[LitCTF 2023]yafu (中级)

题目分数：411

题目标签： 现代密码RSAyafu

题目描述：

你知道yafu分解吗？
出题人：3tefanie丶zhouFlag格式 NSSCTF{}

附件：

from Crypto.Util.number import *
from secret import flag

m = bytes_to_long(flag)
n  = 1
for i in range(15):
    n *=getPrime(32)
e = 65537
c = pow(m,e,n)
print(f'n = {n}')
print(f'c = {c}')
'''
n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307
c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717
'''

解题：

import gmpy2
from Crypto.Util.number import long_to_bytes

n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307

e = 65537
c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717

p1=2151018733
p2=2201440207
p3=2315495107
p4=2585574697
p5=2719600579
p6=2758708999
p7=2767137487
p8=2906576131
p9=2923522073
p10=3354884521
p11=3355651511
p12=3989697563
p13=4021078331
p14=4044505687
p15=4171911923


phi = (p1 - 1) * (p2 - 1) * (p3 - 1) * (p4 - 1) * (p5 - 1) * (p6 - 1) * (p7 - 1) * (p8 - 1) * (p9 - 1) * (p10 - 1) * (p11 - 1) * (p12 - 1) * (p13 - 1) * (p14 - 1) * (p15 - 1)
d = gmpy2.invert(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m))
# LitCTF{Mu1tiple_3m4ll_prim5_fac7ors_@re_uns4f5}