经典的rsa公钥加密，但跟之前的RSA题目不一样，首先flag.enc有base64加密，所以要先解密才能得到c的值
首先可以通过在线解密解出n，e也可以通过代码得到
公钥解析
代码：
# coding=UTF-8
import Crypto.PublicKey.RSA
f = open('D:\\ctf\\buuctf\\kldrsa\\public.key','rb').read()
pub = Crypto.PublicKey.RSA.importKey(f)
n = pub.n
e = pub.e
print(n,'\n',e)可以得到n和e
再通过yafu将n分解
得到p,q这样就可以得到d
这个时候我用之前的方法发现
全是乱码！
看了大佬的wp，用大佬的代码才得到结果
代码：
from Crypto.PublicKey import RSA
import gmpy2
import base64
from Crypto.Util.number import *
from Crypto.Cipher import PKCS1_OAEP
n=79832181757332818552764610761349592984614744432279135328398999801627880283610900361281249973175805069916210179560506497075132524902086881120372213626641879468491936860976686933630869673826972619938321951599146744807653301076026577949579618331502776303983485566046485431039541708467141408260220098592761245010678592347501894176269580510459729633673468068467144199744563731826362102608811033400887813754780282628099443490170016087838606998017490456601315802448567772411623826281747245660954245413781519794295336197555688543537992197142258053220453757666537840276416475602759374950715283890232230741542737319569819793988431443
e=65537
p=3133337
q=25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
text='GVd1d3viIXFfcHapEYuo5fAvIiUS83adrtMW/MgPwxVBSl46joFCQ1plcnlDGfL19K/3PvChV6n5QGohzfVyz2Z5GdTlaknxvHDUGf5HCukokyPwK/1EYU7NzrhGE7J5jPdi0Aj7xi/Odxy0hGMgpaBLd/nL3N8O6i9pc4Gg3O8soOlciBG/6/xdfN3SzSStMYIN8nfZZMSq3xDDvz4YB7TcTBh4ik4wYhuC77gmT+HWOv5gLTNQ3EkZs5N3EAopy11zHNYU80yv1jtFGcluNPyXYttU5qU33jcp0Wuznac+t+AZHeSQy5vk8DyWorSGMiS+J4KNqSVlDs12EqXEqqJ0uA=='
c_bytes = base64.b64decode(text)
rsa_components=(n,e,int(d),p,q)
arsa=RSA.construct(rsa_components)
rsakey=RSA.importKey(arsa.exportKey())
rsakey=PKCS1_OAEP.new(rsakey)
decrypted=rsakey.decrypt(c_bytes)
print(decrypted)得到flag