2023一带一路 crypto

1.

chall.py

from Crypto.Util.number import *
from flag import flag
import gmpy2
assert(len(flag)==38)
flag = bytes_to_long(flag)

p = getPrime(512)
q = getPrime(512)

e = 304
enc = pow(flag,e,p*q)
print(p)
print(q)
print(enc)
#9794998439882070838464987778400633526071369507639213778760131552998185895297188941828281554258704149333679257014558677504899624597863467726403690826271979
#10684338300287479543408040458978465940026825189952497034380241358187629934633982402116457227553161613428839906159238238486780629366907463456434647021345729
#88310577537712396844221012233266891147970635383301697208951868705047581001657402229066444746440502616020663700100248617117426072580419555633169418185262898647471677640199331807653373089977785816106098591077542771088672088382667974425747852317932746201547664979549641193108900510265622890793400796486146522028

gcd(e,phi) ≠ 1
明显的有限域开高次方
由于这里len(flag)==38 即304bit < p,q = 512bit
故可以分别求出modp,q的再用CRT合并加速开方 利用sage自带的CRT_list

from libnum import *
from time import *
p = 9794998439882070838464987778400633526071369507639213778760131552998185895297188941828281554258704149333679257014558677504899624597863467726403690826271979
q = 10684338300287479543408040458978465940026825189952497034380241358187629934633982402116457227553161613428839906159238238486780629366907463456434647021345729
c = 88310577537712396844221012233266891147970635383301697208951868705047581001657402229066444746440502616020663700100248617117426072580419555633169418185262898647471677640199331807653373089977785816106098591077542771088672088382667974425747852317932746201547664979549641193108900510265622890793400796486146522028
e = 304
n = p*q
t1 = time()
P.<a>=PolynomialRing(Zmod(p),implementation='NTL')
f=a^e-c
mps=f.monic().roots()

P.<a>=PolynomialRing(Zmod(q),implementation='NTL')
g=a^e-c
mqs=g.monic().roots()

flag=[]
for mpp in mps:
    x=mpp[0]
    for mqq in mqs:
        y=mqq[0]
        solution = CRT_list([int(x), int(y)], [p, q])
        flag.append(solution)
for i in flag:
    m=n2s(int(i))
    if b'flag'in m:
        print(m)
print('---------------------------------')
print(time()-t1)
# flag{947b6543117e32730a93d1b43c98bc57}

这样加速处理后只用了5s 而直接用n会出现multiple...错误提示 就算加了...=False也基本上跑不出来