下面是一个加密函数a1是明文a2是密钥a3是计算的结果，我的问题是有没有方法通过a1和a3计算出来a2的密钥是什么？
50只是个起步价，希望尽快解决
//数据加密
byte Calculation(unsigned char *a1, unsigned char *a2, unsigned char *a3)
{
char v25[16];
char v23[16] = { 0 };
unsigned char aa2[16] = { 0 };
for (int i = 0; i < 16; i++)
aa2[i] = a2[i];
for (int i = 0; i < 16; i++)
v25[i] = a1[i] ^ aa2[i];
int v6 = 0;
while (1)
{
int v7 = 1 << v6;
v6++;
for (int j = 15; j > 7; j--)
{
if ((v7 >> j) & 1)
v7 = ((283 << (j - 8)) ^ v7);
}
aa2[0] = aa2[0] ^ byte_8019EA8[(BYTE)aa2[0x0D]] ^ v7;
aa2[1] = aa2[1] ^ byte_8019EA8[(BYTE)aa2[0x0E]];
aa2[2] = aa2[2] ^ byte_8019EA8[(BYTE)aa2[0x0F]];
aa2[3] = aa2[3] ^ byte_8019EA8[(BYTE)aa2[0x0C]];
for (int k = 4; k < 16; k++)
aa2[k] = aa2[k] ^ aa2[k - 4];
for (int l = 0; l < 16; l++)
v23[(-3 * l & 0xF)] = byte_8019EA8[(BYTE)v25[l]];
if (v6 == 10) //==后面是循环次数
break;
for (int m = 0; m < 16; m++)
{
int v12 = 2 * (BYTE) * (&v23[(m & 0xC | (m + 1) & 3)]) ^ 2 * (BYTE) * (&v23[m]) ^ (BYTE) * (&v23[(m & 0xC | (m + 1) & 3)]) ^ (BYTE) * (&v23[(m & 0xC | (m + 2) & 3)]) ^ (BYTE) * (&v23[(m & 0xC | (m + 3) & 3)]);
for (int n = 15; n > 7; n--)
{
if ((v12 >> n) & 1)
v12 = (283 << (n - 8)) ^ v12;
}
v25[m] = aa2[m] ^ v12;
}
}
for (int result = 0; result < 16; result++)
a3[result] = v23[result] ^ aa2[result];
return 0;
}
unsigned char byte_8019EA8[256] =
{
0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01, 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,
0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75,
0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84,
0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF,
0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8,
0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5, 0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2,
0xCD, 0x0C, 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, 0x0B, 0xDB,
0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 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, 0x08,
0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6, 0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A,
0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 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, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16
};
你这个加密是自定义的加密方式？
可以参照这个帖子
https://bbs.csdn.net/topics/606212026
估计不行，你这个是不可逆加密算法，我发实现解密。