Road_Safety_Association

That’s what it stands for, right?

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题目给了c,p,q,e的值,直接求d解密即可 代码:

import gmpy2
from Crypto.Util.number import long_to_bytes

c = 34709089913401150635163820358938916881993556790698827096314474131695180194656373592831158701400832173951061153349955626770351918715134102729180082310540500929299260384727841272328651482716425284903562937949838801126975821205390573428889205747236795476232421245684253455346750459684786949905537837807616524618
p = 7049378199874518503065880299491083072359644394572493724131509322075604915964637314839516681795279921095822776593514545854149110798068329888153907702700969
q = 11332855855499101423426736341398808093169269495239972781080892932533129603046914334311158344125602053367004567763440106361963142912346338848213535638676857
e = 65537

n = p * q
phi = (p-1) * (q-1)
d = gmpy2.invert(e, phi)
x = pow(c, d, n)
print(x)
print(long_to_bytes(x))

flag:kqctf{y0uv3_6r4du473d_fr0m_r54_3l3m3n74ry_5ch00l_ac8770bdcebc}