2022 网鼎杯初赛（青龙组） Writeup By Xp0int

xp0int Posted on Aug 26 2022

## 1. Crypto
### 1.1 crypto405
`Author: k1rit0`

利用flag头 flag{ 列出五个关于K的方程，用sagemath一个一个的解出k来，然后爆破所有可能的p，直到找到flag

```python
k0 = var('k0')
k1 = var('k1')
k2 = var('k2')
k3 = var('k3')
k4 = var('k4')
K = [k0,k1,k2,k3,k4]
flag = [ord('f'),ord('l'),ord('a'),ord('g'),ord('{')]
F = []
for i in range(5):
    grasshopper = flag[i]
    for j in range(5):
        grasshopper = grasshopper * K[j]
        K[j] = grasshopper 
    F.append(K[-1] - known[i])
F[0].solve(k0)
# F[1].solve(k1) 需要解出上一条把k0代回去
```

上面这个脚本能解出所有的k

```python
import re
P = [...] # 利用gmpy2的next_prime把所有素数列出来，太多了渲染会卡所以用...省略了
output = '''Grasshopper#00:2066
Grasshopper#01:a222
Grasshopper#02:cbb1
Grasshopper#03:dbb4
Grasshopper#04:deb4
Grasshopper#05:b1c5
Grasshopper#06:33a4
Grasshopper#07:c051
Grasshopper#08:3b79
Grasshopper#09:6bf8
Grasshopper#10:2131
Grasshopper#11:2c40
Grasshopper#12:91ba
Grasshopper#13:7b44
Grasshopper#14:5f25
Grasshopper#15:0208
Grasshopper#16:7edb
Grasshopper#17:62b5
Grasshopper#18:cec5
Grasshopper#19:5ab3
Grasshopper#20:3c46
Grasshopper#21:c272
Grasshopper#22:714b
Grasshopper#23:9e0b
Grasshopper#24:48ee
Grasshopper#25:44cc
Grasshopper#26:05a0
Grasshopper#27:3da3
Grasshopper#28:11b1
Grasshopper#29:259f
Grasshopper#30:899d
Grasshopper#31:a130
Grasshopper#32:e58f
Grasshopper#33:23f3
Grasshopper#34:5829
Grasshopper#35:6beb
Grasshopper#36:3681
Grasshopper#37:0054
Grasshopper#38:a189
Grasshopper#39:2765
Grasshopper#40:c63d
Grasshopper#41:bc68'''.split('\n')
known = [int(re.findall(r'\:(.*)',i)[0],16) for i in output]

k0 = (396091111351853546729322486681828125/16752324664863145168406201892534453994427707392)
k1 = 41892628622080799/1618763711678683806046238092129608000/k0^4
k2 = 216245315/118532050972475400824098480128/(k0^6*k1^3)
k3 = 20753/48479538609216/(k0^4*k1^3*k2^2)
k4 = 4147/51/(k0*k1*k2*k3)
K = [k0,k1,k2,k3,k4]
for p in P:
    _K = [i % p for i in K]
    flag = ''
    # solve x
    for i in range(len(known)):
        inv = 1
        for j in range(5):
            inv = inv * inverse_mod(Integer(_K[j]),Integer(p))
        x = (known[i] * inv) % p
        if 0x20 < x < 0x7e:
            flag+=chr(x)
            for j in range(5):
                x = x * _K[j] % p
                _K[j] = x 
    if len(flag) == 42:
        print(flag)
'''
flag{749d39d4-78db-4c55-b4ff-bca873d0f18e}
'''
```

### 1.2 crypto091
`Author: k1rit0`

盲猜sha256

```python
from hashlib import sha256
HASH = 'c22a563acc2a587afbfaaaa6d67bc6e628872b00bd7e998873881f7c6fdc62fc'
phone_base = '86170'
for i in range(10 ** 8):
    phone = phone_base + str(i).zfill(8)
    if sha256(phone.encode()).hexdigest() == HASH:
        print(phone)
'''
8617091733716
'''
```

### 1.3 crypto162
`Author: k1rit0`

好好的一个递推式子要弄成递归...直接换成递推，题目只需要后2000位，所以只需要计算后2000位，这样就能快速的算了（还是还要两三分钟

```python
cof_t = [...]
S = 0
for V in cof_t:
    print(cof_t.index(V))
    res = [1,2,3]
    for _ in range(3,200000+1):
        res.append((V[2] * res[-3] + V[1] * res[-2] + V[0] * res[-1]) % 10 ^ 2002)
    S += res[-1]

s=str(S)[-2000:-1000]
key = md5(s).hexdigest().decode('hex')
check = sha256(key).hexdigest()
verify = '2cf44ec396e3bb9ed0f2f3bdbe4fab6325ae9d9ec3107881308156069452a6d5'
print(check)
assert(check == verify)
aes = AES.new(key,AES.MODE_ECB)
c = long_to_bytes(0x4f12b3a3eadc4146386f4732266f02bd03114a404ba4cb2dabae213ecec451c9d52c70dc3d25154b5af8a304afafed87)
print (aes.decrypt(c))
'''
flag{519427b3-d104-4c34-a29d-5a7c128031ff}
'''
```

## 2. Pwn
### 2.1 pwn135 / BabyV8
`Author: xf1les`

非预期

![title](https://leanote.com/api/file/getImage?fileId=630892beab644108379e7fe5)

### 2.2 pwn497 / minishell
`Author: xf1les`

非预期

![title](https://leanote.com/api/file/getImage?fileId=630892d2ab644108379e7fe7)

### 2.3 pwn349
`Author: xf1les`

还是非预期（估计没完全修好 pwn497 的非预期解）

![title](https://leanote.com/api/file/getImage?fileId=630893b0ab644108379e7ffe)

## 3. Reverse
### 3.1 re693
`Author: JANlittle`

go源文件直接给，按题目要求直接vscode硬搜，第一个函数是`ZlXDJkH3OZN4Mayd`，第二个是`UhnCm82SDGE0zLYO`，然后直接按逻辑梭出flag就行

### 3.2 re694
`Author: JANlittle`

魔改upx，懒得找改了哪里，x64dbg一路F8，最后有个长跳转，跳到这里：

![title](https://leanote.com/api/file/getImage?fileId=6308934aab644108379e7ff6)

然后scylla直接dump出程序丢IDA，逻辑差不多是检测flag长度是不是20，然后异或0x66、加10、异或0x50，然后比较。

```python
>>> c=[0x0000004B, 0x00000048, 0x00000079, 0x00000013, 0x00000045, 0x00000030, 0x0000005C, 0x00000049, 0x0000005A, 0x00000079, 0x00000013, 0x00000070, 0x0000006D, 0x00000078, 0x00000013, 0x0000006F, 0x00000048, 0x0000005D, 0x00000064, 0x00000064]
>>> for i in range(len(c)):
...     c[i]^=0x50
...     c[i]-=10
...
>>> bytes(c)
b'\x11\x0e\x1f9\x0bV\x02\x0f\x00\x1f9\x163\x1e95\x0e\x03**'
>>> for i in range(len(c)):
...     c[i]^=0x66
...
>>> bytes(c)
b'why_m0dify_pUx_SheLL'
```

打赏还是打残，这是个问题