[黑盾杯 2020]Factor

这道题目的n也是可以分解的，分解出来有三个数，需要找出来与e互质的数。再去求解

脚本（找打大佬的）：

from gmpy2 import *
from Crypto.Util.number import *

n = 3454083680130687060405946528826790951695785465926614724373
e = 3
c = 1347530713288996422676156069761604101177635382955634367208

# 对n进行分解，然后取与e互质的因数，得到新的phi和n。然后求得私钥和m
p = [17172929050033177661,11761833764528579549,17100682436035561357]
phi = 1
n = 1
for i in p:
    if gcd(e, i-1) == 1:
        phi *= i-1
        n *= i

d = invert(e, phi)
print(long_to_bytes(powmod(c, d, n)))

运行可得flag