AES的S-BOX构造优化

之前写过SBOX的构造，后来看到别人的优秀思路，借鉴过来重新改了一点。

原文地址：http://www.cnblogs.com/7hat/p/3383546.html

主要是将矩阵运算改为列运算之和，提高了效率。

#include<iostream>
#include<fstream>
#include <iomanip>
using namespace std;
unsigned char exp[256], log[256], inv[256];
unsigned char GFmul(unsigned char a, unsigned char b){
    //GF(2^8) 乘法
    unsigned char result = 0;
    if((b&1) == 1)result = a;
    b >>= 1;
    for(int i = 1; i < 8; i ++){
        if(a > 127){
            a = (a << 1) ^ 0x1b;
        }
        else{
            a <<= 1;
        }
        if((b&1) == 1){
            result ^= a;
        }
        b >>= 1;
    }
    return result;
}
void generateMulTab(){
    //选择生成元3作为构造乘法表的基础
    const int N = 3;
    exp[1] = N;
    log[N] = 1;
    unsigned char tmp = N;
    for(int i = 2; i < 256; i ++){
        tmp = GFmul(tmp, N);
        exp[i] = tmp;
        log[tmp] = i;
    }
}
void generateMulInverse(){
    //利用exp来构造乘法逆元
    inv[0] = 0;
    inv[1] = 1;
    //若3^m * 3^n = 1 = 3^255，则 m + n = 255
    for(int i = 1; i < 255; i ++){
        inv[exp[i]] = exp[255-i];
    }
}
unsigned char SBoxValue(unsigned char x){
    //返回SBOX对应的值
    unsigned char y = inv[x];
    unsigned char result = 0;
    unsigned char c = 0x1f;         //常数矩阵的第一列
    while(y){
        //将矩阵运算转化为列运算之和，因为y只是一个"一维"的矩阵 
        if(y&1){
            //yi 表示是否要加上对应ci的列
            result ^= c;
        }
        y >>= 1;
        //循环左移为下一列元素
        c = (c<<1) | (c>>7);
    }
    result ^= 0x63;
    return result;
}
int main(){
    //单元测试，输出SBOX的全部值
    generateMulTab();
    generateMulInverse();
    unsigned char SBox[16][16];
    for(int i = 0; i < 16; i ++){
        for(int j = 0; j < 16; j ++){
            unsigned char tmp = i*16 + j;
            SBox[i][j] = SBoxValue(tmp);
        }
    }
    ofstream write("Test.txt");
    for(int i = 0; i < 16; i ++){
        for(int j = 0; j < 16; j ++){
            write<<setiosflags(ios::left)<<setw(4)<<hex<<(int)SBox[i][j];
        }
        write<<endl;
    }
    write.close();
    return 0;
}