密码学学习笔记之连分数 | Van1sh的小屋 (jayxv.github.io)

勒让德定理，满足定理可用连分数近似

c = 
N1 = 
N2 = 
e = 
cf = continued_fraction(Integer(N1) / Integer(N2))
i = 1
while 1:
    q1 = cf.numerator(i)
    q2 = cf.denominator(i)
    if N1 % q1 == 0 and q1 != 1:
        print(q1)
        p1 = N1 // q1
        d = gmpy2.invert(e,(p1-1)*(q1-1))
        m = pow(c,d,N1)
        flag = long_to_bytes(int(m))
        if b"ISCTF" in flag:
            print(flag)
            break
        
    i += 1