知识点：共享素数 n1，n2的最大公约数是q
注意点：注意加密先后 脚本先使用n1加密 再使用n2
所以解密时先用n2解密 再用n1
`from Crypto.Util.number import *
n1 = 103835296409081751860770535514746586815395898427260334325680313648369132661057840680823295512236948953370895568419721331170834557812541468309298819497267746892814583806423027167382825479157951365823085639078738847647634406841331307035593810712914545347201619004253602692127370265833092082543067153606828049061
n2 = 115383198584677147487556014336448310721853841168758012445634182814180314480501828927160071015197089456042472185850893847370481817325868824076245290735749717384769661698895000176441497242371873981353689607711146852891551491168528799814311992471449640014501858763495472267168224015665906627382490565507927272073
e = 65537
c = 60406168302768860804211220055708551816238816061772464557956985699400782163597251861675967909246187833328847989530950308053492202064477410641014045601986036822451416365957817685047102703301347664879870026582087365822433436251615243854347490600004857861059245403674349457345319269266645006969222744554974358264
q = GCD(n1,n2)
p1 = n1//q
p2 = n2//q
phi_n1 = (p1-1)(q-1)
phi_n2 = (p2-1)(q-1)
d1 = inverse(e,phi_n1)
d2 = inverse(e,phi_n2)
c2 = pow(c,d2,n2)
c1 = pow(c2,d1,n1)
print(long_to_bytes(c1))`
标签：Bigrsa,pow,phi,羊城,2021,n1,n2,d2,d1 From： https://www.cnblogs.com/sauy/p/18525969