实现s_box

密码学

思路

　　使用笨方法，按照书上所说的三步走，还没想出更加高效的实现方法。输入一个十六进制数并输出相应的s_box中的值，进行简单的查表操作就可以了。具体代码如下。

代码：

#include <iostream>
#include <iomanip>
using namespace std;
int C = 0X63;
int s_box[16][16];
int x, y, d;
int q = 0, r = 0;
//定义G(2^8)中的 x*2
int mul2(int x) {
    if (x & 0x80) return ((x << 1) ^ 27) & 255;     //&255表示 只取一个字节
    else return (x << 1) & 255;
}
//G(2^8)乘法
int mul(int x, int y) {
    int r = 0, tmp = x;
    while (y) {
        if (y % 2) r ^= tmp;     //加法取模相当于亦或
        tmp = mul2(tmp);
        y /= 2;
    }
    return r;
}
//获取最高位1的位置
int hight_index(int x) {
    if (!x) return 0;   //x等于0，返回0。表示不存在最高位的1
    int index = 1;
    while (x >> 1) {
        x = x >> 1;
        ++index;
    }
    return index;
}
//G(2^n)除法
void divi(int a, int b) {
    int tmp = 0;
    r = 0; q = 0;
    while (b <= a) {
        tmp = 1;
        int indexB = hight_index(b), indexA = hight_index(a);
        int dis = indexA - indexB;
        tmp = tmp << dis;
        q += tmp;
        a = a ^ (tmp*b);
    }
    r = a;
}
//ax + by = d = gcd(a,b), 其中a>b
int egcd(int a, int b) {
    int x1 = 1, y1 = 0, x2 = 0, y2 = 1;
    int tmp_x, tmp_y;
    while (b) {
        divi(a, b);
        tmp_x = x2;
        tmp_y = y2;
        x2 = x1 ^ mul(x2, q);    //1式 减去 (2式 * q), 接着轮换系数
        y2 = y1 ^ mul(y2, q);
        x1 = tmp_x;
        y1 = tmp_y;
        a = b;
        b = r;
    }
    return y1;
}
// ---------------------------------------- 分割线 -------------------------------
//第二步：对s_box中的每个元素求逆元
void inverseSBox() {
    for (int i = 0; i < 16; ++i)
        for (int j = 0; j < 16; ++j)
            s_box[i][j] = egcd(283, s_box[i][j]);
}
int getIndexBit(int x, int index) {
    return (x >> index) & 1;
}
int transform(int x) {
    int trans_x = 0;
    for (int i = 0; i < 8; ++i) {
        int tmp = 0;
        tmp = getIndexBit(x, i) ^ getIndexBit(x, (i + 4) % 8) ^ getIndexBit(x, (i + 5) % 8) ^ getIndexBit(x, (i + 6) % 8) ^ getIndexBit(x, (i + 7) % 8) ^ getIndexBit(C, i);
        trans_x = trans_x | (tmp << i);
    }
    return trans_x;
}
//第三步：对s_box中的元素进行变化
void transformSBox() {
    for (int i = 0; i < 16; ++i) {
        for (int j = 0; j < 16; ++j) {
            s_box[i][j] = transform(s_box[i][j]);
        }
    }
}
void printSBox() {
    for (int i = 0; i < 16; ++i) {
        for (int j = 0; j < 16; ++j)
            cout << hex << setw(3) << s_box[i][j] << "    ";
        cout << endl;
    }
}
int main() {
    for (int i = 0; i < 16; ++i)
        for (int j = 0; j < 16; ++j)
            cin >> hex >> s_box[i][j];
    inverseSBox();
    transformSBox();
    printSBox();
    //输入一个值并输出相应的s_box值
    int xy;
    cin >> hex >> xy;
    int xx = (xy >> 4) & 15;
    int yy = xy & 15;
    cout << hex << xy << "对应的s_box值为" << s_box[xx][yy] << endl;
    system("pause");
    return 0;
}