一串代码求解

[ling1980]

signed int __fastcall Calculation(unsigned char *a1, unsigned char *a2, unsigned char *a3)
{
	char v15[16]; 
	char v25[16]; 
	char v23[16]; 

	for (int i = 0; i < 16; i++)
	{
		v15[i] = a2[i];
		v25[i] = a1[i] ^ a2[i];
	}

	int v6 = 0;
	LOBYTE(v6 = 0);
	while (1)
	{
		int v7 = 1 << v6;
		v6++;
		for (int j = 15; j > 7; j = (unsigned __int8)(j - 1))
		{
			if ((v7 >> j) & 1)
				v7 = (unsigned __int16)((283 << (j - 8)) ^ v7);
		}
		v15[0] = v15[0] ^ SurfaceGrid[(BYTE)v15[0x0D]] ^ v7;
		v15[1] = v15[1] ^ SurfaceGrid[(BYTE)v15[0x0E]];
		v15[2] = v15[2] ^ SurfaceGrid[(BYTE)v15[0x0F]];
		v15[3] = v15[3] ^ SurfaceGrid[(BYTE)v15[0x0C]];
		for (int k = 4; k < 16; k = k++)
			v15[k] = v15[k] ^ v15[k - 4];
		for (int l = 0; l < 16; l = l++)
			v23[(-3 * l & 0xF)] = SurfaceGrid[(BYTE)v25[l]];
		if (v6 == 10)
			break;
		for (int m = 0; m < 16; m = 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 = n--)
			{
				if ((v12 >> n) & 1)
					v12 = (283 << (n - 8)) ^ v12;
			}
			v25[m] = v15[m] ^ v12;
		}
	}
	for (int result = 0; result < 16; result = result++)
		a3[result] = v23[result] ^ v15[result];
	return 0;
}

unsigned char aa1[16] = { 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11 };
unsigned char aa2[16] = { 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22, 0x22 };
unsigned char aa3[16] = { 0x1A, 0x20, 0x0A, 0x8C, 0x16, 0x70, 0xE1, 0xDD, 0x43, 0x19, 0x8E, 0x69, 0xAB, 0x83, 0x08, 0x4F };
Calculation(aa1, aa2, aa3);
上面的程序是用第一个参数和第二个参数计算得出来第三个参数，假如我不知道第二个参数的内容，能不能用第一个和第三个参数得出第二个参数的内容
请高手指点

[schlafenhamster]

unsigned char *a2
去红的

[ling1980]

有没有方法？