2023 柏鹭杯 --- Crypto fractrsa wp

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题目

#python3
import sys 
sys.path.append("..") 
from Crypto.Util.number import *
from random import *
from sage.all import *
from  secret import flag1 as flagnum1 = 3
num2 = 5
while(num1<num2):num1 = getPrime(512)num2 = getPrime(512)
pt = bytes_to_long(flag) + num2ring = RealField(1100)
num3 = ring(num1) / ring(num2)
print("num3 = ", num3)while True:p = randint(2**511, num1)q = randint(2**511, num2)if isPrime(p) and isPrime(q) and p!=q:breakN = p*q
e = 65537
leak = pow(p-q, num1, num1*num2)
ct = pow(pt, e, N)print("ct = ", ct)
print("N = ", N)
print("leak = ", leak)"""
num3 = 1.23389923415003373900567515471436168841941584796842188964423737295914869304653496800649965063081353720701415762591488370228399019899893688681309320356016722276295236528757306976510687729729934668311830828756908988350841843676900575414367123810470585198055372776278588638204471298838884740198056387082949710435502826460830711429956
ct = 31011170589632318837149853165664224847925206003567781692767655474759523146503572164952138829336342836023903919700264739071138739105931471740973631326608186969523753119546323993892359278563753903149741128282349467136720827132122619177620866305659196267641453819504766216964516467658995724859657544518337771393
N = 61860727516406742636690805639158184396057779906729165734489212939937929906456706343476469874085504076991779041906401043694401076841639925611957258119417559980829238154105119701407722069260962772947894516879731956778127512764229384957918619863998939985369399189275568362193066167855420897196095587732512368673
leak = 23213363443983005040318061737977092634638640953366787443691593387275645092922646169818923792205696350020369122807136306157118385984272980615310163206933078119776935167207473544453080959202803743994251355133953187110546017667004996272367137522351606700447920805532616096125523674597551449412004735397779511371
"""

解题过程

将num3转为连分数形式,$num3=\frac{num1}{num2}$，再遍历其收敛值列表，其值满足512bit且是素数的则为num1和num2

#sage
num3 = 1.23389923415003373900567515471436168841941584796842188964423737295914869304653496800649965063081353720701415762591488370228399019899893688681309320356016722276295236528757306976510687729729934668311830828756908988350841843676900575414367123810470585198055372776278588638204471298838884740198056387082949710435502826460830711429956
cf = continued_fraction(num3)
frac_list = cf.convergents()
for i in frac_list:tmp = str(i).split('/')if len(tmp)==2:num1 = int(tmp[0])num2 = int(tmp[1])if num1.bit_length()==512 and num2.bit_length()==512 and is_prime(num1) and is_prime(num2):print(num1)print(num2)

得到num1和num2

num1 = 11167377337790397338811417806698264734026040696284907854286100186126887838302430726803014418419121360514985339992064951270502853852777225947659429837569693num2 = 9050477566333038464101590216458863799039754468566791821195736389139213194857548339787600682491327798736538059818887575696704421576721592454156775006222517

利用$leak$解出$p-q$，且已知$n=p\ast q$，可构建一个方程组，解方程得到p,q。解出$pt$之后减去$num2$即可得到flag

from Crypto.Util.number import *
import gmpy2
from sympy import *num1 = 11167377337790397338811417806698264734026040696284907854286100186126887838302430726803014418419121360514985339992064951270502853852777225947659429837569693
num2 = 9050477566333038464101590216458863799039754468566791821195736389139213194857548339787600682491327798736538059818887575696704421576721592454156775006222517
num3 = 1.23389923415003373900567515471436168841941584796842188964423737295914869304653496800649965063081353720701415762591488370228399019899893688681309320356016722276295236528757306976510687729729934668311830828756908988350841843676900575414367123810470585198055372776278588638204471298838884740198056387082949710435502826460830711429956
ct = 31011170589632318837149853165664224847925206003567781692767655474759523146503572164952138829336342836023903919700264739071138739105931471740973631326608186969523753119546323993892359278563753903149741128282349467136720827132122619177620866305659196267641453819504766216964516467658995724859657544518337771393
n = 61860727516406742636690805639158184396057779906729165734489212939937929906456706343476469874085504076991779041906401043694401076841639925611957258119417559980829238154105119701407722069260962772947894516879731956778127512764229384957918619863998939985369399189275568362193066167855420897196095587732512368673
leak = 23213363443983005040318061737977092634638640953366787443691593387275645092922646169818923792205696350020369122807136306157118385984272980615310163206933078119776935167207473544453080959202803743994251355133953187110546017667004996272367137522351606700447920805532616096125523674597551449412004735397779511371
e = 65537
phi1 = (num2-1)*(num1-1)
d1 = gmpy2.invert(num1,phi1)
p_q = pow(leak,d1,num1*num2)p,q = symbols('p q')
eq = [p*q-n,p-q-p_q]
result = list(nonlinsolve(eq,[p,q]))
p,q = int(result[1][0]),int(result[1][1])
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = pow(ct,d,n)
flag = long_to_bytes(m-num2)
print(flag)

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