攻防世界 cr4-poor-rsa

題目

給了一個壓縮包，內有文件：flag.b64（打開發現是base64編碼的）和key.pub

解題過程

1. 首先利用Crypto.PublicKey的RSA模塊從key.pub中獲取公鑰信息，這個RSA相關知識點中已有介紹
2. 得到n后可以拿到大數分解，得到

p = 863653476616376575308866344984576466644942572246900013156919
q = 965445304326998194798282228842484732438457170595999523426901

3. 由p,q,e得到d（RSA相關知識點中已有介紹）
4. rsa庫的PrivateKey生成私鑰
5. 打開flag.b64並用base64解碼，再利用key解密即可得到flag

from Crypto.PublicKey import RSA
from gmpy2 import invert
import rsa
from base64 import b64decode

f = open("F:\\ChromeCommon\\key.pub","rb").read()
pub = RSA.importKey(f)
n = pub.n
e = pub.e

p = 863653476616376575308866344984576466644942572246900013156919
q = 965445304326998194798282228842484732438457170595999523426901
d = int(invert(e, (p-1)*(q-1)))

key = rsa.PrivateKey(n,e,d,p,q)
f = open("F:\\ChromeCommon\\flag.b64",'r').read()
c = b64decode(f)
flag = rsa.decrypt(c,key)
print(flag)

ALEXCTF{SMALL_PRIMES_ARE_BAD}