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BUUCTF-RSA roll

2022-07-27 21:07:00 【[email protected]】

1.题目

{920139713,19}

704796792
752211152
274704164
18414022
368270835
483295235
263072905
459788476
483295235
459788476
663551792
475206804
459788476
428313374
475206804
459788476
425392137
704796792
458265677
341524652
483295235
534149509
425392137
428313374
425392137
341524652
458265677
263072905
483295235
828509797
341524652
425392137
475206804
428313374
483295235
475206804
459788476
306220148

2.复现

直接分解n:

import gmpy2
import  libnum
n=920139713
e=19
c=[704796792,752211152,274704164,18414022,368270835,483295235
   ,263072905
   ,459788476
   ,483295235
   ,459788476
   ,663551792
   ,475206804
   ,459788476
   ,428313374
   ,475206804
   ,459788476
   ,425392137
   ,704796792
   ,458265677
   ,341524652
   ,483295235
   ,534149509
   ,425392137
   ,428313374
   ,425392137
   ,341524652
   ,458265677
   ,263072905
   ,483295235
   ,828509797
   ,341524652
   ,425392137
   ,475206804
   ,428313374
   ,483295235
   ,475206804
   ,459788476
   ,306220148]
p=18443
q=49891
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
flag=''
for i in c:
    m=pow(i,d,n)
    flag=flag+chr(m)
print(flag)
#flag{13212je2ue28fy71w8u87y31r78eu1e2}

原网站

版权声明
本文为[[email protected]]所创，转载请带上原文链接，感谢
https://blog.csdn.net/qq_61774705/article/details/124647420

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