• [原创]强网杯2024 solve2-apk WriteUp

solve2-apk

我们首先使用jeb进行分析，搜索关键词success定位到逻辑
image-20241129233403123

外层函数是一个魔改tea，过了前32字节检测才能进入下一个函数进行判断
image-20241129233648640

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#include <iostream>
#include <cstdio>
#include <stdint.h>  // For uint32_t
using namespace std;
 
void tea_decrypt(uint32_t* v) {
    uint32_t v0 = v[0], v1 = v[1], sum = 0xC6EF3720, i;
    uint32_t delta = 0x9e3779b9;
    uint32_t k[5] = { 598323648, 1213115916, 970832168, 274853062};
     
    for (i = 0; i < 32; i++) {
 
        v1 -= (((v0 << 4) + k[2] ^ v0) + (sum ^ (v0 >> 5)) + k[3]);
        v0 -= (((v1 << 4) + k[0] ^ v1) + (sum ^ (v1 >> 5)) + k[1]);
        sum -= delta;
    }
    v[0] = v0;
    v[1] = v1;
}
 
uint32_t switchEndian(uint32_t num) {
    return ((num >> 24) & 0x000000FF) | // 取最高字节
           ((num >> 8) & 0x0000FF00) | // 取第二字节
           ((num << 8) & 0x00FF0000) | // 取第三字节
           ((num << 24) & 0xFF000000); // 取最低字节
}
 
int main() {
    uint32_t key[] = { 598323648, 1213115916, 970832168, 274853062 };
 
    uint32_t data[] = {
        0x5E5440B0, 2057046228, 0x4A1ED228, 0x233FE7C, 0x96461450, 0x88A670ED, 0xF79BFC89, 0x20C3D75F,0
    };
 
    for (int i = 0; i < 8; i += 2) {
        tea_decrypt(&data[i]);
    }
     
    for (int i = 0; i < 8; ++i) {
         data[i] = switchEndian(data[i]);
    }
    printf("%s",data);
     
    return 0;
}
// Come on you are about to get it>

即可得到前32位的正确数据，将后面的测试数据放在>后继续在H0.a.successWithString()中进行二轮check
进入这个函数即可看到两个[256]的sbox，将部分数据搜索即可知为twofish算法
image-20241130153432307

image-20241130153245125

找到源码与jeb里的很相似

[link]:https://android.googlesource.com/platform/tools/base/+/master/jobb/src/main/java/Twofish/Twofish_Algorithm.java

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/**
 * Use (12, 8) Reed-Solomon code over GF(256) to produce a key S-box
 * 32-bit entity from two key material 32-bit entities.
 *
 * @param  k0  1st 32-bit entity.
 * @param  k1  2nd 32-bit entity.
 * @return  Remainder polynomial generated using RS code
 */
private static final int RS_MDS_Encode( int k0, int k1) {
   int r = k1;
   for (int i = 0; i < 4; i++) // shift 1 byte at a time
      r = RS_rem( r );
   r ^= k0;
   for (int i = 0; i < 4; i++)
      r = RS_rem( r );
   return r;
}

image-20241130154603346

我们在H0.a.c处下断点动调获取key

image-20241130164840825

image-20241130165101102

即可得到twofish的key
根据代码可知有两段data[16]，我们可以对v2[15]下断点得到所有的data
image-20241130165748281

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import twofish
 
key = bytes.fromhex("000102030405060708090a0b0c0d0e0f")  # key
tf = twofish.Twofish(key)
data1 = bytes([159, 46, 128, 211, 56, 34, 22, 223, 236, 150, 252, 143, 26, 34, 136, 115])
decrypted1 = tf.decrypt(data1)
print(decrypted1)
#flag{iT3N0t7H@tH

即可得到前半段flag，我们将前半部分flag输入进去再进行check即可得到part2的check

Come on you are about to get it>flag{iT3N0t7H@tH111111111111111}

之后有对我们传入的测试值的flag的part2的异或数据提取出来（这第二段算法是rc4，直接将加密后的值异或回去即可得到）
image-20241130174040159

将这段数据异或我们的输入再异或data2[16]即可还原得到第二段flag

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data2 = [169, 217, 118, 189, 119, 187, 86, 154, 49, 179, 222, 168, 101, 142, 26, 50]
enc1 = bytes([0xD8, 0xAD, 0x71, 0xC8, 0x76, 0xD3, 0x28, 0xFD, 0x37, 0xEA, 0xA6, 0xF7, 0x3F, 0xEC, 0x1B, 0x32])
enc2 = b'111111111111111}'
dec2 = ''.join(chr(data2[i] ^ enc1[i] ^ enc2[i]) for i in range(len(data2)))
print(dec2)
#@E6D0YOV7hInkS0}

这wp应该是出题人的预期解，本人走了许多弯路最终写出这份wp供大家学习