PWHhub2022冬季赛密码学习

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from gmpy2 import *
from Crypto.Util.number import *
from secret import flag
m = bytes_to_long(flag)

R = getPrime(256)
S = getPrime(512)
A = getPrime(1024)

N = R * S * A
c = pow(m, 0x10001, N) # c = m^e mod N

RA = R & A
print('RSA1',hex(RA * S))
print('RSA2',hex(RA | S))

print('c', hex(c))
print('N',hex(N))

# RSA1 0x97be543979cb98c109103fa118c1c930ff13a6b2562166417021afd6e46cb0837a5cc5f4094fcea5fcc33efdfa495050e0fb8269922b3ee2d403210ed1ba339af2dc3d4e8952f0c784fcc655436cf255b98cdaf8080df47f6c28bc0bae68c713
# RSA2 0xa887aa84f3a0bd8b79ed59a7bb98d8e58a85414f85cf2ddf53ff4bd9294bfdadf7d6d6adfe7fbed55fc71b5a6bfcfe79ced27e2f41e7546a8679daf5b63dda37
# c 0x2f62fb7e7e8e27823193119f8412050ade9084ade25261a5875da23a07d5d5145e72d460697984d8aa668a25822009a4fdc85df2b208941cd3219b312f21c3c7bc4ef7aa8c18b4f91a0e815fe1892fca0f72406e571fbd0fea2c4710c601165ccd7e8a5a828721a5e2c956b732223d683d1413ef393b5f80a431c52bf9099e22b8e27daafb9d3e055242b89b5419b8925744ccf348e1bea519225af8efe7dbcc202425251039cbfe6b892a7fcf7e9d72224ea9381e3fb32ab837139af4b4112a3c7a6571c88e7d6c5db4c3f91e25edd15eb5544ef2f29a9e1bb1062ec86f1902
# N 0x58a7ff25292651e1a8d82656d64fe3b458d6e688405e85aa6c02e0c33469ad3dbaef6c6eaf8faf22f2d15e80856ab7b90a40fd50c36f7b59932bc94e6fb4fabefa87b11bf4ef74df4ccf8d254f0c6812628df3c5b3786af35e3dde9c87b462d1a565af6f100750718ccb7235174947f00cec5836765150f1680d0c58a5f9ea2473a6033c218c75664dc53377dde9386f37e1a89d77e61a716129d290c5a41f81cd3490bab6fe51f232ab27cb1ac9c8eb88e908c12109a125b7439c25b6879283a17a3467823fbb089709eb836cfd03386cc4bf186eb45401472ab0bdec605fd7

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from Crypto.Util.number import *
from z3 import *
from gmpy2 import *
# from sage.all import *
from math import gcd

rsa1 = 0x97be543979cb98c109103fa118c1c930ff13a6b2562166417021afd6e46cb0837a5cc5f4094fcea5fcc33efdfa495050e0fb8269922b3ee2d403210ed1ba339af2dc3d4e8952f0c784fcc655436cf255b98cdaf8080df47f6c28bc0bae68c713
rsa2 = 0xa887aa84f3a0bd8b79ed59a7bb98d8e58a85414f85cf2ddf53ff4bd9294bfdadf7d6d6adfe7fbed55fc71b5a6bfcfe79ced27e2f41e7546a8679daf5b63dda37
c = 0x2f62fb7e7e8e27823193119f8412050ade9084ade25261a5875da23a07d5d5145e72d460697984d8aa668a25822009a4fdc85df2b208941cd3219b312f21c3c7bc4ef7aa8c18b4f91a0e815fe1892fca0f72406e571fbd0fea2c4710c601165ccd7e8a5a828721a5e2c956b732223d683d1413ef393b5f80a431c52bf9099e22b8e27daafb9d3e055242b89b5419b8925744ccf348e1bea519225af8efe7dbcc202425251039cbfe6b892a7fcf7e9d72224ea9381e3fb32ab837139af4b4112a3c7a6571c88e7d6c5db4c3f91e25edd15eb5544ef2f29a9e1bb1062ec86f1902
N = 0x58a7ff25292651e1a8d82656d64fe3b458d6e688405e85aa6c02e0c33469ad3dbaef6c6eaf8faf22f2d15e80856ab7b90a40fd50c36f7b59932bc94e6fb4fabefa87b11bf4ef74df4ccf8d254f0c6812628df3c5b3786af35e3dde9c87b462d1a565af6f100750718ccb7235174947f00cec5836765150f1680d0c58a5f9ea2473a6033c218c75664dc53377dde9386f37e1a89d77e61a716129d290c5a41f81cd3490bab6fe51f232ab27cb1ac9c8eb88e908c12109a125b7439c25b6879283a17a3467823fbb089709eb836cfd03386cc4bf186eb45401472ab0bdec605fd7
e = 0x10001

# S = gcd(N, rsa2) 因为 rsa1 和 N 都有相同的因子 (S == Because both rsa1 and N has same factor S. by BaakingDog
S = gcd(N, rsa1)
# 这里想尝试分解pq, p*q太大未遂
# print(N//S)
# 非预期解 常规解法应该求出N的全部因子 (R, A) 然后d = e^-1 mod phi(N); m = c^d mod N (明文就求出来了)
# 而这里因为m比s小, 所以在模n和模s下求出来的m是一样的, 于是把s当n算
d = inverse(e, S - 1) # d = e^-1 mod phi(N) 求逆元
m = pow(c, d, S) # 把s当n算 m = c^d mod N
print(S > m) # True 加以验证
print(bytes.fromhex(hex(m)[2:])) # 转为字符

# print(long_to_bytes(m)) # 另一种转为字符的方式