G(2^8)有限域

G(2^8)有限域上的运算
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本原多项式P(x) = x8+x4+x3+x2+1        [1 0001 1101]
定义 P(x) = 0 则 x8 = x4+x3+x2+1    [0 0001 1101]
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区间[0,255]可以表示成各种多项式。例如：
    0   = 00000000 = 0
    1   = 00000001 = x0 = 1
    2   = 00000010 = x2
    129 = 10000001 = x7+x
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加法：按位异或
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乘法：
    3*7        = (x+1)*(x2+x+1)
            = x*x2+x*x+x+x2+x+1
            = x3+1
            = 00001001
            = 9
    129*5    = (x7+1)*(x2+1)
            = x9+x7+x2+1
            = x5+x4+x3+x1 + x7+x2+1    ( 因为: x9 = x5+x4+x3+x1 )
            = x7+x5+x4+x3+x2+x1+1
            = 10111111
            = 191
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x的指数运算：可以得出x^i(i=0,1,2……)，用于乘法运算。
    x0  = 00000001 = 1
    x1  = 00000010 = x1
    x2  = 00000100 = x2
    x3  = 00001000 = x3
    x4  = 00010000 = x4
    x5  = 00100000 = x5
    x6  = 01000000 = x6
    x7  = 10000000 = x7
    x8  = 00011101 = x4+x3+x2+1
    x9  = 00111010 = x1*x8 = x1*(x4+x3+x2+1) = x5+x4+x3+x1
    x10 = 01110100 = x1*x9 = x1*(x5+x4+x3+x1) = x6+x5+x4+x2
    x11 = 11101000
    x12 = 11001101 = x1*(x7+x6+x5+x3) = x8+x7+x6+x4
                                      = x4+x3+x2+1+x7+x6+x4
                                      = x7+x6+x3+x2+1
    x13 = 10000111
    ……

    x8 = x4+x3+x2+1 = 29 = 00111010
    x12= x7+x6+x3+x2+1 = 205 = 11001101
    则： 29*205=x20
    29/205 = x8/x12 = x251
    205/29 = x12/x8 = x4

/*-----------------------------------------------------
	用有限域实现加解密
	本原多项式:	 0x11B( x8+x4+x3+x+1 )
	明文:	C[i]
	密文:	M[i]
	DEF :	加密乘法因子 或 解密被除数. 注：DEF不能为0
	加密:	M[i] = C[i] * DEF
	解密:	C[i] = M[i] / DEF
*/
int XN[256], NX[256];
void genlt()
{
	int i, j;
	
	XN[0] = 1;
	for (i = 1; i < 256; i++){
		j = ( XN[i-1] << 1) ^ XN[i-1];
		if ((j & 0x100) != 0)	j ^= 0x11B;        
		XN[i] = j;
	}
	
	NX[0] = NX[1] = 0;
	for (i = 1; i < 255; i++)	NX[XN[i]] = i;
}
int mul( int a, int b )
{
	if( a==0 || b==0 )	return 0;
	else	return XN[ (NX[a]+NX[b])%255 ];
}
int div( int a, int b )
{
	int x;
	if( a==0 )	return 0;
	if( b==0 )	return -1;
	x = NX[a]-NX[b] < 0 ? NX[a]-NX[b]+255 : NX[a]-NX[b];
	return XN[x];
}
void main()
{
	genlt();
	int i;
	unsigned char DEF = 2;
	unsigned char C[256], M[256], O[256];
	sprintf( (char*)C, "zhongguo shi shufa de haiyang !" );
	printf( "%s\n", C );
	for( i=0; i<256; i++ ){
		M[i] = mul( DEF, C[i] );
	}
	printf( "%s\n", M );
	for( i=0; i<256; i++ ){
		O[i] = div( M[i], DEF );
	}
	printf( "%s\n", O );
}