SM4加解密算法的c++实现

#include<bits/stdc++.h>
using  namespace std;
typedef long long ll;
#define Reverse(_x, _y) (((_x) << (_y)) | ((_x) >> (32 - (_y))))  
#define SboxTrans(_A) (Sbox[(_A) >> 24 & 0xFF] << 24 ^ Sbox[(_A) >> 16 & 0xFF] << 16 ^ Sbox[(_A) >>  8 & 0xFF] <<  8 ^ Sbox[(_A) & 0xFF])
#define L1(_B) ((_B) ^ Rotl(_B, 2) ^ Rotl(_B, 10) ^ Rotl(_B, 18) ^ Rotl(_B, 24))  
#define L2(_B) ((_B) ^ Rotl(_B, 13) ^ Rotl(_B, 23))    
const unsigned int CK[32] = {  
    0x00070e15, 0x1c232a31, 0x383f464d, 0x545b6269,  
    0x70777e85, 0x8c939aa1, 0xa8afb6bd, 0xc4cbd2d9,  
    0xe0e7eef5, 0xfc030a11, 0x181f262d, 0x343b4249,  
    0x50575e65, 0x6c737a81, 0x888f969d, 0xa4abb2b9,  
    0xc0c7ced5, 0xdce3eaf1, 0xf8ff060d, 0x141b2229,  
    0x30373e45, 0x4c535a61, 0x686f767d, 0x848b9299,  
    0xa0a7aeb5, 0xbcc3cad1, 0xd8dfe6ed, 0xf4fb0209,  
    0x10171e25, 0x2c333a41, 0x484f565d, 0x646b7279 };
const unsigned int RK[4]={0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC}; 
const unsigned char Sbox[256] = {  
    0xd6,0x90,0xe9,0xfe,0xcc,0xe1,0x3d,0xb7,0x16,0xb6,0x14,0xc2,0x28,0xfb,0x2c,0x05,  
    0x2b,0x67,0x9a,0x76,0x2a,0xbe,0x04,0xc3,0xaa,0x44,0x13,0x26,0x49,0x86,0x06,0x99,  
    0x9c,0x42,0x50,0xf4,0x91,0xef,0x98,0x7a,0x33,0x54,0x0b,0x43,0xed,0xcf,0xac,0x62,  
    0xe4,0xb3,0x1c,0xa9,0xc9,0x08,0xe8,0x95,0x80,0xdf,0x94,0xfa,0x75,0x8f,0x3f,0xa6,  
    0x47,0x07,0xa7,0xfc,0xf3,0x73,0x17,0xba,0x83,0x59,0x3c,0x19,0xe6,0x85,0x4f,0xa8,  
    0x68,0x6b,0x81,0xb2,0x71,0x64,0xda,0x8b,0xf8,0xeb,0x0f,0x4b,0x70,0x56,0x9d,0x35,  
    0x1e,0x24,0x0e,0x5e,0x63,0x58,0xd1,0xa2,0x25,0x22,0x7c,0x3b,0x01,0x21,0x78,0x87,  
    0xd4,0x00,0x46,0x57,0x9f,0xd3,0x27,0x52,0x4c,0x36,0x02,0xe7,0xa0,0xc4,0xc8,0x9e,  
    0xea,0xbf,0x8a,0xd2,0x40,0xc7,0x38,0xb5,0xa3,0xf7,0xf2,0xce,0xf9,0x61,0x15,0xa1,  
    0xe0,0xae,0x5d,0xa4,0x9b,0x34,0x1a,0x55,0xad,0x93,0x32,0x30,0xf5,0x8c,0xb1,0xe3,  
    0x1d,0xf6,0xe2,0x2e,0x82,0x66,0xca,0x60,0xc0,0x29,0x23,0xab,0x0d,0x53,0x4e,0x6f,  
    0xd5,0xdb,0x37,0x45,0xde,0xfd,0x8e,0x2f,0x03,0xff,0x6a,0x72,0x6d,0x6c,0x5b,0x51,  
    0x8d,0x1b,0xaf,0x92,0xbb,0xdd,0xbc,0x7f,0x11,0xd9,0x5c,0x41,0x1f,0x10,0x5a,0xd8,  
    0x0a,0xc1,0x31,0x88,0xa5,0xcd,0x7b,0xbd,0x2d,0x74,0xd0,0x12,0xb8,0xe5,0xb4,0xb0,  
    0x89,0x69,0x97,0x4a,0x0c,0x96,0x77,0x7e,0x65,0xb9,0xf1,0x09,0xc5,0x6e,0xc6,0x84,  
    0x18,0xf0,0x7d,0xec,0x3a,0xdc,0x4d,0x20,0x79,0xee,0x5f,0x3e,0xd7,0xcb,0x39,0x48  
};
const unsigned int Rotl(unsigned int n,int b){ return n<<b|n>>(32-b); } 
unsigned int xx[32];
void SM4KeyExt(unsigned int *key,unsigned int *rk, unsigned int CryptFlag){//秘钥扩展算法,flag为1代表解密 
    unsigned int r,tmp,k0,k1,k2,k3; 
    k0=key[0]^RK[0];  
    k1=key[1]^RK[1];  
    k2=key[2]^RK[2];  
    k3=key[3]^RK[3];
    for(r=0;r<32;r+=4){  
        /*rk(i) = k(4+i) = k(i) xor T[k(i+1) xor k(i+2) xor k(i+3) xor CK(i)]*/  
        /*合成置换T的过程包括非线性变换（ByteSub函数，从SBox中查找）和线性变换（L2函数，移位和异或运算）*/  
        tmp=k1^k2^k3^CK[r+0];  
        tmp=SboxTrans(tmp);  
        k0=k0^L2(tmp);  
        rk[r+0]=k0;  
  
        tmp=k2^k3^k0^CK[r + 1];  
        tmp=SboxTrans(tmp);  
        k1=k1^L2(tmp);  
        rk[r+1]=k1;  
          
        tmp= k3^k0^k1^CK[r+2];  
        tmp=SboxTrans(tmp);  
        k2=k2^L2(tmp);  
        rk[r+2]=k2;  
  
        tmp=k0^k1^k2^CK[r + 3];  
        tmp=SboxTrans(tmp);  
        k3=k3^L2(tmp);  
        rk[r+3]=k3;  
    }  
    if(CryptFlag==1){
        for(r=0;r<16;r++) swap(rk[r],rk[31-r]);  
    }
}

void SM4Crypt(unsigned int *Input, unsigned int *Output, unsigned int *rk){  
    unsigned int r, tmp, x0, x1, x2, x3, *y;  
    y=(unsigned int *)Input;  
    x0=y[0];  
    x1=y[1];  
    x2=y[2];  
    x3=y[3];  
    for (r=0;r<32;r+=4){  
        /*x4 = x0 ^ T(x1 ^ x2 ^ x3 ^ rk[0])*/  
        tmp=x1^x2^x3^rk[r+0];  
        tmp=SboxTrans(tmp);  
        x0^=L1(tmp);  
          xx[r+0]=x0;
        tmp=x2^x3^x0^rk[r+1];  
        tmp=SboxTrans(tmp);  
        x1^=L1(tmp);  
          xx[r+1]=x1;
        tmp=x3^x0^x1^rk[r+2];  
        tmp=SboxTrans(tmp);  
        x2^=L1(tmp);  
          xx[r+2]=x2;
        tmp=x0^x1^x2^rk[r+3];  
        tmp=SboxTrans(tmp);  
        x3^=L1(tmp);
        xx[r+3]=x3;  
    }  
    y=(unsigned int *)Output;  
    /*(y0,y1,y2,y3) = (x35,x34,x33,x32)*/ 
    y[0]=x3;  
    y[1]=x2;  
    y[2]=x1;  
    y[3]=x0;  
}

unsigned int key[4]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};
unsigned int miwen[4]={0x595298c7,0xc6fd271f,0x0402f804,0xc33d3f66};
unsigned int mingwen[4]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};
unsigned int rk[32];
unsigned int output[4]={};
void solve1(){
    printf("题目1:\n");
    SM4KeyExt(key,rk,0);
    printf("rk数组:\n"); 
    for(int i=0;i<32;i++) printf("%08x\n",rk[i]);
    printf("\n");
    printf("x数组：\n");
    SM4Crypt(mingwen,output,rk);
    for(int i=0;i<32;i++) printf("%08x\n",xx[i]);
    printf("结果：\n");
    for(int i=0;i<4;i++) printf("%08x ",output[i]);
    printf("\n");
}
void solve2(){
    printf("题目2：\n");
    for(int i=0;i<1000000;i++){
        SM4KeyExt(key,rk,0);
        SM4Crypt(mingwen,mingwen,rk);
    }
    for(int i=0;i<4;i++) cout<<hex<<mingwen[i]<<" ";
    cout<<"\n";
}
int main(){
    //freopen("out.txt","w",stdout);
    solve1();
    solve2();
    return 0;
}