引言

今天也是元气满满的一天（打哈欠）

childRSA

题目给了 n和c
照例碰运气爆破 n：

啊这
真让我爆破出来了
然后就是 RSA 常规操作
代码如下：

from Crypto.Util.number import *

# p, q, c, n
e = int('0x10001', 16)
phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

结果为：NCTF{Th3r3_ar3_1ns3cure_RSA_m0duli_7hat_at_f1rst_gl4nce_appe4r_t0_be_s3cur3}

bbbbbbrsa

附件 python2 加密代码：

from base64 import b64encode as b32encode
from gmpy2 import invert,gcd,iroot
from Crypto.Util.number import *
from binascii import a2b_hex,b2a_hex
import random

flag = "******************************"

nbit = 128

p = getPrime(nbit)
q = getPrime(nbit)
n = p*q

print p
print n

phi = (p-1)*(q-1)

e = random.randint(50000,70000)

while True:
	if gcd(e,phi) == 1:
		break;
	else:
		e -= 1;

c = pow(int(b2a_hex(flag),16),e,n)

print b32encode(str(c))[::-1]

# 2373740699529364991763589324200093466206785561836101840381622237225512234632

这个注释就很魔性
一猜就知道是 c
另一个附件内容为：
p = 177077389675257695042507998165006460849
n = 37421829509887796274897162249367329400988647145613325367337968063341372726061
c = ==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4UzM2cTM5kDN2MTOyUTO5YDM0czM3MjM

根据加密代码，c 经过了 base64 加密并且倒序输出了
为了保险起见，还是先解码给出的 c
代码如下：

import base64
import math
from Crypto.Util.number import *

c = '==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4UzM2cTM5kDN2MTOyUTO5YDM0czM3MjM'[::-1]
base64_bytes = c.encode('ascii')
message_bytes = base64.b64decode(base64_bytes)
message = message_bytes.decode('ascii')
print(message)
c = int(message)

c = 2373740699529364991763589324200093466206785561836101840381622237225512234632
啊这
跟注释一模一样
解密代码：

p = 177077389675257695042507998165006460849
n = 37421829509887796274897162249367329400988647145613325367337968063341372726061
q = n//p
phi = (p-1)*(q-1)

for e in range(50000, 70000):
    if math.gcd(e, phi) == 1:
        d = inverse(e, phi)
        m = str(long_to_bytes(pow(c, d, n)))
        if '{' in m and '}' in m:
            print(m)

结果如下：

结语

啊。。。好像也没什么好说的
希望继续坚持