52pojie【2022春节】解题领红包WP

从哪里跌倒就从哪里起来。还记得去年罚坐的痛苦，今年来填坑了。

Android题花了一点时间学习ollvm，大概也算是初步入门了。

另外几题的难度相当友好，考察基础和技巧。

封面引自52pojie论坛，侵删。

Windows 简单题 2/2

比对数据自增

arr = bytearray([0x00000031, 0x0000002F, 0x00000031, 0x00000031, 0x00000047, 0x00000060, 0x0000006F, 0x0000006F, 0x00000078, 0x0000004D, 0x00000064, 0x00000076, 0x00000058, 0x00000064, 0x00000060, 0x00000071, 0x00000034, 0x00000031, 0x0000004F, 0x0000006E, 0x00000049, 0x00000068, 0x00000064])
for i in range(len(arr)):
    arr[i] += 1
print(arr)

Windows 中等题 2/3

魔改upx，OEP脱掉

输入UID、Key，一堆单字符变换后对比数据

简单看了下变换，分成a~z、A~Z、其他字符。

规定

• v11 = UID % 25
• v10来自函数sub_401110不作分析

有下面处理规则

• a~z：flag[i] = v10 * (flag[i] - v11 - 97) % 26 + 'a'
• A~Z：flag[i] = v10 * (flag[i] - v11 - 65) % 26 + 'A'

而后flag[i] += 26

• 其他字符：不作处理

实际上是仿射密码（Affine Cipher），其加解密公式如下

C = (a * P + b) mod 26
P = (a ^ -1 * (C - b)) mod 26

具体可以前往Affine cipher – Wikipedia了解

偷懒方法

最简单的办法，直接映射

调试拿到表

屑python跑下

raw = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" + "0123456789_{}"
new = "kjihgfedcbazyxwvutsrqponmlKJIHGFEDCBAZYXWVUTSRQPONML" + "0123456789_{}"


fake = "flag{Happy_New_Year_52Pojie_2022}"
flag = fake.translate(fake.maketrans(new, raw))
print(flag)

正常方法

之前提到的v10来自函数sub_401110，这个函数其实是求了模逆元

糊了下完整的算法

from gmpy2 import invert

CH_A = ord('A')
CH_Z = ord('Z')
CH_a = ord('a')
CH_z = ord('z')
PRIME = [1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25]
FAKE = b"flag{Happy_New_Year_52Pojie_2022}"


def cmod(_v: int, _m: int) -> int:
    signed = 1
    if _v < 0:
        signed = -1
    _v = abs(_v)
    return signed * (_v % _m)


def round(text: bytearray or bytes, _a: int, _b: int):
    out = bytearray(text)
    for i in range(len(out)):
        v = out[i]
        if CH_A <= v <= CH_Z:
            v = cmod(_a * (v - _b - CH_A), 26) + CH_A
            if _a * (out[i] - _b - CH_A) < 0:
                v += 26

        elif CH_a <= v <= CH_z:
            v = cmod(_a * (v - _b - CH_a), 26) + CH_a
            if _a * (out[i] - _b - CH_a) < 0:
                v += 26
        out[i] = v
    return out


def encrypt(text: bytearray or bytes, _a: int, _b: int):
    return round(text, _a, _b)


def decrypt(text: bytearray or bytes, _a: int, _b: int):
    return round(text, invert(_a, 26), _b)


if __name__ == '__main__':
    uid = 1150835
    k1 = PRIME[uid % 12]
    k2 = uid % 25
    txt = encrypt(FAKE, k1, k2)
    print(txt)
    # print(decrypt(fake, k1, k2))

Android 中级题 2/4

看到ndk调用checkSn，输入长度限制16位。

通过导出函数表中的compare定位到主要函数

手动创建函数

有经验的能看出是ollvm，加了bcf和sub

不透明谓词

能看到形如下图的不透明谓词（相邻数字相乘恒为偶数）

处理办法：

利用HexRays Decomplier的DCE

• Edit Segment，取消Write勾选
• .bss段patch 0

处理效果如下图

另一种不透明谓词形式为*(off_xxx + xxx) - xxx)

利用相同的办法修改.data，取消勾选Write，手动转换为具体值

形如MOV W9, #(off_36318 - 0xFFFFFFFF85C92A64)，选中0xFFFFFFFF85C92A64按Q

BCF

显而易见，之前创建的函数并不完整。

手动选中创建后Decomplier显示形似__asm { BR X8 }

动调起来，trace后patch

修正后的checkSN大概长这样

特征及解析

逐步分析各个函数，大量的BCF。

调试过程中能发现一些算法特征

FK

找份源码对着看

SMx/sm4.c at master · NEWPLAN/SMx (github.com)

反复调试，观察函数调用，调用形式如下所示

分析后得到的一些流程和数据

关键调用定位

• sm4F：+0x0AB9C
• sm4Sbox：+0x0AB44
• sm4Lt：+0x0ACEC
• sm4_setkey：+0x081BC

• 执行流程：

one round:
out[0:16] = sm4(xor(input[0:16] ^ iv[0:16]))

two round:
out[16:32] = sm4(xor(input[16:32] ^ out[0:16]))

实际上后16字节没什么用，只是CBC的填充：输入字符的截断（\x00）以及15字节的\10

• 对比数据：6E6649305BAF80C49B1B063C0500C80346CCFD42B3063AE7312B52A21CD334D8
• iv：022B26284A337015F04DE065E05FC094
• sk：

static const QWORD sk[32] = {
            0x003C3B90E65E1F82, 0x0072C025EC20D5A2, 0x004145992B8EE3EF, 0x0069CFF9C3113C5E,
            0x0073A0BDC7E35BB5, 0x006587F6717F5417, 0x0033F15055CC7021, 0x0054D89634FD983E,
            0x007FB7780A7FADBB, 0x004F4F92EF4F948E, 0x005F72AC76224645, 0x0064FEDB2AE2D1AC,
            0x005AA8969E2AC4BC, 0x00680FEA99EF2053, 0x006C1E634F07843E, 0x000677673B06C935,
            0x003D1B6BE4557920, 0x001B04F5E1299F9C, 0x00326A513916A44C, 0x001FF10831D1D12E,
            0x00072FB7BDB5A88E, 0x006C3867E930366F, 0x004F884102F7796B, 0x0007005C6149330C,
            0x000284C586C9B290, 0x0019CA128D08D882, 0x00544B18A95A6BDE, 0x006A14600A6179ED,
            0x001D4BE54074B82B, 0x007581763C59130D, 0x004BB3C753B8931B, 0x00668BD94D4ABAA7
    };

• sbox未作改变
• 改变了PUT_ULONG_BE/GET_ULONG_BE，具体为1、2字节换位

#ifndef MODIFY_GET_ULONG_BE
#define MODIFY_GET_ULONG_BE(n,b,i)                      \
{                                                       \
    (n) = ( (unsigned long) (b)[(i)    ] << 24 )        \
        | ( (unsigned long) (b)[(i) + 1] <<  8 )        \
        | ( (unsigned long) (b)[(i) + 2] << 16 )        \
        | ( (unsigned long) (b)[(i) + 3]       );       \
}
#endif

#ifndef MODIFY_PUT_ULONG_BE
#define MODIFY_PUT_ULONG_BE(n,b,i)                      \
{                                                       \
    (b)[(i)    ] = (unsigned char) ( (n) >> 24 );       \
    (b)[(i) + 1] = (unsigned char) ( (n) >>  8 );       \
    (b)[(i) + 2] = (unsigned char) ( (n) >> 16 );       \
    (b)[(i) + 3] = (unsigned char) ( (n)       );       \
}
#endif

完整解题脚本如下：

#include <stdio.h>
#include <stdint.h>
#include <string.h>

static const unsigned char SboxTable[16][16] =
        {
                {0xd6, 0x90, 0xe9, 0xfe, 0xcc, 0xe1, 0x3d, 0xb7, 0x16, 0xb6, 0x14, 0xc2, 0x28, 0xfb, 0x2c, 0x05},
                {0x2b, 0x67, 0x9a, 0x76, 0x2a, 0xbe, 0x04, 0xc3, 0xaa, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99},
                {0x9c, 0x42, 0x50, 0xf4, 0x91, 0xef, 0x98, 0x7a, 0x33, 0x54, 0x0b, 0x43, 0xed, 0xcf, 0xac, 0x62},
                {0xe4, 0xb3, 0x1c, 0xa9, 0xc9, 0x08, 0xe8, 0x95, 0x80, 0xdf, 0x94, 0xfa, 0x75, 0x8f, 0x3f, 0xa6},
                {0x47, 0x07, 0xa7, 0xfc, 0xf3, 0x73, 0x17, 0xba, 0x83, 0x59, 0x3c, 0x19, 0xe6, 0x85, 0x4f, 0xa8},
                {0x68, 0x6b, 0x81, 0xb2, 0x71, 0x64, 0xda, 0x8b, 0xf8, 0xeb, 0x0f, 0x4b, 0x70, 0x56, 0x9d, 0x35},
                {0x1e, 0x24, 0x0e, 0x5e, 0x63, 0x58, 0xd1, 0xa2, 0x25, 0x22, 0x7c, 0x3b, 0x01, 0x21, 0x78, 0x87},
                {0xd4, 0x00, 0x46, 0x57, 0x9f, 0xd3, 0x27, 0x52, 0x4c, 0x36, 0x02, 0xe7, 0xa0, 0xc4, 0xc8, 0x9e},
                {0xea, 0xbf, 0x8a, 0xd2, 0x40, 0xc7, 0x38, 0xb5, 0xa3, 0xf7, 0xf2, 0xce, 0xf9, 0x61, 0x15, 0xa1},
                {0xe0, 0xae, 0x5d, 0xa4, 0x9b, 0x34, 0x1a, 0x55, 0xad, 0x93, 0x32, 0x30, 0xf5, 0x8c, 0xb1, 0xe3},
                {0x1d, 0xf6, 0xe2, 0x2e, 0x82, 0x66, 0xca, 0x60, 0xc0, 0x29, 0x23, 0xab, 0x0d, 0x53, 0x4e, 0x6f},
                {0xd5, 0xdb, 0x37, 0x45, 0xde, 0xfd, 0x8e, 0x2f, 0x03, 0xff, 0x6a, 0x72, 0x6d, 0x6c, 0x5b, 0x51},
                {0x8d, 0x1b, 0xaf, 0x92, 0xbb, 0xdd, 0xbc, 0x7f, 0x11, 0xd9, 0x5c, 0x41, 0x1f, 0x10, 0x5a, 0xd8},
                {0x0a, 0xc1, 0x31, 0x88, 0xa5, 0xcd, 0x7b, 0xbd, 0x2d, 0x74, 0xd0, 0x12, 0xb8, 0xe5, 0xb4, 0xb0},
                {0x89, 0x69, 0x97, 0x4a, 0x0c, 0x96, 0x77, 0x7e, 0x65, 0xb9, 0xf1, 0x09, 0xc5, 0x6e, 0xc6, 0x84},
                {0x18, 0xf0, 0x7d, 0xec, 0x3a, 0xdc, 0x4d, 0x20, 0x79, 0xee, 0x5f, 0x3e, 0xd7, 0xcb, 0x39, 0x48}
        };

#define  SHL(x,n) (((x) & 0xFFFFFFFF) << n)
#define ROTL(x,n) (SHL((x),n) | ((x) >> (32 - n)))

/*
 * 32-bit integer manipulation macros (big endian)
 */
#ifndef GET_ULONG_BE
#define GET_ULONG_BE(n,b,i)                             \
{                                                       \
    (n) = ( (unsigned long) (b)[(i)    ] << 24 )        \
        | ( (unsigned long) (b)[(i) + 1] << 16 )        \
        | ( (unsigned long) (b)[(i) + 2] <<  8 )        \
        | ( (unsigned long) (b)[(i) + 3]       );       \
}
#endif

#ifndef PUT_ULONG_BE
#define PUT_ULONG_BE(n,b,i)                             \
{                                                       \
    (b)[(i)    ] = (unsigned char) ( (n) >> 24 );       \
    (b)[(i) + 1] = (unsigned char) ( (n) >> 16 );       \
    (b)[(i) + 2] = (unsigned char) ( (n) >>  8 );       \
    (b)[(i) + 3] = (unsigned char) ( (n)       );       \
}
#endif

#ifndef MODIFY_GET_ULONG_BE
#define MODIFY_GET_ULONG_BE(n,b,i)                      \
{                                                       \
    (n) = ( (unsigned long) (b)[(i)    ] << 24 )        \
        | ( (unsigned long) (b)[(i) + 1] <<  8 )        \
        | ( (unsigned long) (b)[(i) + 2] << 16 )        \
        | ( (unsigned long) (b)[(i) + 3]       );       \
}
#endif

#ifndef MODIFY_PUT_ULONG_BE
#define MODIFY_PUT_ULONG_BE(n,b,i)                      \
{                                                       \
    (b)[(i)    ] = (unsigned char) ( (n) >> 24 );       \
    (b)[(i) + 1] = (unsigned char) ( (n) >>  8 );       \
    (b)[(i) + 2] = (unsigned char) ( (n) >> 16 );       \
    (b)[(i) + 3] = (unsigned char) ( (n)       );       \
}
#endif

typedef unsigned char BYTE;
typedef unsigned long DWORD;
typedef unsigned short WORD;
typedef unsigned long long QWORD;

/*
 * private function:
 * look up in SboxTable and get the related value.
 * args:    [in] inch: 0x00~0xFF (8 bits unsigned value).
 */
static unsigned char sm4Sbox(unsigned char inch)
{
    unsigned char *pTable = (unsigned char *)SboxTable;
    unsigned char retVal = (unsigned char)(pTable[inch]);
    return retVal;
}


static QWORD sm4Lt(QWORD ka)
{
    QWORD bb = 0;
    QWORD c = 0;
    BYTE a[4];
    BYTE b[4];
    PUT_ULONG_BE(ka, a, 0)
    b[0] = sm4Sbox(a[0]);
    b[1] = sm4Sbox(a[1]);
    b[2] = sm4Sbox(a[2]);
    b[3] = sm4Sbox(a[3]);
    GET_ULONG_BE(bb, b, 0)
    c = bb ^ (ROTL(bb, 2)) ^ (ROTL(bb, 10)) ^ (ROTL(bb, 18)) ^ (ROTL(bb, 24));
    return c;
}

static QWORD sm4F(QWORD x0, QWORD x1, QWORD x2, QWORD x3, QWORD rk)
{
    return (x0 ^ sm4Lt(x1 ^ x2 ^ x3 ^ rk));
}

/*
 * SM4 standard one round processing
 *
 */
static void sm4_one_round( QWORD sk[32],
                           unsigned char input[16],
                           unsigned char output[16] )
{
    unsigned long i = 0;
    QWORD ulbuf[36];

    memset(ulbuf, 0, sizeof(ulbuf));
    MODIFY_GET_ULONG_BE( ulbuf[0], input, 0 )
    MODIFY_GET_ULONG_BE( ulbuf[1], input, 4 )
    MODIFY_GET_ULONG_BE( ulbuf[2], input, 8 )
    MODIFY_GET_ULONG_BE( ulbuf[3], input, 12 )

    while (i < 32)
    {
        ulbuf[i + 4] = sm4F(ulbuf[i], ulbuf[i + 1], ulbuf[i + 2], ulbuf[i + 3], sk[i]);

        printf("rk(%02d) = 0x%016llx,  X(%02d) = 0x%016llx \n",i,sk[i], i+4, ulbuf[i+4] );

        i++;
    }

    MODIFY_PUT_ULONG_BE(ulbuf[35], output, 0);
    MODIFY_PUT_ULONG_BE(ulbuf[34], output, 4);
    MODIFY_PUT_ULONG_BE(ulbuf[33], output, 8);
    MODIFY_PUT_ULONG_BE(ulbuf[32], output, 12);
}


int  main()
{
    static const QWORD sk[32] = {
            0x003C3B90E65E1F82, 0x0072C025EC20D5A2, 0x004145992B8EE3EF, 0x0069CFF9C3113C5E,
            0x0073A0BDC7E35BB5, 0x006587F6717F5417, 0x0033F15055CC7021, 0x0054D89634FD983E,
            0x007FB7780A7FADBB, 0x004F4F92EF4F948E, 0x005F72AC76224645, 0x0064FEDB2AE2D1AC,
            0x005AA8969E2AC4BC, 0x00680FEA99EF2053, 0x006C1E634F07843E, 0x000677673B06C935,
            0x003D1B6BE4557920, 0x001B04F5E1299F9C, 0x00326A513916A44C, 0x001FF10831D1D12E,
            0x00072FB7BDB5A88E, 0x006C3867E930366F, 0x004F884102F7796B, 0x0007005C6149330C,
            0x000284C586C9B290, 0x0019CA128D08D882, 0x00544B18A95A6BDE, 0x006A14600A6179ED,
            0x001D4BE54074B82B, 0x007581763C59130D, 0x004BB3C753B8931B, 0x00668BD94D4ABAA7
    };
    unsigned char iv[16] = {2, 43, 38, 40, 74, 51, 112, 21, 240, 77, 224, 101, 224, 95, 192, 148};
    unsigned char out[16] = {0};
    unsigned char input[16] = {110, 102, 73, 48, 91, 175, 128, 196, 155, 27, 6, 60, 5, 0, 200, 3};

    // decrypt
    QWORD inv_sk[32] = {
        0x00668BD94D4ABAA7, 0x004BB3C753B8931B, 0x007581763C59130D, 0x001D4BE54074B82B,
0x006A14600A6179ED, 0x00544B18A95A6BDE, 0x0019CA128D08D882, 0x000284C586C9B290, 
0x0007005C6149330C, 0x004F884102F7796B, 0x006C3867E930366F, 0x00072FB7BDB5A88E, 
0x001FF10831D1D12E, 0x00326A513916A44C, 0x001B04F5E1299F9C, 0x003D1B6BE4557920, 
0x000677673B06C935, 0x006C1E634F07843E, 0x00680FEA99EF2053, 0x005AA8969E2AC4BC, 
0x0064FEDB2AE2D1AC, 0x005F72AC76224645, 0x004F4F92EF4F948E, 0x007FB7780A7FADBB, 
0x0054D89634FD983E, 0x0033F15055CC7021, 0x006587F6717F5417, 0x0073A0BDC7E35BB5, 
0x0069CFF9C3113C5E, 0x004145992B8EE3EF, 0x0072C025EC20D5A2, 0x003C3B90E65E1F82
    }; // SK[::-1]

    sm4_one_round(inv_sk, input, out);

    for (int i = 0; i < 16; ++i) {
        out[i] ^= iv[i];
    }

    for (int i = 0; i < 16; ++i) {
        printf("%02x", out[i]);
    }
    printf("\n");


    return 0;
}

得到答案

Web 中级题 2/5

fidder载入saz，dump出live_*.ts

看m3u8，加密方式为AES-128

#EXT-X-KEY:METHOD=AES-128,URI="key://live"

简单分析下script.bundle.js

key和iv的构造方式如下

• 随机字符SHA-256生成32字节取前16字节，作为参数h调用drm接口
• 响应返回32字节，前16字节与h异或后与后16字节异或得到key
• iv通过r.iv = this.createInitializationVector(e)获得，实际上e = 0，简单运行下，其实是b'\x00' * 16

var t = 0
for (var e = new Uint8Array(16), r = 12; r < 16; r++)
    e[r] = t >> 8 * (15 - r) & 255;
console.log(e)

脚本批量解密后合并

# -*- coding:utf-8 -*-
"""
@Author: Mas0n
@File: 52pojie_5.py
@Time: 2022-02-06 1:11
@Desc: It's all about getting better.
"""

import os

from Crypto.Cipher import AES

h = b''.fromhex("7b10311e6e310f0df068d9ede10475a8")
res = bytearray([0x08, 0xA5, 0xE6, 0xC2, 0xC2, 0x61, 0xA8, 0xAC, 0xB4, 0xD7, 0x9C, 0x49, 0xAF, 0x16, 0x0A, 0x3A, 0xDA, 0x4E, 0x5C, 0xEA, 0xE1, 0x6F, 0xED, 0x46, 0xEB, 0x6F, 0x49, 0x8C, 0x9B, 0x63, 0xD5, 0x3B])
tmp = res[0:16]
tmp2 = res[16:32]

a = [h[i] ^ tmp[i] for i in range(16)]
b = bytearray([a[i] ^ tmp2[i] for i in range(16)])

ctx = AES.new(key=b, mode=AES.MODE_CBC, iv=b"\x00"*16)

data = b""
for path in os.listdir(r"tsdec"):
    test = open(r"tsdec\\" + path, "rb").read()
    dec = ctx.decrypt(test)
    data += dec

open(r"out\\" + "merge.ts", "wb").write(data)

参考

• Affine Ciphers
• android so文件攻防实战-百度加固免费版libbaiduprotect.so反混淆
• Hex-Rays: 十步杀一人，两步秒OLLVM-BCF
• The SM4 Blockcipher Algorithm And Its Modes Of Operations
• 加密ts文件解密