[{"c": 7366067574741171461722065133242916080495505913663250330082747465383676893970411476550748394841437418105312353971095003424322679616940371123028982189502042, "e": 10, "n": 25162507052339714421839688873734596177751124036723831003300959761137811490715205742941738406548150240861779301784133652165908227917415483137585388986274803},
{"c": 21962825323300469151795920289886886562790942771546858500842179806566435767103803978885148772139305484319688249368999503784441507383476095946258011317951461, "e": 10, "n": 23976859589904419798320812097681858652325473791891232710431997202897819580634937070900625213218095330766877190212418023297341732808839488308551126409983193},
{"c": 6569689420274066957835983390583585286570087619048110141187700584193792695235405077811544355169290382357149374107076406086154103351897890793598997687053983, "e": 10, "n": 18503782836858540043974558035601654610948915505645219820150251062305120148745545906567548650191832090823482852604346478335353784501076761922605361848703623},
{"c": 4508246168044513518452493882713536390636741541551805821790338973797615971271867248584379813114125478195284692695928668946553625483179633266057122967547052, "e": 10, "n": 23383087478545512218713157932934746110721706819077423418060220083657713428503582801909807142802647367994289775015595100541168367083097506193809451365010723},
{"c": 22966105670291282335588843018244161552764486373117942865966904076191122337435542553276743938817686729554714315494818922753880198945897222422137268427611672, "e": 10, "n": 31775649089861428671057909076144152870796722528112580479442073365053916012507273433028451755436987054722496057749731758475958301164082755003195632005308493},
{"c": 17963313063405045742968136916219838352135561785389534381262979264585397896844470879023686508540355160998533122970239261072020689217153126649390825646712087, "e": 10, "n": 22246342022943432820696190444155665289928378653841172632283227888174495402248633061010615572642126584591103750338919213945646074833823905521643025879053949},
{"c": 1652417534709029450380570653973705320986117679597563873022683140800507482560482948310131540948227797045505390333146191586749269249548168247316404074014639, "e": 10, "n": 25395461142670631268156106136028325744393358436617528677967249347353524924655001151849544022201772500033280822372661344352607434738696051779095736547813043},
{"c": 15585771734488351039456631394040497759568679429510619219766191780807675361741859290490732451112648776648126779759368428205194684721516497026290981786239352, "e": 10, "n": 32056508892744184901289413287728039891303832311548608141088227876326753674154124775132776928481935378184756756785107540781632570295330486738268173167809047},
{"c": 8965123421637694050044216844523379163347478029124815032832813225050732558524239660648746284884140746788823681886010577342254841014594570067467905682359797, "e": 10, "n": 52849766269541827474228189428820648574162539595985395992261649809907435742263020551050064268890333392877173572811691599841253150460219986817964461970736553},
{"c": 13560945756543023008529388108446940847137853038437095244573035888531288577370829065666320069397898394848484847030321018915638381833935580958342719988978247, "e": 10, "n": 30415984800307578932946399987559088968355638354344823359397204419191241802721772499486615661699080998502439901585573950889047918537906687840725005496238621}]
題目直接給了10個c,n,e其监，并且e都是一樣的造烁，所以可以知道這題是RSA廣播攻擊盗痒，需要利用中國剩余定理去解出m，直接參考別人的代碼
# -*- coding:utf8 -*-
import gmpy # gmpy 是一個在 python 提供類似 GMP 的高等算術(shù)的 module份招，這裡用它算模反元素跟次方根
import json, binascii # json 是 python 下的 json parsing 工具，binascii 則是用來做 binary 跟 ascii 之間轉(zhuǎn)換的工具
from functools import reduce
def chinese_remainder(n, a): # Rosetta Code 上的 CRT Solver code圆丹，就只是把中國餘數(shù)定理 code 化而已
sum = 0
prod = reduce(lambda a, b: a*b, n)
for n_i, a_i in zip(n, a):
p = prod // n_i
sum += a_i * modinv(p, n_i) * p
return int(sum % prod)
def modinv(a, m): return int(gmpy.invert(gmpy.mpz(a), gmpy.mpz(m))) # 用 gmpy 算模反元素界逛，回傳轉(zhuǎn)成 int 的結(jié)果
with open("mayday.json") as dfile: #題目給的n,e,c
data = json.loads(dfile.read()) # 打開檔案，讀成 json
data = {k:[d.get(k) for d in data] for k in {k for d in data for k in d}} # 從 [{c1, e1, n1}, {c2, e2, n2}] 轉(zhuǎn)成 {"c": [c1, c2], "e": [e1, e2], "n": [n1, n2]}
t_to_e = chinese_remainder(data['n'], data['c']) # 用中國餘式定理解同餘方程組派继，推出原先的 t^e
t = int(gmpy.mpz(t_to_e).root(10)[0]) # 算 t^e 的 10 次方根（因為 e=10）宾袜，推回原本的 t
print(binascii.unhexlify(hex(t)[2:-1])) # 把結(jié)果從數(shù)字先轉(zhuǎn)成 hex 再轉(zhuǎn)成字串
參考資料：https://github.com/pcchou/ctf-writeups/tree/master/2016-tw-edu-ctf/mayday
