ecc1

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from Crypto.Util.number import *
import gmpy2
c_hex = '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'
n_hex = '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'
e_hex = '10001'

c = int(c_hex, 16)
n = int(n_hex, 16)
e = int(e_hex, 16)

p = gmpy2.gcd(c,n)
q = n//p
phi = (q-1)*(p-1)
d = gmpy2.invert(e,phi)
M = pow(c,d,n)
m = M//(2022*1011*p)
print(long_to_bytes(m))

16进制转码

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c = M^e mod n
c = M^e + kqp
c mod p = 0
n = p*q
p是c的因子,所以求最大公因数得p,剩余正常rsa解密
p = gcd(c,p)

结果:b’977260aae9b5}’