Problem: [LitCTF 2024]真·EasyRSA
思路
有点猜谜了属于是
EXP
from Crypto.Util.number import * import gmpy2 c1= 78995097464505692833175221336110444691706720784642201874318792576886638370795877665241433503242322048462220941850261103929220636367258375223629313880314757819288233877871049903331061261182932603536690216472460424869498053787147893179733302705430645181983825884645791816106080546937178721898460776392249707560 c2= 3784701757181065428915597927276042180461070890549646164035543821266506371502690247347168340234933318004928718562990468281285421981157783991138077081303219 n = 111880903302112599361822243412777826052651261464069603671228695119729911614927471127031113870129416452329155262786735889603893196627646342615137280714187446627292465966881136599942375394018828846001863354234047074224843640145067337664994314496776439054625605421747689126816804916163793264559188427704647589521 e=65537 p = gmpy2.iroot(n, 4)[0] phi = p**3*(p-1) d = inverse(e, phi) m1 = pow(c1, d, n) m2 = pow(c2, d, n) print(long_to_bytes(m1)) #print(long_to_bytes(m2)) #b'LitCTF{HeRe_1s_Weak_F1aG}hahahaha_____hint_is_93492332457019255141294502555555489582661562346262162342211605562996217352449' p = 93492332457019255141294502555555489582661562346262162342211605562996217352449 phi = p-1 d = inverse(e, phi) m = pow(c2, d, p) print(long_to_bytes(m))
总结
- 对该题的考点总结