rsa-apprentice

一只路过的迅猛龙 2026-04-12 22:28:19 2 0 返回题目详情


分解n得到

p=1033247481589406269253
q=1177043968824330681533

参考代码:

from Crypto.Util.number import *

from gmpy2 import *
n = 1216177716507739302616478655910148392804849
c1 = 257733734393970582988408159581244878149116
c2 = 843105902970788695411197846605744081831851
e = 65537
p=1033247481589406269253
q=1177043968824330681533
phi=(p-1)*(q-1)
d=pow(e,-1,phi)
m1=pow(c1,d,n)
m2=pow(c2,d,n)
flag1=long_to_bytes(m1)
flag2=long_to_bytes(m2)
print(flag1+flag2)

分类:Crypto
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作者:一只路过的迅猛龙

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