68:4b:71:ba:
87:44:57:c1:97:a0:8a:2d:53:93:f3:0a:60:87:a3:
85:c8:45:e6:0a:88:85:b5:ff:c7:09:9a:76:03:fe:
99:b6:fb:8a:1e:9f:a8:42:3a:0a:c9:a9:bf:1c:87:
2c:c4:99:10:db:46:e3:a9:a5:79:93:8c:75:71:ec:
c6:3b:af:44:dc:60:c4:53:f6:3c:e8:73:2f:50:10:
38:e7:6f:d0:a5:4b:ae:e3:1e:43:11:42:2c:a2:38:
e6:3f:0b:13:54:63:e8:2f:9e:61:ab:08:65:97:e0:
27:30:19:fd:a7:fe:5c:d8:11:b8:34:87:ad:02:c2:
bc:cd:73:d3:86:be:fd:2a:b4:fe:7d:7e:d3:64:bb:
6f:63:ed:a6:1d:ee:f2:80:da:9d:7a:23:7f:c1:39:
b0:98:0c:85:8f:d0:4b:9f:e4:1a:26:fc:44:d1:67:
03:32:03:0c:91:61:23:4c:81:6f:42:18:88:41:dc:
27:55:a3:07:7c:a1:ad:f3:58:4d:91:07:65:f1:63:
f2:34:d5:17:0e:59:c6:bb:b6:6d:7d:0c:d2:64:4b:
b9:9c:52:59:03:8e:2a:43:23:76:33:c3:e8:72:3b:
1c:e0:40:97:36:5f:ae:00:d7:e3:09:eb:df:55:44:
22:b4:09:00:b5:09:41:70:6c:5c:3b:98:d3:34:7e:
60:a2:b8:93:bd:af:32:77:48:48:8a:a5:9c:0e:6a:
a1:79:36:86:8c:e9:3f:b1:a2:a7:4a:3a:d8:d6:f6:
dd:62:d8:ae:9e:13:bb:0c:6b:b1:65:68:0d:7e:58:
3f:68:1e:91:49:13:19:68:2b:fd:3c:5e:52:fa:76:
b0:57:fc:0e:35:d8:71:56:41:06:ef:50:99:56:dd:
d4:9a:1f:d3:46:26:12:9c:15:4b:43:fc:1b:de:c9:
06:ad:82:56:63:c8:a4:83:32:d2:35:05:23:15:52:
d9:0a:73:85:5e:c9:c2:56:af:69:d2:5f:77:04:28:
c8:4c:b9:a6:d4:15:15:b5:15:99:13:ef:a9:a5:de:
5a:74:b1:03:cf:32:a5:03:69:f8:e9:bb:7e:16:31:
5e:43:e7:02:51:ac:c5:f6:bf:ef:1c:74:f7:13:0c:
19:ad:b7""")
p_ranges, pmask_msk, pmask_val = msk("""00: :05:89: :bd:35: : :23: : : : :84:
: :ed: :70:14: : : :10: : :87: :51:
ea: :97:69: :52: : : : : :ea: : :15:
: :34: :be:11:23: : :34:14: :94: :10:
: :74:87:37:ee:81:62:ee:95: : :dc:49:dd:
: :35: :81: :fa: : : :86: : : :fb:
:93: : :12: :14: :ab:76: :96: : :27:
:21: :04:01:41: :98: :ff: : :12:dc: :
cd: :39:95:30: :47: :fa:ff: :34: :ad: :
:52:02:fa:bc:14:22:22:48:61:62:bd:53: : :
72:08:cb:41:88: : : :63:91:30:fe: : :42:
87: :18:52: :39:dd: :68: :fe:06:88:81: :
: : :ae:fd: : :fb:21:37:59: :53: :fa:
:07:40:eb:33:77:51:64:10:dd: :73: :86:62:
:bf: :79: :34: :bb: :44:ff: :46:fe:90:
ef: :52:ad: : :fe: :69:18:89:bd:cd:09:46:
: :74:71: : : :41:66: : :11: :25: :
39:8b""")
q_ranges, qmask_msk, qmask_val = msk(""" 00:ce:43:ef: :76:58:17:43:31: : :32:70: :
89: : :36:55:06: :79:66:78: : : : : :
:85: : : : : :33:bb: : :56: :66:cb:
:08: : :90:cb: : :24:fa:ca:47: : : :
:88: :83:01: :62: : : : : : :ad:ae:
: : :58: :ec: : : :09:04:86: :05:00:
:df:50:84:81:80: :ae: :24: :94:da: :04:
ce: :ef: : :ed:be:bf:43:78: : :05:93: :
08:52:05: : : : :ae: : : : :ab: : :
:76:ce: : : : :19:bd:22: :ef:dc:bf:ea:
ab:78:01: : :85: : : :ea: : :fb: : :
92:66:19: : :ab: : :82: : :31: : :da:
82: :13:82:43: : :94:13:41: : : :37: :
:04:56:02:87:dd: :58:27: : :24: : : :
28: : :09:14:89: : : :49:59: :16:eb:65:
:01:22: : :dd: :78: : :db:90: :ac: :
:fd: :03:74: : : : :92: :00:ba: : :
:05""")
_, dmask_msk, dmask_val = msk("""11: : :69:62:64: : : : :15: :13:de:de:
cf: : :17: : :75: :98:42:fc: :12:15:08:
: : : : :36: :be:25:48: : :19: : :
:47:11:19: :03: :49:fc:da: :96:45:eb: :
: : :91: :ea: : :55:ff: :37:58: : :
19: : :73:40: :91:15:01:da:91:22:fd:32: :
: :50: : :66: : : :42: : :ef: : :
df:42: :97:30: :39: : : : : : :dc: :
: : : : :38: : : :88:28: :05: : :
78:59:fa: :86: :19:24: : : : :da:cf:15:
39: : : : :ef:55: :ce:47: :58:89: :fb:
:24: : : :92: : :ee: : :db:67:31:ce:
:28: :72:ec:89: :04: : :50: : : : :
:37: :44: : : : :56: :38: :bb:47:bb:
66:83:99:22:07:72: : :48:52:02: : : :29:
:82:56: :67: :95: : :56:94: : :71: :
bf:27:98: : :54:98:26:06:87: :ae: :53:be:
: :80:37:60:61:ea:ef:de: : :df:90:81: :
70: :06:33:26: :75:fe:95: :92: :78:cd:05:
64:cc:68: : :36:54: :bd:16:90:ee:60: : :
: :41: : :91: :79:58:06:50: :46: : :
45: :09:ca:ac:16: :27:98: : :ba:82: :77:
93:98:ad: :15: :67:53:97:ad:ee:50:44: :31:
07: :ff:01: :09: : : : : :46: : :42:
15: :db:df:42:be: : : :78: :41: : : :
:14: : :25:fc: :84: : : : : : :20:
da:46:01:eb:87: :12:57: : :56:af: :87:93:
60: :02: :18:89:63:72:ad: :ed:cf: : :84:
:22: :13: : :dd: :ff: : : :de:62:37:
:19:66: : :86:02: :38: : : : :ec:14:
12: :43:93:19:65:98: : :03: : : :ef: :
: :ca:07:92:22: : :bb:15:eb: : : :35:
:72:29:cd: : :99: : : : :41:06: : :
:43:33: :32: : :54:be:92:62: :78:59:42:
79:89""")
_, dpmask_msk, dpmask_val = msk(""" :39: :28:16:02:89:ce:11:fe: : : : :af:
: : :ed:97: : :11:20:ba:ae:98:ad: : :
:10:87:ac:07: : : : :50: : :70:50:52:
df:89:eb:02: : : : :93:11: : :12: :56:
:08: : :ea: :10:fa:19: : : :54:45:07:
: :bc:ff:33: :db:63:49:fe:52: :33: : :
bf:cd:45:91: :10: : :92:81:40:03: :80: :
29: :30: :ed:43:64:ca: :bf:64: : :bf: :
: : :24:72:84: : :ff: : :24: :81:27:
db:23: :64: :67: :ba: : :bc: : : : :
:ae:88: : : : : :91: : :14: :ba:ef:
:89: : : : : : : : :05: :75:52: :
: : :be:ad:df: :02:88:00: : :15:45: :
cf:32: :ca: :93: :32: :40: :27:dd: :19:
73:dc: : : : : :cf: : :dd: : :ca: :
ee: :ca: : : :49: :27: :58:53: :64:25:
:22:06:16:ff:62:bc: : : : :24:fc: : :
df""")
_, dqmask_msk, dqmask_val = msk("""02: :bd: :19:25:98:75: :65: :55:28:33:bc:
34:84:91:01:96: : :08: :32:45: :27: : :
:fe: :bb:63:32:68: :51:bd:75:40: :52:52:
: : :78:85:fc:94: :07: :14: : : : :
15:dd: : :93: :01: : :77:ca: :40: :da:
:89:bc:87:62:dc:ac:61:88: : :70: :69: :
:36: : :21:08: :dc:73: :ad:da:ee:fe: :
96: :58: : :46: :29:ff:97:ce: : : :cb:
51: : :81: :22: : :19: :10:69:41:36:ca:
:22:49: :cc:cf:06: : :08: :76: : :45:
98: : :45: : : :69:13:65: : :da:54: :
19: :ee:24: :73: : : : : : :18:53:40:
21:25: : :84:52:cd: :49:33:78: : :ed: :
25:27: : : :ca: : : :ca: : :bc: :02:
31:70: :10:ca:84:59: : : :52: :27:76: :
47: :66:bf:ff: :03: :99:ff: :df: : : :
:46:27:45: :65:07: :48:da:dc: :80: : :
f9""")
def search(K, Kp, Kq, check_level, break_step):
max_step = 0
cands = [0]
for step in range(1, break_step + 1):
#print " ", step, "( max =", max_step, ")"
max_step = max(step, max_step)
mod = 1 << (4 * step)
mask = mod - 1
cands_next = []
for p, new_digit in product(cands, p_ranges[-step]):
pval = (new_digit << ((step - 1) * 4)) | p
if check_level >= 1:
qval = solve_linear(pval, N & mask, mod)
if qval is None or not check_val(qval, mask, qmask_msk, qmask_val):
continue
if check_level >= 2:
val = solve_linear(E, 1 + K * (N - pval - qval + 1), mod)
if val is None or not check_val(val, mask, dmask_msk, dmask_val):
continue
if check_level >= 3:
val = solve_linear(E, 1 + Kp * (pval - 1), mod)
if val is None or not check_val(val, mask, dpmask_msk, dpmask_val):
continue
if check_level >= 4:
val = solve_linear(E, 1 + Kq * (qval - 1), mod)
if val is None or not check_val(val, mask, dqmask_msk, dqmask_val):
continue
if pval * qval == N:
print "Kq =", Kq
print "pwned"
print "p =", pval
print "q =", qval
p = pval
q = qval
d = invmod(E, (p - 1) * (q - 1))
coef = invmod(p, q)
from Crypto.PublicKey import RSA
print RSA.construct(map(long, (N, E, d, p, q, coef))).exportKey()
quit()
cands_next.append(pval)
if not cands_next:
return False
cands = cands_next
return True
def check_val(val, mask, mask_msk, mask_val):
test_mask = mask_msk & mask
test_val = mask_val & mask
return val & test_mask == test_val
# K = 4695
# Kp = 15700
# Kq = 5155
for K in range(1, E):
if K % 100 == 0:
print "checking", K
if search(K, 0, 0, check_level=2, break_step=20):
print "K =", K
break
for Kp in range(1, E):
if Kp % 1000 == 0:
print "checking", Kp
if search(K, Kp, 0, check_level=3, break_step=30):
print "Kp =", Kp
break
for Kq in range(1, E):
if Kq % 100 == 0:
print "checking", Kq
if search(K, Kp, Kq, check_level=4, break_step=9999):
print "Kq =", Kq
break
Nous obtenons p, q, c, nous pouvons facilement résoudre le texte brut
import gmpy2
def shuchu(mingwenstr):
if mingwenstr[len(mingwenstr)-1]=='L':
mingwenstr=mingwenstr[2:len(mingwenstr)-1]
else:
mingwenstr=mingwenstr[2:len(mingwenstr)]
if not len(mingwenstr)%2==0:
mingwenstr='0'+mingwenstr
i=len(mingwenstr)
mingwen=""
while i>=1:
str1=mingwenstr[i-2:i]
if int(str1,16)>33 and int(str1,16)<126:
mingwen=chr(int(str1,16))+mingwen
else :
mingwen=" "+mingwen
i=i-2
print mingwen
p = 30804877236372761296348297513767908130120426767441642194038947059431749919743933282721728129660558520306627781991434638545287122418576024822599938752655436891429241798416041881441469038271460545196755187872022209260074336340748692939443634393492611052850561312058115000234467417922716845989845380178291512893577636848676778152648705150749219629638913963012345388388992649857974643758097581431795569765569985118215469798809551704275008726932734117893757436777110974529289423114881289423038562352073193732977840168067817149865622380253870276206212656648830136975036452877460473463818007722056777837507566352911184181643
q = 26038591288856688238001759665609016744197175469090080494077820415283745172609947555684568450035539489682168553390403854805974969118763740560638548072896648612347287461822059996717273680094814363090434263883250281614203478279438635312321752371517752177819983938115532573238089291708699056464231184039223531822571471611431921747169774540943776543504663419138030516108434288911593973010680364553026970545232818747951718950151516127319881685156986937644295056292836729469548074713781625918117631575942194589642230959265894967721587381648790905383499092379075578245308113268969812469233669312409066969648987454629639842309
n=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
assert n==p*q
e=65537
fi=open("flag.txt.en","rb")
miwen=fi.read()
fi.close()
miwenhex=miwen.encode("hex")
miwenint=int(miwenhex,16)
d=gmpy2.invert(e,(p-1)*(q-1))
mingwen=pow(miwenint,d,n)
print len(hex(mingwen)[2:])
print shuchu(hex(mingwen))
4. Le secret du pharaon
Selon le titre, nous supposons qu'il s'agit d'un problème de partage clé
from secretsharing import PlaintextToHexSecretSharer
flags=['1-fddc7d57594928fb74a507ab9cba0b28b92bb6e7b36a9925a105eeddac020e64','3-84f82314003c9690eeacd823b22680ccbe93ac098cabdd0a992c095dde0031cf','5-b0e2e8d2cadc91f8f2f357a42e26aeabaccbfa7731437298ca23d8a4a5424ce4','7-810e7545213971a3c7c2dce3d0998764d0bc1e3b866b15ad0deebaa7abcf64c5','9-b4da0bd03394e4bdfef92f16365e8811d9614f11b99111bcf8a4e68ba79626a2','b-661069e7d491719759a3199be1f65ffb6db92d1b014abb4e33ca7e32f85ee276','d-1f84ab9b467a4ec4de4451ed187987785b567bbdde0126d0722e3335a5307d68','f-9001dc36dd28c5c5dd7333968e7263986f55dd79cd9be286d21f45e46f53c399']
print PlaintextToHexSecretSharer.recover_secret(flags[0:8])5. Erreur parfaite
Selon le titre "Encodage pour éviter la confusion des caractères", je suppose qu'il s'agit d'un encodage en base58, mais la différence est l'ordre des caractères d'encodage de base58. L'ordre des caractères de cette question est "ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz123456789", donc nous mettons directement l'alphabet dans base5yth8. " Changez en alphabet = 'ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz123456789', puis déchiffrez le texte chiffré 'RJv9mjS1bM9MZafGV77uTyDaapNLSk6t358j2Mdf1pbCByX'
