40 points，732 solves

題目給的程式碼

from Crypto.Util.number import getStrongPrime f = [REDACTED] m = int.from_bytes(f,'big') p = getStrongPrime(512) q = getStrongPrime(512) n = p*q e = 65537 d = pow(e,-1,(p-1)*(q-1)) c = pow(m,e,n) print("n =",n) print("p =",p) print("q =",q) print("e =",e) print("c =",c)

題目給的 n，p，q，e，c 值

n = 113138904645172037883970365829067951997230612719077573521906183509830180342554841790268134999423971247602095979484887092205889453631416247856139838680189062511282674134361726455828113825651055263796576482555849771303361415911103661873954509376979834006775895197929252775133737380642752081153063469135950168223

p = 11556895667671057477200219387242513875610589005594481832449286005570409920461121505578566298354611080750154513073654150580136639937876904687126793459819369

q = 9789731420840260962289569924638041579833494812169162102854947552459243338614590024836083625245719375467053459789947717068410632082598060778090631475194567

e = 65537

c = 108644851584756918977851425216398363307810002101894230112870917234519516101802838576315116490794790271121303531868519534061050530562981420826020638383979983010271660175506402389504477695184339442431370630019572693659580322499801215041535132565595864123113626239232420183378765229045037108065155299178074809432

叫我們解RSA加密

phi = (p - 1) * (q - 1) n = p * q d = modinv( e, phi ) m = fastExpMod(c, d, n) print(long_to_bytes(m))

就能解出答案了 b'actf{old_but_still_good_well_at_least_until_quantum_computing}'