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一、Crypto

NumberTheory

题目源码：

from Crypto.Util.number import *
import hint

flag=b'xxx'
e=65537
p=getPrime(512)
q=getPrime(512)
n=p*q
m=bytes_to_long(flag)
c=pow(m,e,n)
k=getPrime(1024)
assert hint + 233 * k == 233 * k * p
print(n)
print(c)
print(hint)

# 84099006955126261966925371456202769943592466221370095794235167154956697927281125181449320270460637820908574232493978429962263974458426503598700104493216727535451616752760724333653967152401716945549285008242019874215196489846481143398374860288545040874468108191037481101604627874268575884573685952474988256841
# 28098063654079651384124474197746356824080585622155888018279898490747561415908220072536298610509681898119018709183606442183944207485940115624047842734359988590155403601250406116023121958193303908964857108526965815235457652033182982467968474248778435731228104089366239566977364311197776651102290796373095167764
# 411245630228311610573345621334618725748702407327926883063919892785851166202383809662483938501531987094884084543300939673794551515912845363503988032311234800260819110323258416786417746444373651130257247926678135654564298408894174083333804257126735899220917359603430399033328133462456659839525671074605146583034398735379485362144932899212206419889556154825755723979850750847762362288223441051219637465296077020565435562941976546609555729574021362954126496825972439730

解题exp：

from random import randint
from Cryptodome.Util.number import long_to_bytes
from gmpy2 import gcd, invert
e=65537
n=84099006955126261966925371456202769943592466221370095794235167154956697927281125181449320270460637820908574232493978429962263974458426503598700104493216727535451616752760724333653967152401716945549285008242019874215196489846481143398374860288545040874468108191037481101604627874268575884573685952474988256841
c=28098063654079651384124474197746356824080585622155888018279898490747561415908220072536298610509681898119018709183606442183944207485940115624047842734359988590155403601250406116023121958193303908964857108526965815235457652033182982467968474248778435731228104089366239566977364311197776651102290796373095167764
hint=411245630228311610573345621334618725748702407327926883063919892785851166202383809662483938501531987094884084543300939673794551515912845363503988032311234800260819110323258416786417746444373651130257247926678135654564298408894174083333804257126735899220917359603430399033328133462456659839525671074605146583034398735379485362144932899212206419889556154825755723979850750847762362288223441051219637465296077020565435562941976546609555729574021362954126496825972439730
a=2
p = gcd(pow(a, hint, n)-1, n)
q = n // p
phi = (p-1) * (q-1)
d = invert(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

easy_lwe

题目源码：

from Crypto.Util.number import *
from secrets import flag
assert len(flag) == 38

p = getPrime(512)
m = getPrime(512)
while m > p:
    m = getPrime(512)

aa = []
cc = []
bb = []
for i in range(30):
    a = getPrime(512)
    b = getPrime(400)
    c = (a * m + b) % p
    aa.append(a)
    cc.append(c)
    bb.append(b)

enc = pow(m,flag,p)
print(f'p = {p}')
print(f'aa = {aa}')
print(f'cc = {cc}')
print(f'enc = {enc}')

解题exp：

from Crypto.Util.number import *
p= 0x83b05d231fd40ff8ca26b4fb8136dc920754c14412960ce2ec700457861d48fe74f3958fc3a153f77a23fb850ecf0ac1e9722c71b6cc8a104b372cc17bf1528f
rs =[0xf53f440f2e76b60380e68e96508f5dd716b2c3df2ed8265ced83a93fd61a708eeff31fbee9efa22fa7b441cbc406c210de6273f81eb7d093561d5c6394ef2abd,0xe9ecbd5457dcf1bdbc1a852b625c7a8ae6f530e348c2dc0416afd6a375aeb06d4800cc6471ae7d29681715d0407aa8726c32cd35e54960f56b0d9b47a2eed9ed,0x86a261c1590a774ced6c7db439e53e4068a8dbc0ad111a0e0371e8731fb939a068348d035da04afb9a3914a011574e35cefc4d5c5740f7cbc27459d944f51d15,0xba99a871d8f805e3c0dcd4e04cee66ec5e213a7902a65f2faa8e86368e56c42d09fde536b07471fff8f72db922725a24d6288d1bfe9edc2cd76b756eb464e1eb,0xa99efeb79377db6baa8787b5d2d2ebf123d8ec77e820d1b88644883a07c38498aa08df82cb9802c7fa128a5fa08b66c56a0805f70a78b7cb45b1a74bde095165,0xe6341800304b0a6f2de941432bb253f53a3c73c7b0f0382fd1ec4c882da5fd1b5151c619f9279de3767ae03af387c495c50c8c0ff79fd6c8acb51bd0b16afc23,0xa8670f7142e5d90e781e335d5e870188b94288defd8302c1b183bf20b5a3720be1ae1afe0ad937bc4678727e1daf262194c430086f3810447dfce721c8fb36a5,0xe9d150d4e7a7f82e89d26fcaa003579c8dfc25f4a50794175ae0a4407dd33e87eb5fded328155009548c002d2a3198afa356e1692a6bb820d8ae71ca6e506b35,0xd7ec72dd643449f5d21a32a11e2458d292c524b91b4b1a8515a5f8351717813f51a55c5675aa0eaa5c80a32d26aa385d7425e7d3e1e50936b2744a1534e3c7b7,0xaf5d8062f6bc03875b5fb10d0888d586cb08fa62709910cd7d931201d4d833d31d935003b801cac4fd51ef1e4db3bc13e41d740d7881560c0942faf9acca55d5,0xf63c4a5add0474bf6c4fb6617a2965e2474a2902c59bcc4243d5dcffa0f1aa0a6136e9c9093ea05d84a26888aaad63ec652602200141abdfcdc0d1912d208dbb,0xd65c8ef9b01085819f2f5f2dfbb7641521966600ddd7a03b886320328144ec42ed5585206744c8e1b5beb5502aff6f0cd01934890926bbb5387363b321dbb6bd,0x9bd7a0e9d4940126c9668de7e29e198ac38010d505b90e92ee560307b8a134545bbd0277f14e7651e91c5a9362207097e17888b9a49a889c672f76681f41962f,0xff387c4e94f47cba242936e80d620d3eeb203a1c7f365ae178b33b29ff3d8c2f932733d0605876c23942bae7f096b3ee457d02758796bda01bbcd3bd2e1229a1,0xbd6c24205cf5b9b2acc909296369ca77ef34ceac5f9900742a7ec37c00b94a56418d168d576d33383d9f782386524c4cecebc3a9fa68c81a5a6867de564d07e3,0x8850ac2051b6a7450f228e676abbfd510ce08a43a0b791182aedcfb6b3f1d478e3dc953e59eb99fe370f71d52af3a1625e0700078927f8a5919becbfa60af96b,0xfa474444b5372681c5b7316871cf9a93306b4d6a3eac2492c71780eb6903bd4ff77b1c4a28608dfa10c0c8b8bfc23942fc0a8ec64d3967504621e692eafcf4af,0xfd7f772c719929d33f15ce0122f6efc278b728d75dbd16a343c649e49118a79084d169d6db1b6b859e1f4c82694a850622dcdff23c8fbd7d0a736a409b94f471,0xedf6566ecd7f0a0beaa2f45a2a509358c9537dbf772e6f63ccb0e21b5c5c3dae0939e9e15c9bcdf80f7875e74ebfba16f0660737719c8ad435981b5b8df89a23,0xd7e718dafb753084cb7c1deee139f6e158a0228bbf9b0ae5ed3a8b3d2508ce62e4e32cd43a3135540b31e052e6e3a0dba3ae623c69d74da34387c3429b6f0487,0xbff5ab718c54909a34ba8ab5787cbfb0c9c57394350b71744d5751f577cefda87cc8263b21dd2c21a8f19fff8362c3371fcb054dddb1293df7d6efe2d661e4ab,0xb946ad47cf1f9186c1eeaeed12f21b41a29f9fb74b14f577c97731004b372c5f41023aafc9b93557c984ff87dd263b293aa500c3a2ce1f0815319263cf7c42d5,0x9b93853dc5f4c052fdda339f69c625bc98ffdbde439078458b84278e34542f9110863e5beef166a766f565f7a815ec4a462510b42f81454b65e648622c610b27,0xae050aaa86b9580747efa33561973597d7d7a0684a45219eed3224ebc37fdf067ed61a29ecaa501266c9bcdd16b850c0dd40da3f964a5b03ae60c3967fd55913,0xea57b0482a876d49a3e21bb885bda0aae21f5fcf2521f7e1db771998b6b639d0833c17d7c8c67c12ce9b60e4210e068f98344a8cabc5faa17a048100bab797bb,0x9f866ad27c9ea2fb29168ec6db0f7755221484d0c89bfef0cd70fc2ed14fda15f7b59bbca2e23f40b5effeb9e53ad821e06cf986b34da4407c85bfbf78d2920b,0xab0939605dcff5d1ec8405b6daeb65eccc6e3b5956601dddea95c6310ac32bfeec1bab6b83e85371078a16ea9489050175098d39488aaed2a190c647fe2b1b69,0xe85698e04d8bedbac2c7884c914b7026dbb1dd5134c4a5a8e7541b07c8a94ab3d2f12eb7f1171ccd564054e1dc63d5ac044e5c5552870b419fef35a572199239,0xea94663e103e4354fb7feb80b11d06c7e16feb7265f69ee882180baba70fa075df24e3fe1ad12a99f054cabf4a3f5e4823416d4c4daa02ca51ce3926034186db,0x93662718294005f4ac8f79b0799e240a05eea871ef07d623ea7c68ef818b3b55fb4f9b6f06c399726e59cec03389053b448f187404cddb93cb3c55e3d12cebc9]
cs =[0x4aae29aaaab89fbc672db400c41d1ad3ffce937e7810065bed552c12101fb778046d22b00c05bbc3f61825b5af3c3e57f1abaafe3d9a58a573a905e2a1cecde7,0x502a1c23ef44f174aaedb9a49705eb72f805bab13e82d599525ea7484cc11f2e7c4475526b4be344390e46bcb8bcdaba2768c6321f8ca5482666171eee498f28,0x546f2276b59b7186ef5b9a04e0ba2691648d005fc780303411a1082ea3b05be127a10e26921a4b84b14d8acc45a6c32a0142ac6eb396415ecb5841a01b775b5d,0x43986eb7208b66dd86c12f953e10b2d3907873151170278393b6a4f7ea518ab5745e2db5f4dbec84087c4817c5df10e743f35ed1190515aba34832b1b274bf2e,0x7b2e448342bc7409ed891cdbf5137014f417866097297302085d8800458495e374fba8398d069f1c1c7792a9f03194e36921c378308de18313fb9b62db45cc0a,0x41ac8f265bc96ae868256fb08caf0a43a547346522b5d90cb9489c87cd5d726447d20354332dd3cb771003eddfa9c4bfa6923ae45ac8c0994f7cce46a3302eea,0x51b59ad39b388fdb2056279e2de02d32d36b52da1cc1fe4f6843964273b4585704e21405e2528e5894bd4f

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