buu [BJDCTF2020]easyrsa
下载附件是一个py文件,打开之后,发现是常规的rsa,不过有几个函数不知道。
这里记录一下,
Fraction(a,b) 相当于 a/b
Derivative(f(x),x) : 当x='x’时,f(x)的导数值
from Crypto.Util.number import getPrime,bytes_to_long
from sympy import Derivative
from fractions import Fraction
from secret import flag
p=getPrime(1024)
q=getPrime(1024)
e=65537
n=p*q
z=Fraction(1,Derivative(arctan(p),p))-Fraction(1,Derivative(arth(q),q))
m=bytes_to_long(flag)
c=pow(m,e,n)
print(c,z,n)
'''
output:
7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441
'''
那这里Derivative(artan§,p)相当于是1/1+p^2,而另一边是1/1-p ^2 ,再倒一下,那么这个z实际上就是p ^ 2+q ^2
n又是p*q
这里写脚本用到一些gmpy2的库函数,这里也记录一下
Encoding=UTF-8
import gmpy2
# gmpy2.mpz(x)
# 初始化一个大整数x
gmpy2.mpfr(x)
# 初始化一个高精度浮点数x
C = gmpy2.powmod(M,e,n)
# 幂取模,结果是 C = (M^e) mod n
d = gmpy2.invert(e,n) # 求逆元,de = 1 mod n
gmpy2.is_prime(n) # 判断n是不是素数
gmpy2.gcd(a,b) # 欧几里得算法
gmpy2.gcdext(a,b) # 扩展欧几里得算法
gmpy2.iroot(x,n) # x开n次根
from Crypto.Util.number import getPrime,bytes_to_long
from sympy import Derivative
from fractions import Fraction
from gmpy2 import *
def num2str(n):
tmp=str(hex(n))[2:]
if len(tmp)%2==0:
pass
else:
tmp='0'+tmp
s=''
for i in range(0,len(tmp),2):
temp=tmp[i]+tmp[i+1]
s+=chr(int(temp,16))
return s
c=7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
z=32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
n=15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441
p_plus_q=iroot(z+2*n,2)[0]
p_sub_q=iroot(z-2*n,2)[0]
e=65537
p=(p_plus_q+p_sub_q)//2
q=(p_plus_q-p_sub_q)//2
d=invert(e,((p-1)*(q-1)))
m=pow(c,int(d),n)
print(num2str(m))
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