ciscn 2020 Crypto bd

Start

简单RSA

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

m = bytes_to_long(flag)

p = getPrime(512)
q = getPrime(512)
N = p * q
phi = (p-1) * (q-1)
while True:
    d = getRandomNBitInteger(200)
    if GCD(d, phi) == 1:
        e = inverse(d, phi)
        break

c = pow(m, e, N)

print(c, e, N, sep='\n')

# 37625098109081701774571613785279343908814425141123915351527903477451570893536663171806089364574293449414561630485312247061686191366669404389142347972565020570877175992098033759403318443705791866939363061966538210758611679849037990315161035649389943256526167843576617469134413191950908582922902210791377220066
# 46867417013414476511855705167486515292101865210840925173161828985833867821644239088991107524584028941183216735115986313719966458608881689802377181633111389920813814350964315420422257050287517851213109465823444767895817372377616723406116946259672358254060231210263961445286931270444042869857616609048537240249
# 86966590627372918010571457840724456774194080910694231109811773050866217415975647358784246153710824794652840306389428729923771431340699346354646708396564203957270393882105042714920060055401541794748437242707186192941546185666953574082803056612193004258064074902605834799171191314001030749992715155125694272289

这里看到输出的e很大,维纳攻击Wiener-attack
照着脚本改就行了:

exp.py:

'''
Created on Dec 14, 2011

@author: pablocelayes
'''
from Crypto.Util.number import *
import ContinuedFractions, Arithmetic, RSAvulnerableKeyGenerator

def hack_RSA(e,n):
    '''
    Finds d knowing (e,n)
    applying the Wiener continued fraction attack
    '''
    frac = ContinuedFractions.rational_to_contfrac(e, n)
    convergents = ContinuedFractions.convergents_from_contfrac(frac)
    
    for (k,d) in convergents:
        
        #check if d is actually the key
        if k!=0 and (e*d-1)%k == 0:
            phi = (e*d-1)//k
            s = n - phi + 1
            # check if the equation x^2 - s*x + n = 0
            # has integer roots
            discr = s*s - 4*n
            if(discr>=0):
                t = Arithmetic.is_perfect_square(discr)
                if t!=-1 and (s+t)%2==0:
                    print("Hacked!")
                    return d


if __name__ == "__main__":
    c = 37625098109081701774571613785279343908814425141123915351527903477451570893536663171806089364574293449414561630485312247061686191366669404389142347972565020570877175992098033759403318443705791866939363061966538210758611679849037990315161035649389943256526167843576617469134413191950908582922902210791377220066
    e = 46867417013414476511855705167486515292101865210840925173161828985833867821644239088991107524584028941183216735115986313719966458608881689802377181633111389920813814350964315420422257050287517851213109465823444767895817372377616723406116946259672358254060231210263961445286931270444042869857616609048537240249
    N = 86966590627372918010571457840724456774194080910694231109811773050866217415975647358784246153710824794652840306389428729923771431340699346354646708396564203957270393882105042714920060055401541794748437242707186192941546185666953574082803056612193004258064074902605834799171191314001030749992715155125694272289
    d = hack_RSA(e, N)
    m = pow(c, d, N)
    flag = long_to_bytes(m).decode()
    print(flag)


# flag{d3752538-90d0-c373-cfef-9247d3e16848}

End

你可能感兴趣的:(CTF,Reverse,国赛)