当前位置：首页 > news >正文

2026 吾爱破解【春节】解题领红包之六 Windows 中级题

news 2026/3/27 11:09:54

upx 解压，得到 CM1.exe 是一个 DialogBoxParam 窗口程序

很容易找到 67 行的 sub_140008720 函数

sub_140008720 中，很多函数都加了会互相抵消的偏移，用来干扰分析。把它们 patch 掉即可

首先根据输入的 password 计算一个 64 位的 crc

v16 是 crc 的上下文，下面的结构体可能不准确，但其实不重要

struct crc64_ctx // sizeof=0x810
{unsigned __int64 crc;unsigned __int64 table[256];int count;int field_80C;
};

接下来是解密

此时 v16 变成了解密的上下文。经过亿点点时间分析，结构体如下

struct decrypt_ctx // sizeof=0x130
{__int64 key1;__int64 field_8;__int64 key2;__int64 field_18;BYTE sbox[256];unsigned int crc32;unsigned int expected_crc32;int field_128;int field_12C;
};

decrypt_data 每轮循环解密 8 个字节

每一轮使用的 key2 来自上一轮未解密的数据，key1 来自上一轮 key1 的变换。因此只要找到 key1 和 key2 的初始值即可。由 png 文件前 8 个字节确定，可以直接解出 key1 初始值。

#include <stdint.h>
#include <stdio.h>uint64_t ror64(uint64_t x, uint8_t n)
{n &= 63;return ((x >> n) | (x << (64 - n))) & 0xffffffffffffffff;
}uint64_t rol64(uint64_t x, uint8_t n)
{n &= 63;return ((x << n) | (x >> (64 - n))) & 0xffffffffffffffff;
}int main()
{uint8_t key1[8] = {};uint8_t key2[8] = {0xf5, 0x69, 0x73, 0x60, 0x1, 0xcb, 0x35, 0xbc};uint8_t sbox[256] = {0x63, 0x7c, 0x77, 0x7b, 0xf2, 0x6b, 0x6f, 0xc5, 0x30, 0x1, 0x67, 0x2b, 0xfe, 0xd7, 0xab, 0x76, 0xca, 0x82, 0xc9, 0x7d, 0xfa, 0x59, 0x47, 0xf0, 0xad, 0xd4, 0xa2, 0xaf, 0x9c, 0xa4, 0x72, 0xc0, 0xb7, 0xfd, 0x93, 0x26, 0x36, 0x3f, 0xf7, 0xcc, 0x34, 0xa5, 0xe5, 0xf1, 0x71, 0xd8, 0x31, 0x15, 0x4, 0xc7, 0x23, 0xc3, 0x18, 0x96, 0x5, 0x9a, 0x7, 0x12, 0x80, 0xe2, 0xeb, 0x27, 0xb2, 0x75, 0x9, 0x83, 0x2c, 0x1a, 0x1b, 0x6e, 0x5a, 0xa0, 0x52, 0x3b, 0xd6, 0xb3, 0x29, 0xe3, 0x2f, 0x84, 0x53, 0xd1, 0x0, 0xed, 0x20, 0xfc, 0xb1, 0x5b, 0x6a, 0xcb, 0xbe, 0x39, 0x4a, 0x4c, 0x58, 0xcf, 0xd0, 0xef, 0xaa, 0xfb, 0x43, 0x4d, 0x33, 0x85, 0x45, 0xf9, 0x2, 0x7f, 0x50, 0x3c, 0x9f, 0xa8, 0x51, 0xa3, 0x40, 0x8f, 0x92, 0x9d, 0x38, 0xf5, 0xbc, 0xb6, 0xda, 0x21, 0x10, 0xff, 0xf3, 0xd2, 0xcd, 0xc, 0x13, 0xec, 0x5f, 0x97, 0x44, 0x17, 0xc4, 0xa7, 0x7e, 0x3d, 0x64, 0x5d, 0x19, 0x73, 0x60, 0x81, 0x4f, 0xdc, 0x22, 0x2a, 0x90, 0x88, 0x46, 0xee, 0xb8, 0x14, 0xde, 0x5e, 0xb, 0xdb, 0xe0, 0x32, 0x3a, 0xa, 0x49, 0x6, 0x24, 0x5c, 0xc2, 0xd3, 0xac, 0x62, 0x91, 0x95, 0xe4, 0x79, 0xe7, 0xc8, 0x37, 0x6d, 0x8d, 0xd5, 0x4e, 0xa9, 0x6c, 0x56, 0xf4, 0xea, 0x65, 0x7a, 0xae, 0x8, 0xba, 0x78, 0x25, 0x2e, 0x1c, 0xa6, 0xb4, 0xc6, 0xe8, 0xdd, 0x74, 0x1f, 0x4b, 0xbd, 0x8b, 0x8a, 0x70, 0x3e, 0xb5, 0x66, 0x48, 0x3, 0xf6, 0xe, 0x61, 0x35, 0x57, 0xb9, 0x86, 0xc1, 0x1d, 0x9e, 0xe1, 0xf8, 0x98, 0x11, 0x69, 0xd9, 0x8e, 0x94, 0x9b, 0x1e, 0x87, 0xe9, 0xce, 0x55, 0x28, 0xdf, 0x8c, 0xa1, 0x89, 0xd, 0xbf, 0xe6, 0x42, 0x68, 0x41, 0x99, 0x2d, 0xf, 0xb0, 0x54, 0xbb, 0x16};uint8_t enc_data[8] = {0xfb, 0xdd, 0x1b, 0x92, 0x2b, 0xef, 0xe3, 0x23};uint8_t dec_data[8] = {0x89, 0x50, 0x4E, 0x47, 0x0D, 0x0A, 0x1A, 0x0A}; for(int i = 0; i < 8; i++){key1[i] = enc_data[i] ^ dec_data[i] ^ key2[i];for(int j = 0;j<256;j++){if(sbox[j] == key1[i]){key1[i] = j;break;}}}*(uint64_t*)key1 = ror64(*(uint64_t*)key1, 3);printf("%016llX\n", *(uint64_t*)key1);// 55A4F867BA4475DD
}

下断点，修改 crc64_final 返回值，得到 flag.png