[MoeCTF 2022]一次就好

1、题目信息

from Crypto.Util.strxor import strxor
from Crypto.Util.number import *
from gmpy2 import powmod,next_prime
from FLAG import flag
import codecs

c = b'Just once,I will accompany you to see the world'
flag = flag.ljust(len(c),'#')
key = strxor(flag.encode(), c)
m = bytes_to_long(key)

p = getPrime(512)
q = next_prime(p)
N = p*q
e = 0x10001

gift = powmod(m, e, N)

print(gift)
print(N)

# gift = 127749242340004016446001520961422059381052911692861305057396462507126566256652316418648339729479729456613704261614569202080544183416817827900318057127539938899577580150210279291202882125162360563285794285643498788533366420857232908632854569967831654923280152015070999912426044356353393293132914925252494215314
# N = 164395171965189899201846744244839588935095288852148507114700855000512464673975991783671493756953831066569435489213778701866548078207835105414442567008315975881952023037557292470005621852113709605286462434049311321175270134326956812936961821511753256992797013020030263567313257339785161436188882721736453384403

2、解码

#https://kbthe0phi1us.github.io/
#p，q接近，费马分解
from math import isqrt
import gmpy2
from Crypto.Util.number import *
from Crypto.Util.strxor import strxor


def fermat(n):
    a = isqrt(n)
    b2 = a * a - n
    b = isqrt(n)
    count = 0
    while b * b != b2:
        a = a + 1
        b2 = a * a - n
        b = isqrt(b2)
        count += 1
    p = a + b
    q = a - b
    assert n == p * q
    return p, q

if __name__ == '__main__':
    gift = 127749242340004016446001520961422059381052911692861305057396462507126566256652316418648339729479729456613704261614569202080544183416817827900318057127539938899577580150210279291202882125162360563285794285643498788533366420857232908632854569967831654923280152015070999912426044356353393293132914925252494215314
    n = 164395171965189899201846744244839588935095288852148507114700855000512464673975991783671493756953831066569435489213778701866548078207835105414442567008315975881952023037557292470005621852113709605286462434049311321175270134326956812936961821511753256992797013020030263567313257339785161436188882721736453384403
    e=0x10001
    c = b'Just once,I will accompany you to see the world'
    p, q = fermat(n)
    print("p=", p)
    print("q=", q)
    phi = (q - 1) * (p - 1)
    d = gmpy2.invert(e, phi)
    m = gmpy2.powmod(gift, d, n)
    key = long_to_bytes(m)
    flag = strxor(key, c)
    print(flag)
#moectf{W0w_y02_k5ow_w6at_1s_one_t1m3_pa7}