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[原创][Rev赛题复现]DASCTF Apr X FATE 2022
2022-5-7 16:02 9209

[原创][Rev赛题复现]DASCTF Apr X FATE 2022

t0hka1 活跃值
1
2022-5-7 16:02
9209

目录

[Rev赛题复现]DASCTF Apr X FATE 2022

总共4题，贴了3题，还有一道go逆向的直接看这位师傅的吧

Crackme

几个关键点：mfc逆向，win32 加密api的识别，ZwSetInformationThread反调试

 

看程序图标是个mfc的程序，先打开看看，随便输入一点东西，看到弹窗弹出

 

 

直接拖进ida搜索Wrong!!!字符串，借此通过查看引用跳转到主函数

 

先简单的修复一下变量名

 

 

大概可以看到key的前4位经过一次sub_403510，后4位也经过一次sub_403510，整个key8位又经过一次sub_403510

 

随后进入sub_403510函数看一看，看到有一堆为win32 加密api的函数

 

image-20220430094801764

 

那就对照着MSDN一个个函数查阅一下

CryptAcquireContext 函数用于获取特定加密服务提供程序 （CSP） 中特定密钥容器的句柄。此返回的句柄用于调用使用所选 CSP 的 CryptoAPI 函数。

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BOOL CryptAcquireContextA(
[out] HCRYPTPROV *phProv,
[in]  LPCSTR     szContainer,
[in]  LPCSTR     szProvider,
[in]  DWORD      dwProvType,
[in]  DWORD      dwFlags
);

CryptCreateHash 函数启动数据流的哈希。它创建加密服务提供程序 （CSP） 哈希对象的句柄并将其返回给调用应用程序。此句柄用于对 CryptHashData 和 CryptHashSessionKey 的后续调用，以哈希会话密钥和其他数据流。

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BOOL CryptCreateHash(
[in]  HCRYPTPROV hProv,
[in]  ALG_ID     Algid,
[in]  HCRYPTKEY  hKey,
[in]  DWORD      dwFlags,
[out] HCRYPTHASH *phHash
);

注意这里Algid是标识要使用的哈希算法的参数，通过不同的值的传入选择不同的hash算法，可查下面的链接

ALG_ID (Wincrypt.h) - Win32 apps | Microsoft Docs

CryptHashData 函数将数据添加到指定的哈希对象。此函数和CryptHashSessionKey可以多次调用，以计算长数据流或不连续数据流的哈希值。

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BOOL CryptHashData(
[in] HCRYPTHASH hHash,
[in] const BYTE *pbData,
[in] DWORD      dwDataLen,
[in] DWORD      dwFlags
);

CryptGetHashParam 函数检索控制哈希对象操作的数据。可以使用此函数检索实际的哈希值。

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BOOL CryptGetHashParam(
[in]      HCRYPTHASH hHash,
[in]      DWORD      dwParam,
[out]     BYTE       *pbData,
[in, out] DWORD      *pdwDataLen,
[in]      DWORD      dwFlags
);

CryptEncrypt 函数对数据进行加密。用于加密数据的算法由 CSP 模块持有的密钥指定，并由 hKey 参数引用。

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BOOL CryptEncrypt(
[in]      HCRYPTKEY  hKey,
[in]      HCRYPTHASH hHash,
[in]      BOOL       Final,
[in]      DWORD      dwFlags,
[in, out] BYTE       *pbData,
[in, out] DWORD      *pdwDataLen,
[in]      DWORD      dwBufLen
);
 

程序大概的逻辑就是这样

 

 

然后我们就通过动调去拿数据，这里有两种方式，一种是通过ida patch反调试函数的方式，一种是通过od 的sharp od插件直接绕过

绕反调试方法一：先用ida绕过反调试

对比各种反调试和去IAT表找导入函数没有找到，后面在strings界面可以发现ZwSetInformationThread反调试的特征ZwSetInformationThread - CTF Wiki (ctf-wiki.org)

ZwSetInformationThread通过为线程设置 ThreadHideFromDebugger，可以禁止线程产生调试事件

绕过: ZwSetInformationThread 函数的第 2 个参数为 ThreadHideFromDebugger，其值为 0x11。调试执行到该函数时，若发现第 2 个参数值为 0x11，跳过或者将 0x11 修改为其他值即可

 

 

看来是自己实现调用dll导入的

 

 

类似于这种写法

 

我们可以先在调用处下一个断点，跑起来之后再修改patch 0x11 改掉，我这里patch成了0x9

 

 

然后继续下断点拿密文

 

 

 

 

然后再扔到md5解密网站解密

 

 

类似的拿到sha1解密后的key后四位 https://crackstation.net/

 

 

key:NocTuRne

 

md5(key):C0804C74E05B4C7440AC4D7480954C74

绕反调试方法二：OD odsharp插件直接绕

 

之后就是模拟调用win32 的aes解密api来解密的过程了

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#include <Windows.h>
#include <stdio.h>
#include <wincrypt.h>
 
int main(){
    BYTE pbData[] = {0x5c,0x53,0xa4,0xa4,0x1d,0x52,0x43,0x7a,0x9f,0xa1,0xe9,0xc2,0x6c,0xa5,0x90,0x90,0x0};  //key_buf
    BYTE flag_encrypt[] = {0x5B, 0x9C, 0xEE, 0xB2, 0x3B, 0xB7, 0xD7, 0x34, 0xF3, 0x1B, 0x75, 0x14, 0xC6, 0xB2, 0x1F, 0xE8, 0xDE, 0x33, 0x44, 0x74, 0x75, 0x1B, 0x47, 0x6A, 0xD4, 0x37, 0x51, 0x88, 0xFC, 0x67, 0xE6, 0x60, 0xDA, 0x0D, 0x58, 0x07, 0x81, 0x43, 0x53, 0xEA, 0x7B, 0x52, 0x85, 0x6C, 0x86, 0x65, 0xAF, 0xB4,0x0};
    DWORD dwDataLen = 0x10;
    DWORD ddwDataLen;
    DWORD* pdwDataLen = &ddwDataLen;
    *pdwDataLen = 0x20;
 
 
    BOOL v6; // [esp+4h] [ebp-18h]
    HCRYPTKEY phKey; // [esp+Ch] [ebp-10h] BYREF
    HCRYPTPROV phProv; // [esp+10h] [ebp-Ch] BYREF
    HCRYPTHASH phHash; // [esp+14h] [ebp-8h] BYREF
 
    phProv = 0;
    phHash = 0;
    phKey = 0;
    v6 = CryptAcquireContextA(&phProv, 0, 0, 0x18u, 0xF0000000);
    if (v6)
    {
        v6 = CryptCreateHash(phProv, 0x8003u, 0, 0, &phHash);
        if (v6)
        {
            v6 = CryptHashData(phHash, pbData, dwDataLen, 0);
            if (v6)
            {
                v6 = CryptDeriveKey(phProv, 0x660Eu, phHash, 1u, &phKey);// key的md5值再生成aes密钥
                if (v6)
                    v6 = CryptDecrypt(phKey, 0, 1, 0, flag_encrypt, pdwDataLen);
                    printf("%s", flag_encrypt);
            }
        }
    }
    if (phKey)
        CryptDestroyKey(phKey);
    if (phHash)
        CryptDestroyHash(phHash);
    if (phProv)
        CryptReleaseContext(phProv, 0);
    return v6;
}

拿到flag！

 

补充方法：Hook Windows API 求解

另外经mas0n师傅补充，附上frida来hook求解的方法

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  var baseAddr = Process.findModuleByName('Crackme_1.exe');
 
// input 32 length flag, e.g. 11111111111111111111111111111111
// key: NocTuRne
// frida attach -p 48964 -l agent\hook_win.js
 
 
// memcmp
var hookAddr = ptr(0x0109D4BC);
Interceptor.attach(hookAddr, {
    onEnter: function(args) {
        let Buf1 = args[0];
        let Buf2 = args[1];
        let Size = args[2];
        console.log("-----\n[Size]\n", Size);
        let size = Size.toInt32();
        console.log("-----\n[Buf1]\n", Buf1.readByteArray(size));
        console.log("-----\n[Buf2]\n", Buf2.readByteArray(size));
        console.log("---------------------------");
    },
    onLeave: function(arg) {
        return arg;
    }
})
 
var libAddr = Process.findModuleByName('ADVAPI32.dll');
var fn_CryptEncrypt = libAddr.getExportByName("CryptEncrypt");
var fn_CryptDecrypt = libAddr.getExportByName("CryptDecrypt");
 
var flag = null;
Interceptor.replace(fn_CryptEncrypt, fn_CryptDecrypt);
Interceptor.attach(fn_CryptDecrypt, {
    onEnter: function(args) {
        args[4].writeByteArray([0x5b,0x9c,0xee,0xb2,0x3b,0xb7,0xd7,0x34,0xf3,0x1b,0x75,0x14,0xc6,0xb2,0x1f,0xe8,0xde,0x33,0x44,0x74,0x75,0x1b,0x47,0x6a,0xd4,0x37,0x51,0x88,0xfc,0x67,0xe6,0x60,0xda,0x0d,0x58,0x07,0x81,0x43,0x53,0xea,0x7b,0x52,0x85,0x6c,0x86,0x65,0xaf,0xb4]);
        args[5].writeInt(0x40);
        flag = args[4];
        console.log("hook fn_CryptDecrypt");
        return args;
    },
    onLeave: function(arg) {
        console.log(flag.readCString());
        return arg;
    }
})

奇怪的交易

拖进ida里发现经过upx加壳

 

直接upx -d脱壳

 

接着再使用pyinstxtractor解包得到一堆文件(比较坑的地方是本地python环境必须与源程序的python环境相同才能解包PYZ-00.pyz)，下面是彻底解包后的几个关键的文件

 

image-20220424125642694

 

奇怪的交易.pyc文件内容如下

 

python反编译 - 在线工具 (tool.lu)

 

image-20220424125705312

 

这样其实逻辑很明显还是有问题的，题目特意用的Python3.10版本，导致反编译结果会不正确

 

通过pycdump可以dump出opcode，对比进行变量名和代码逻辑的修复

 

可以参考下面的文章

 

Python字节码文档 Python字节码详解(介绍了Python的特有类型以及遍历等操作)

 

image-20220424125724609

 

修复后的 奇怪的交易.py

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from cup import *
from libnum import *
 
if __name__ == '__main__':
    flag = input('请输入flag')
    pub_key = [
        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
    m = libnum.s2n(flag)
    c = str(pow(m, pub_key[1], pub_key[0]))  # 极长的一串东西
    store = []
    cipher = [3532577106, 1472742623, 3642468664, 4193500461, 2398676029, 617653972, 1474514999, 1471783658, 1012864704,
              3615627536, 993855884, 438456717, 3358938551, 3906991208, 198959101, 3317190635, 3656923078, 613157871,
              2398768861, 97286225, 2336972940, 1471645170, 3233163154, 583597118, 2863776301, 3183067750, 1384330715,
              2929694742, 3522431804, 2181488067, 3303062236, 3825712422, 145643141, 2148976293, 2940910035, 506798154,
              994590281, 2231904779, 3389770074, 2814269052, 1105937096, 1789727804, 3757028753, 2469686072, 1162286478,
              680814033, 2934024098, 2162521262, 4048876895, 2121620700, 4240287315, 2391811140, 3396611602, 3091349617,
              3031523010, 2486958601, 3164065171, 1285603712, 798920280, 2337813135, 4186055520, 3523024366, 1077514121,
              1436444106, 2731983230, 1507202797, 500756149, 198754565, 2382448647, 880454148, 1970517398, 3217485349,
              1161840191, 560498076, 1782600856, 2643721918, 1285196205, 788797746, 1195724574, 4061612551, 103427523,
              2502688387, 4147162188, 617564657, 978211984, 1781482121, 2205798970, 3939973102, 3826603515, 659557668,
              2582884932, 1561884856, 2217488804, 1189296962, 169145316, 2781742156, 1323893433, 824667876, 408202876,
              3759637634, 4094868412, 1508996065, 162419237, 3732146944, 3083560189, 3955940127, 2393776934, 2470191468,
              3620861513, 481927014, 2756226070, 3154651143, 1261069441, 2063238535, 2222237213, 101459755, 3159774417,
              1721190841, 1078395785, 176506553, 3552913423, 1566142515, 1938949000, 1499289517, 3315102456, 829714860,
              3843359394, 952932374, 1283577465, 2045007203, 3957761944, 3767891405, 2917089623, 3296133521, 482297421,
              1734231412, 3670478932, 2575334979, 2827842737, 3413631016, 1533519803, 4008428470, 3890643173, 272960248,
              317508587, 3299937500, 2440520601, 27470488, 1666674386, 1737927609, 750987808, 2385923471, 2694339191,
              562925334, 2206035395]
 
    i = 0
    # rsa 生成的密文遍历加密
    while i < len(c):  # i<155
        index = 0
        for ii in c[i:i + 4]:
            index = (index << 8) + ord(ii)
        store.append(index)
 
        i += 4
        if not i < len(c):
            key = [54, 54, 54, 54]
            store_len = len(store)
            res = encrypt(store_len, store, key)
            if store == cipher:
                print('You are right!')
                input('')
                quit()
            else:
                print('Why not drink a cup of tea and have a rest?')
 
        continue

发现从cup包导入了一个encrypt函数

 

以下是对经key加密后的cup.pyc.encrypted的解密脚本[原创]Python逆向——Pyinstaller逆向-软件逆向-看雪论坛-安全社区|安全招聘|bbs.pediy.com

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      #!/usr/bin/env python3
import tinyaes
import zlib
 
CRYPT_BLOCK_SIZE = 16
 
# 从crypt_key.pyc获取key，也可自行反编译获取
key = bytes('0000000000000tea', 'utf-8')
 
inf = open('cup.pyc.encrypted', 'rb') # 打开加密文件
outf = open('output.pyc', 'wb') # 输出文件
 
# 按加密块大小进行读取
iv = inf.read(CRYPT_BLOCK_SIZE)
 
cipher = tinyaes.AES(key, iv)
 
# 解密
plaintext = zlib.decompress(cipher.CTR_xcrypt_buffer(inf.read()))
 
# 补pyc头(最后自己补也行)
outf.write(b'\x6f\x0d\x0d\x0a\0\0\0\0\0\0\0\0\0\0\0\0')
 
# 写入解密数据
outf.write(plaintext)
 
inf.close()
outf.close()

解密得到发现是一个python实现的xxtea加密，最基础的版本，甚至连key都没变

 

python实现xxtea加解密参考链接

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#!/usr/bin/env python
# visit https://tool.lu/pyc/ for more information
import libnum
from ctypes import *
 
def MX(z, y, total, key, p, e):
    temp1 = (z.value >> 5 ^ y.value << 2) + (y.value >> 3 ^ z.value << 4)
    temp2 = (total.value ^ y.value) + (key[p & 3 ^ e.value] ^ z.value)
    return c_uint32(temp1 ^ temp2)
 
 
def encrypt(ᘗ, ᘖ, ᘘ):
    ᘜ = 0x9E3779B9L
    ᘛ = 6 + 52 // ᘗ
    total = c_uint32(0)
    ᘔ = c_uint32(ᘖ[ᘗ - 1])
    ᘕ = c_uint32(0)
    if ᘛ > 0:
        total.value += ᘜ
        ᘕ.value = total.value >> 2 & 3
        ᘚ = c_uint32(ᘖ[0])
        ᘖ[ᘗ - 1] = c_uint32(ᘖ[ᘗ - 1] + MX(ᘔ, ᘚ, total, ᘘ, ᘗ - 1, ᘕ).value).value
        ᘔ.value = ᘖ[ᘗ - 1]
        ᘛ -= 1
        if not ᘛ > 0:
            return ᘖ

先解密xxtea得到结果rsa加密的密文c

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from ctypes import *
 
 
def MX(z, y, total, key, p, e):
    temp1 = (z.value>>5 ^ y.value<<2) + (y.value>>3 ^ z.value<<4)
    temp2 = (total.value ^ y.value) + (key[(p&3) ^ e.value] ^ z.value)
 
    return c_uint32(temp1 ^ temp2)
 
 
def encrypt(n, v, key):
    delta = 0x9e3779b9
    rounds = 6 + 52//n
 
    total = c_uint32(0)
    z = c_uint32(v[n-1])
    e = c_uint32(0)
 
    while rounds > 0:
        total.value += delta 
        e.value = (total.value >> 2) & 3
        for p in range(n-1):
            y = c_uint32(v[p+1])
            v[p] = c_uint32(v[p] + MX(z,y,total,key,p,e).value).value
            z.value = v[p]
        y = c_uint32(v[0])
        v[n-1] = c_uint32(v[n-1] + MX(z,y,total,key,n-1,e).value).value
        z.value = v[n-1]
        rounds -= 1
 
    return v
 
 
def decrypt(n, v, key):
    delta = 0x9E3779B9
    rounds = 6 + 52//n
 
    total = c_uint32(rounds * delta)
    y = c_uint32(v[0])
    e = c_uint32(0)
 
    while rounds > 0:
        e.value = (total.value >> 2) & 3
        for p in range(n-1, 0, -1):
            z = c_uint32(v[p-1])
            v[p] = c_uint32((v[p] - MX(z,y,total,key,p,e).value)).value
            y.value = v[p]
        z = c_uint32(v[n-1]) 
        v[0] = c_uint32(v[0] - MX(z,y,total,key,0,e).value).value
        y.value = v[0] 
        total.value -= delta
        rounds -= 1
 
    return v
 
 
#  test 
if __name__ == "__main__":
    # 该算法中每次可加密不只64bit的数据，并且加密的轮数由加密数据长度决定
 
    k = [54, 54, 54, 54]
    n = 155
 
    res=[3532577106, 1472742623, 3642468664, 4193500461, 2398676029, 617653972, 1474514999, 1471783658, 1012864704, 3615627536, 993855884, 438456717, 3358938551, 3906991208, 198959101, 3317190635, 3656923078, 613157871, 2398768861, 97286225, 2336972940, 1471645170, 3233163154, 583597118, 2863776301, 3183067750, 1384330715, 2929694742, 3522431804, 2181488067, 3303062236, 3825712422, 145643141, 2148976293, 2940910035, 506798154, 994590281, 2231904779, 3389770074, 2814269052, 1105937096, 1789727804, 3757028753, 2469686072, 1162286478, 680814033, 2934024098, 2162521262, 4048876895, 2121620700, 4240287315, 2391811140, 3396611602, 3091349617, 3031523010, 2486958601, 3164065171, 1285603712, 798920280, 2337813135, 4186055520, 3523024366, 1077514121, 1436444106, 2731983230, 1507202797, 500756149, 198754565, 2382448647, 880454148, 1970517398, 3217485349, 1161840191, 560498076, 1782600856, 2643721918, 1285196205, 788797746, 1195724574, 4061612551, 103427523, 2502688387, 4147162188, 617564657, 978211984, 1781482121, 2205798970, 3939973102, 3826603515, 659557668, 2582884932, 1561884856, 2217488804, 1189296962, 169145316, 2781742156, 1323893433, 824667876, 408202876, 3759637634, 4094868412, 1508996065, 162419237, 3732146944, 3083560189, 3955940127, 2393776934, 2470191468, 3620861513, 481927014, 2756226070, 3154651143, 1261069441, 2063238535, 2222237213, 101459755, 3159774417, 1721190841, 1078395785, 176506553, 3552913423, 1566142515, 1938949000, 1499289517, 3315102456, 829714860, 3843359394, 952932374, 1283577465, 2045007203, 3957761944, 3767891405, 2917089623, 3296133521, 482297421, 1734231412, 3670478932, 2575334979, 2827842737, 3413631016, 1533519803, 4008428470, 3890643173, 272960248, 317508587, 3299937500, 2440520601, 27470488, 1666674386, 1737927609, 750987808, 2385923471, 2694339191, 562925334, 2206035395]
 
    res = decrypt(n, res, k)
 
    # print(res)
    for i in res:
        print(chr(i>>24),end="")
        print(chr((i&0x00ff0000)>>16),end="")
        print(chr((i&0x0000ff00)>>8),end="")
        print(chr(i&0x000000ff),end="")
 
 #c= 10610336534759505889607399322387179316771488492347274741918862678692508953185876570981227584004676580623553664818853686933004290078153620168054665086468417541382824708104480882577200529822968531743002301934310349005341104696887943182074473298650903541494918266823037984054778903666406545980557074219162536057146090758158128189406073809226361445046225524917089434897957301396534515964547462425719205819342172669899546965221084098690893672595962129879041507903210851706793788311452973769358455761907303633956322972510500253009083922781934406731633755418753858930476576720874219359466503538931371444470303193503733920039

接下来是一个低解密指数 rsa 就可以得到flag了

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import gmpy2
from Crypto.PublicKey import RSA
import ContinuedFractions, Arithmetic
from Crypto.Util.number import long_to_bytes
 
 
def wiener_hack(e, n):
    # firstly git clone https://github.com/pablocelayes/rsa-wiener-attack.git !
    frac = ContinuedFractions.rational_to_contfrac(e, n)
    convergents = ContinuedFractions.convergents_from_contfrac(frac)
    for (k, d) in convergents:
        if k != 0 and (e * d - 1) % k == 0:
            phi = (e * d - 1) // k
            s = n - phi + 1
            discr = s * s - 4 * n
            if (discr >= 0):
                t = Arithmetic.is_perfect_square(discr)
                if t != -1 and (s + t) % 2 == 0:
                    return d
    return False
 
 
def main():
    pub_key = [
    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
    0x647327833ACFEF1F9C83E74E171FC300FA347D4A6769476C33DA82C95120ACB38B62B33D429206FE6E9BB0BB7AB748A1036971BEA36EC47130B749C1C9FF6FE03D0F7D9FC5346EB0E575BDFA6C530AA57CD676894FC080D2DD049AB59625F4B9C78BCFD95CDCD2793E440E26E189D251121CB6EB177FEDB596409034E8B0C5BBD9BD9342235DBB226C9170EFE347FF0FD2CFF9A1F7B647CC83E4D8F005FD7125A89251C768AFE70BDD54B88116814D5030F499BCAC4673CCCC342FB4B6AC58EA5A64546DC25912B6C430529F6A7F449FD96536DE269D1A1B015A4AC6B6E46EE19DCE8143726A6503E290E4BAE6BD78319B5878981F6CFFDB3B818209341FD68B]
    # 0->n,1->e
 
    n = pub_key[0]
    e = pub_key[1]
    c = 10610336534759505889607399322387179316771488492347274741918862678692508953185876570981227584004676580623553664818853686933004290078153620168054665086468417541382824708104480882577200529822968531743002301934310349005341104696887943182074473298650903541494918266823037984054778903666406545980557074219162536057146090758158128189406073809226361445046225524917089434897957301396534515964547462425719205819342172669899546965221084098690893672595962129879041507903210851706793788311452973769358455761907303633956322972510500253009083922781934406731633755418753858930476576720874219359466503538931371444470303193503733920039
    d = wiener_hack(e, n)
    m = pow(c, d, n)
    print(long_to_bytes(m)) #  flag{You_Need_Some_Tea}
 
 
if __name__ == "__main__":
    main()

FakePica

先用BlackDex脱个壳，然后把解密后的dex文件pull到自己的电脑上后拖进jadx

 

看到主要是一个登录逻辑

 

 

题目逻辑非常简单，可以直接Aes解密，但由于主类里有了解密的方法，肯定是要选择更加有意思的方法来玩玩，我这里的做法是直接采用frida来hook

 

先hook绕过认证

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console.log("Script loaded successfully");
Java.perform(function x(){
    console.log("inside java perform function");
    //定位类
    var my_class = Java.use("com.pica.picapica.MainActivity");
    console.log("Java.use Successfully");
    my_class.check.implementation = function(x,y){
        return true;
    }
})

返回页面如下

 

 

看到有两个解密相关的方法，果断继续hook

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console.log("Script loaded successfully");
Java.perform(function x(){
    console.log("inside java perform function");
    //定位类
    var my_class = Java.use("com.pica.picapica.MainActivity");
    console.log("Java.use Successfully");
    my_class.check.implementation = function(x,y){
        var email=this.decryptByHexString(this.bytesConvertHexString(this.content0.value),this.key.value);
        var password=this.decryptByHexString(this.bytesConvertHexString(this.content1.value),this.key.value);
        console.log("flag{"+email+password+"}"); 
        return true;
    }
})

直接拿到flag

 

 

奇怪的交易这题附件过大，这里就直接贴我的链接了
奇怪的交易


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最后于 2022-6-6 19:22 被t0hka1编辑 ，原因：
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sunfishi
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sunfishi 4 2022-5-10 13:06
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给一个Hook过的

  var baseAddr = Process.findModuleByName('Crackme_1.exe');

// input 32 length flag, e.g. 11111111111111111111111111111111
// key: NocTuRne
// frida attach -p 48964 -l agent\hook_win.js


// memcmp
var hookAddr = ptr(0x0109D4BC);
Interceptor.attach(hookAddr, {
    onEnter: function(args) {
        let Buf1 = args[0];
        let Buf2 = args[1];
        let Size = args[2];
        console.log("-----\n[Size]\n", Size);
        let size = Size.toInt32();
        console.log("-----\n[Buf1]\n", Buf1.readByteArray(size));
        console.log("-----\n[Buf2]\n", Buf2.readByteArray(size));
        console.log("---------------------------");
    },
    onLeave: function(arg) {
        return arg;
    }
})

var libAddr = Process.findModuleByName('ADVAPI32.dll');
var fn_CryptEncrypt = libAddr.getExportByName("CryptEncrypt");
var fn_CryptDecrypt = libAddr.getExportByName("CryptDecrypt");

var flag = null;
Interceptor.replace(fn_CryptEncrypt, fn_CryptDecrypt);
Interceptor.attach(fn_CryptDecrypt, {
    onEnter: function(args) {
        args[4].writeByteArray([0x5b,0x9c,0xee,0xb2,0x3b,0xb7,0xd7,0x34,0xf3,0x1b,0x75,0x14,0xc6,0xb2,0x1f,0xe8,0xde,0x33,0x44,0x74,0x75,0x1b,0x47,0x6a,0xd4,0x37,0x51,0x88,0xfc,0x67,0xe6,0x60,0xda,0x0d,0x58,0x07,0x81,0x43,0x53,0xea,0x7b,0x52,0x85,0x6c,0x86,0x65,0xaf,0xb4]);
        args[5].writeInt(0x40);
        flag = args[4];
        console.log("hook fn_CryptDecrypt"); 
        return args;
    }, 
    onLeave: function(arg) {
        console.log(flag.readCString());
        return arg;
    }
})
t0hka1
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t0hka1 1 2022-5-11 01:16
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sunfishi 给一个Hook过的&nbsp;&nbsp;var&nbsp;baseAddr&nbsp;=&nbsp;Process.findModuleByName(& ...
感谢师傅提供的脚本，之前比赛的时候确实也想过直接hook，奈何对Windows下hook脚本编写不太熟悉，学习到了！！
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