AES加密算法——字节替代、列混合（c/c++）

AES加密算法

由于笔者最近上密码学这门课，分组密码的作业可谓是折磨。

以下是我自己写的相关计算代码。

列混合

#include <iostream>
using namespace std;

int StateMatrix[4][4];  // 状态矩阵

int muti(int hex1, int hex2){
    switch(hex1){
        case 0x01:
            return hex2;
        case 0x02:
            if(hex2 >= 0x80){
                hex2 = hex2 << 1;
                hex2 = hex2 % 32;
                hex2^= 0x1b;
            }
            else{
                hex2 = hex2 << 1;
                //hex2 = hex2 %16;
            }
            return hex2;
        case 0x03:
            return muti(0x01,hex2) ^ muti(0x02,hex2);
    }
    printf("出错啦！");
    return -1;
}

int main(){
    int input;
    for(int i=0;i<4;i++)
        for(int j=0;j<4;j++){
            cin >> hex >> input;        // 以16进制读取状态矩阵。
            StateMatrix[i][j] = input;
        }
    int d0,d1,d2,d3;
    for(int i=0;i<4;i++){
        d0 = muti(0x02,StateMatrix[0][i]) ^ muti(0x03,StateMatrix[1][i]);
        d0^= muti(0x01,StateMatrix[2][i]) ^ muti(0x01,StateMatrix[3][i]);

        d1 = muti(0x01,StateMatrix[0][i]) ^ muti(0x02,StateMatrix[1][i]);
        d1^= muti(0x03,StateMatrix[2][i]) ^ muti(0x01,StateMatrix[3][i]);

        d2 = muti(0x01,StateMatrix[0][i]) ^ muti(0x01,StateMatrix[1][i]);
        d2^= muti(0x02,StateMatrix[2][i]) ^ muti(0x03,StateMatrix[3][i]);

        d3 = muti(0x03,StateMatrix[0][i]) ^ muti(0x01,StateMatrix[1][i]);
        d3^= muti(0x01,StateMatrix[2][i]) ^ muti(0x02,StateMatrix[3][i]);

        printf("第%d列的结果是：\n%x %x %x %x\n",i,d0,d1,d2,d3);   
        StateMatrix[0][i] = d0;
        StateMatrix[1][i] = d1;
        StateMatrix[2][i] = d2;
        StateMatrix[3][i] = d3;
    } 
    return 0;
}
/*
e5 f3 59 47
e4 b5 24 34
59 c4 62 68
eb 01 8a 84
*/

字节替代

#include <iostream>
using namespace std;
 
unsigned char sbox[]={
		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,  
};
void PrintMatrix(unsigned char m[4][4])
{
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<4;j++)
		{
			printf("%4x",m[i][j]);
			if(j%4==3)
				
			puts("");
		}
	}
}
void SubByte(unsigned char state[][4])
{
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<4;j++)
		{
			state[i][j]=sbox[state[i][j]];      //S盒查表操作
		}
	}
}
int main()
{
	unsigned char state[4][4]={
		0x2a,0x7e,0x15,0x16,
		0x28,0xae,0xd2,0xa6,
		0xab,0xf7,0x15,0x88,
		0x09,0xcf,0x4f,0x3c,
	};
	printf("明文为:\n");PrintMatrix(state);
	SubByte(state);            //调用SubByte函数
	printf("经过S盒后输出:\n");PrintMatrix(state);
	return 0;
}

轮密钥加

我觉得这个东西，你自己稍微写一下就可以~
下面的东西特别乱，我自己都看不太懂那个是那个了。笑死。

#include <iostream>
using namespace std;

int main(){
    unsigned long long W[] = {0x2B7E1516,0x28AED2A6,0xABF71588,0x09CF4F3C};
    unsigned long long WW[] = {0x0,0x0,0x0,0x0};
    unsigned long long w0 = 0x7c636363;
    unsigned long long w1 = 0x63636363;
    printf("%llx ",w0^W[0]);
    for(int i=0;i<3;i++)
        printf("%llx jiehsu\n",w1^WW[i]);
    unsigned long long W1[] = {0x01000000,0x0,0x0,0x0};
    for(int i=0;i<4;i++)
        printf("\n%llx :end\n",W[i]^W1[i]);
    unsigned long long W4[] ={0x54f65544,0x7c34569d,0x363d3ca2,0x3e133486};
    unsigned long long W5[] = {0xA0FAFE17,0x88542CB1,0x23A33939,0x2A6C7605};
    for(int i=0;i<4;i++)
        printf("%llx abc\n",W4[i]^W5[i]);
    return 0;
}

比较差异

这个就异或一下，然后转化成2进制数1就行，

写作业的时候实在是觉得128bit的数据还是人工数1的个数比写程序快~

你有兴趣写完后面的那就更好了~~

#include <cstdio>
using namespace std;
int main(){
    typedef unsigned long long ll;
    ll c1[] ={0xf40cab53,0xf4607a2c,0x159e059b,0x147f4283};
    ll c10[]={0x01000000,0x00000000,0x00000000,0x00000000};
    ll c2[] ={0x63636363,0x63636363,0x63636363,0x63636363};
    ll c20[]={0x00000000,0x00000000,0x00000000,0x00000000};
    printf("\nc1^c10 = \n");
    for(int i=0;i<4;i++)
        printf("%llx\n",c1[i]^c10[i]);

    printf("\nc1^c2 = \n");
    for(int i=0;i<4;i++)
        printf("%llx\n",c1[i]^c2[i]);

    printf("\nc2^c20 = \n");
    for(int i=0;i<4;i++)
        printf("%llx\n",c2[i]^c20[i]);
}