当前位置：网站首页>[2022强网杯] polydiv和gamemaster
[2022强网杯] polydiv和gamemaster
2022-08-03 22:51:00 【石氏是时试】
目录
polydiv
题目先需要先爆破一个4字节的sha256再求40次 模二多项式，成功以后就直接给flag
import socketserver
import os, sys, signal
import string, random
from hashlib import sha256
from secret import flag
from poly2 import *
pad = lambda s:s + bytes([(len(s)-1)%16+1]*((len(s)-1)%16+1))
testCases = 40
class Task(socketserver.BaseRequestHandler):
def _recvall(self):
BUFF_SIZE = 2048
data = b''
while True:
part = self.request.recv(BUFF_SIZE)
data += part
if len(part) < BUFF_SIZE:
break
return data.strip()
def send(self, msg, newline=True):
try:
if newline:
msg += b'\n'
self.request.sendall(msg)
except:
pass
def recv(self, prompt=b'> '):
self.send(prompt, newline=False)
return self._recvall()
def close(self):
self.send(b"Bye~")
self.request.close()
def proof_of_work(self):
random.seed(os.urandom(8))
proof = ''.join([random.choice(string.ascii_letters+string.digits) for _ in range(20)])
_hexdigest = sha256(proof.encode()).hexdigest()
self.send(f"sha256(XXXX+{proof[4:]}) == {_hexdigest}".encode())
x = self.recv(prompt=b'Give me XXXX: ')
if len(x) != 4 or sha256(x+proof[4:].encode()).hexdigest() != _hexdigest:
return False
return True
def guess(self):
from Crypto.Util.number import getPrime
a,b,c = [getPrime(i) for i in [256,256,128]]
pa,pb,pc = [PP(bin(i)[2:]) for i in [a,b,c]]
r = pa*pb+pc
self.send(b'r(x) = '+str(r).encode())
self.send(b'a(x) = '+str(pa).encode())
self.send(b'c(x) = '+str(pc).encode())
self.send(b'Please give me the b(x) which satisfy a(x)*b(x)+c(x)=r(x)')
#self.send(b'b(x) = '+str(pb).encode())
return self.recv(prompt=b'> b(x) = ').decode() == str(pb)
def handle(self):
#signal.alarm(1200)
if not self.proof_of_work():
return
for turn in range(testCases):
if not self.guess():
self.send(b"What a pity, work harder.")
return
self.send(b"Success!")
else:
self.send(b'Congratulations, this is you reward.')
self.send(flag)
class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
pass
#class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
class ForkedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
pass
if __name__ == "__main__":
HOST, PORT = '0.0.0.0', 10000
server = ForkedServer((HOST, PORT), Task)
server.allow_reuse_address = True
server.serve_forever()
运算式子很简单，已知r,a,c求b，其实有个除法就行了
r = pa*pb+pc题目给定了一个 模二多项式环的运算 的类（有删减），里边用到除的部分被删掉了，从网上找原版，发现原版被删了，只有搜索引擎的缓存，但python程序在缓存时压缩了空格并删掉了回车换行这样的程序基本无法读。并且把 [0] 也丢掉。拿到程序根据意思理解也很难，然后用搜狗搜了一下发现一个正常版的。搜狗平常啥都搜不到，但偶尔还会有点用。
def div(self,other):
r,b = self.param[::-1],other.param[::-1]
if len(r) < len(b):
return Polynomial2(),self
q= [0]*(len(r) - len(b) + 1)
for i in range(len(q)):
if len(r)>=len(b):
index = len(r) - len(b) + 1 # 确定所得商是商式的第index位
q[-index] = int(r[0] / b[0]) # 更新被除多项式
b_=b.copy()
b_.extend([0]*(len(r) - len(b)))
b_ = [t*q[i] for t in b_]
r = [(r[t] - b_[t])%2 for t in range(len(r))]
for j in range(len(r)): #除去列表最左端无意义的0
if r[0]==0:
r.remove(0)
else:
break
else:
break
return Polynomial2(q),Polynomial2(r)
同时由于给出的数据是多项式，需要改为可处理的方式（第三种）
def p3(s):
return eval( b'['+s.replace(b'+', b',').replace(b', 1', b', 0').replace(b'x^', b'').replace(b'x', b'1')+ b']' )
然后就可以连后台了
from pwn import *
import string
from hashlib import sha256
from poly2 import *
def get_proof(last, shav):
s = string.ascii_letters+string.digits
for v1 in s:
for v2 in s:
for v3 in s:
for v4 in s:
t = (v1+v2+v3+v4).encode() + last
#print(t, sha256(t).hexdigest(), shav)
#return
if sha256(t).hexdigest() == shav:
return (v1+v2+v3+v4).encode()
context.log_level = 'debug'
p = remote('59.110.212.61', 26882)
#爆破sha256
p.recvuntil(b'sha256(XXXX+')
last = p.recvuntil(b') == ', drop=True)
shav = p.recvline()[:-1]
print('last:', last)
print('sha256:', shav)
found = get_proof(last, shav.decode())
p.sendlineafter(b'Give me XXXX: ', found)
#计算guess
for i in range(40):
p.recvuntil(b'(x) = ')
pr = Poly(p3(p.recvline()[:-1]))
p.recvuntil(b'(x) = ')
pa = Poly(p3(p.recvline()[:-1]))
p.recvuntil(b'(x) = ')
pc = Poly(p3(p.recvline()[:-1]))
pb = (pr - pc).div(pa)
p.sendlineafter(b'> b(x) = ', str(pb[0]))
p.recvuntil(b'Success!\n')
p.recv()
p.recv()
p.recv()
p.recv()
pause()
re_gamemaster
这是个逆向题，结了一个.net写和游戏和一个gamemessage文件
用dnSpy打开找到以下几段
1，BlackjackConsole Program.Main() 里将gamemessage读入memory变量
FileStream fileStream = File.OpenRead("gamemessage");
int num = (int)fileStream.Length;
Program.memory = new byte[num];
fileStream.Position = 0L;
fileStream.Read(Program.memory, 0, num);
2，BlackjackConsole Game.goldFunc 567 将memory的内容按字节与34异或
for (int i = 0; i < Program.memory.Length; i++)
{
byte[] array = Program.memory;
int num = i;
array[num] ^= 34;
}
3，BlackjackConsole Game.goldFunc() 538 将memory 用AES，ECB 解密生成ExploitClass.dll
byte[] key = new byte[]
{
66,
114,
97,
105,
110,
115,
116,
111,
114,
109,
105,
110,
103,
33,
33,
33
};
ICryptoTransform cryptoTransform = new RijndaelManaged
{
Key = key,
Mode = CipherMode.ECB,
Padding = PaddingMode.Zeros
}.CreateDecryptor();
Program.m = cryptoTransform.TransformFinalBlock(Program.memory, 0, Program.memory.Length);
这样就可以恢复这个动态库了
data = open('gamemessage', 'rb').read()
data = bytes([v^0x22 for v in data])
from Crypto.Cipher import AES
key = bytes([0x42, 0x72, 0x61, 0x69, 110, 0x73, 0x74, 0x6f, 0x72, 0x6d, 0x69, 110, 0x67, 0x21, 0x21, 0x21])
aes = AES.new(key,AES.MODE_ECB)
m = aes.decrypt(data)
pos = 0x13eb
m = m[pos:]+m[:pos]
open('temp.dll', 'wb').write(m)
然后用dnSpy再打开这个动态库就可以找到加密部分了
主要函数是check1 在程序中会对3个数字赋值（当然不知道是多少了），然后进行检查，查检通过后把这3个数作成12字节的key与密文异或就是flag
check1把这3个数向左移位并填充尾位（用其它位运算得到）共320次循环，最终这320个1位结果作成40个字节去校验。
private static void Check1(ulong x, ulong y, ulong z, byte[] KeyStream)
{
int num = -1;
for (int i = 0; i < 320; i++)
{
x = (((x >> 29 ^ x >> 28 ^ x >> 25 ^ x >> 23) & 1UL) | x << 1);
y = (((y >> 30 ^ y >> 27) & 1UL) | y << 1);
z = (((z >> 31 ^ z >> 30 ^ z >> 29 ^ z >> 28 ^ z >> 26 ^ z >> 24) & 1UL) | z << 1);
bool flag = i % 8 == 0;
if (flag)
{
num++;
}
KeyStream[num] = (byte)((long)((long)KeyStream[num] << 1) | (long)((ulong)((uint)((z >> 32 & 1UL & (x >> 30 & 1UL)) ^ (((z >> 32 & 1UL) ^ 1UL) & (y >> 31 & 1UL))))));
}
}
解法想来想去最后想到z3,1000多个变量，不过运算比较快1分钟之内就解决。
from z3 import *
first = [101,5,80,213,163,26,59,38,19,6,173,189,198,166,140,183,42,247,223,24,106,20,145,37,24,7,22,191,110,179,227,5,62,9,13,17,65,22,37,5]
array5 = [60,100,36,86,51,251,167,108,116,245,207,223,40,103,34,62,22,251,227]
sec = [0]*320
for i in range(40):
for j in range(8):
sec[i*8+j] = (first[i] >>(7-j))&1
x = [BitVec(f'x{i}',1) for i in range(360+30)]
y = [BitVec(f'y{i}',1) for i in range(360+31)]
z = [BitVec(f'z{i}',1) for i in range(320+32)]
s = Solver()
s.add([x[0]==0,x[1]==0,y[0]==0])
for i in range(320):
x[i+32] = x[i+2]^x[i+3]^x[i+6]^x[i+8]
y[i+32] = y[i+1]^y[i+4]
z[i+32] = z[i]^z[i+1]^z[i+2]^z[i+3]^z[i+5]^z[i+7]
s.add(sec[i] == (z[i]&x[i+2])^((z[i]^1)&y[i+1]))
if s.check() == sat:
d = s.model()
print('x=', end='')
for i in range(32):
print(str(d[x[i]]), end='')
print('\ny=', end='')
for i in range(32):
print(str(d[y[i]]), end='')
print('\nz=', end='')
for i in range(32):
print(str(d[z[i]]), end='')
#x=00001001010100010101010001100101
#y=00110011110000101000100101110011
#z=10111010101000101100011000110011
#156324965,868387187,3131229747
最后通过这3个数字得到flag
n1 = 156324965
n2 = 868387187
n3 = 3131229747
from pwn import p32
key = (p32(n1)+p32(n2)+p32(n3))*2
array5 = [60,100,36,86,51,251,167,108,116,245,207,223,40,103,34,62,22,251,227]
for i,v in enumerate(array5):
print(chr(v^key[i]), end='')
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