百度到的论文已给出算法。

flag为1为解密，flag为0是加密。

#include<bits/stdc++.h>

using  namespace std;

typedef long long ll;

#define Rotl(_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(x) ((x)^Rotl(x,2)^Rotl(x,10)^Rotl(x,18)^Rotl(x,24))

#define L2(x) ((x)^Rotl(x,13)^Rotl(x,23))

const unsigned int CK[] = {

    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[]={0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC};

const unsigned char Sbox[] = {

    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[];

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[]^RK[];

    k1=key[]^RK[];

    k2=key[]^RK[];

    k3=key[]^RK[];

    for(r=;r<;r+=){

        /*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+];

        tmp=SboxTrans(tmp);

        k0=k0^L2(tmp);

        rk[r+]=k0;  

        tmp=k2^k3^k0^CK[r + ];

        tmp=SboxTrans(tmp);

        k1=k1^L2(tmp);

        rk[r+]=k1;  

        tmp= k3^k0^k1^CK[r+];

        tmp=SboxTrans(tmp);

        k2=k2^L2(tmp);

        rk[r+]=k2;  

        tmp=k0^k1^k2^CK[r + ];

        tmp=SboxTrans(tmp);

        k3=k3^L2(tmp);

        rk[r+]=k3;

    }

    if(CryptFlag==){

        for(r=;r<;r++) swap(rk[r],rk[-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[];

    x1=y[];

    x2=y[];

    x3=y[];

    for (r=;r<;r+=){

        /*x4 = x0 ^ T(x1 ^ x2 ^ x3 ^ rk[0])*/

        tmp=x1^x2^x3^rk[r+];

        tmp=SboxTrans(tmp);

        x0^=L1(tmp);

          xx[r+]=x0;

        tmp=x2^x3^x0^rk[r+];

        tmp=SboxTrans(tmp);

        x1^=L1(tmp);

          xx[r+]=x1;

        tmp=x3^x0^x1^rk[r+];

        tmp=SboxTrans(tmp);

        x2^=L1(tmp);

          xx[r+]=x2;

        tmp=x0^x1^x2^rk[r+];

        tmp=SboxTrans(tmp);

        x3^=L1(tmp);

        xx[r+]=x3;

    }

    y=(unsigned int *)Output;

    /*(y0,y1,y2,y3) = (x35,x34,x33,x32)*/

    y[]=x3;

    y[]=x2;

    y[]=x1;

    y[]=x0;

}

unsigned int key[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int miwen[]={0x595298c7,0xc6fd271f,0x0402f804,0xc33d3f66};

unsigned int mingwen[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int rk[];

unsigned int output[]={};

void solve1(){

    printf("题目1:\n");

    SM4KeyExt(key,rk,);

    printf("rk数组:\n");

    for(int i=;i<;i++) printf("%08x\n",rk[i]);

    printf("\n");

    printf("x数组：\n");

    SM4Crypt(mingwen,output,rk);

    for(int i=;i<;i++) printf("%08x\n",xx[i]);

    printf("结果：\n");

    for(int i=;i<;i++) printf("%08x ",output[i]);

    printf("\n");

}

void solve2(){

    printf("题目2：\n");

    SM4KeyExt(key,rk,);

    for(int i=;i<;i++){

        SM4Crypt(mingwen,mingwen,rk);

    }

    for(int i=;i<;i++) cout<<hex<<mingwen[i]<<" ";

    cout<<"\n";

}

int main(){

    //freopen("out.txt","w",stdout);

    solve1();

    solve2();

    return ;

}

将模板缩减了一下，sbox的结果是unsigned char，进行移位运算之后是int，SboxTrans的结果为int，右移是算术右移导致出错，最好分开写或者强转。

#include<bits/stdc++.h>

using  namespace std;

typedef long long ll;

#define Rotl(_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(x) ((x)^Rotl(x,2)^Rotl(x,10)^Rotl(x,18)^Rotl(x,24))

#define L2(x) ((x)^Rotl(x,13)^Rotl(x,23))

const unsigned int CK[] = {

    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[]={0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC};

const unsigned char Sbox[] = {

    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[];

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,k[];

    for(int i=;i<;i++) k[i]=key[i]^RK[i];

    for(int i=;i<;i+=){

        for(int j=;j<;j++){

            tmp=SboxTrans(k[(j+)%]^k[(j+)%]^k[(j+)%]^CK[i+j]);

            rk[i+j]=k[j]^=L2(tmp);

        }

    }

    if(CryptFlag==) for(r=;r<;r++) swap(rk[r],rk[-r]);  

}

void SM4Crypt(unsigned int *Input, unsigned int *Output, unsigned int *rk){

    unsigned int tmp, x[], *y;

    y=(unsigned int *)Input;

    for(int i=;i<;i++) x[i]=y[i];

    for(int i=;i<;i+=){

        for(int j=;j<;j++){

            tmp=SboxTrans(x[(j+)%]^x[(j+)%]^x[(j+)%]^rk[i+j]);//为什么这样就好了？

            x[j]^=L1(tmp);

            xx[i+j]=x[j];

        }

    }

    y=(unsigned int *)Output;

    for(int i=;i<;i++) y[i]=x[-i];

}

unsigned int key[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int miwen[]={0x595298c7,0xc6fd271f,0x0402f804,0xc33d3f66};

unsigned int mingwen[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int rk[];

unsigned int output[]={};

void solve1(){

    printf("题目1:\n");

    SM4KeyExt(key,rk,);

    printf("rk数组:\n");

    for(int i=;i<;i++) printf("%08x\n",rk[i]);

    printf("\n");

    printf("x数组：\n");

    SM4Crypt(mingwen,output,rk);

    for(int i=;i<;i++) printf("%08x\n",xx[i]);

    printf("结果：\n");

    for(int i=;i<;i++) printf("%08x ",output[i]);

    printf("\n");

}

void solve2(){

    printf("题目2：\n");

    SM4KeyExt(key,rk,);

    for(int i=;i<;i++){

        SM4Crypt(mingwen,mingwen,rk);

    }

    for(int i=;i<;i++) cout<<hex<<mingwen[i]<<" ";

    cout<<"\n";

}

int main(){

    //freopen("out.txt","w",stdout);

    solve1();

    solve2();

    return ;

}

坑爹的错误

#include<bits/stdc++.h>

using  namespace std;

typedef long long ll;

#define Rotl(_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(x) ((x)^Rotl(x,2)^Rotl(x,10)^Rotl(x,18)^Rotl(x,24))

#define L2(x) ((x)^Rotl(x,13)^Rotl(x,23))

const unsigned int CK[] = {

    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[]={0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC};

const unsigned char Sbox[] = {

    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[];

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,k[];

    for(int i=;i<;i++) k[i]=key[i]^RK[i];

    for(int i=;i<;i+=){

        for(int j=;j<;j++){

            k[j]^=L2((unsigned int)SboxTrans(k[(j+)%]^k[(j+)%]^k[(j+)%]^CK[i+j]));

            rk[i+j]=k[j];

        }

    }

    if(CryptFlag==) for(r=;r<;r++) swap(rk[r],rk[-r]);  

}

void SM4Crypt(unsigned int *Input, unsigned int *Output, unsigned int *rk){

    unsigned int tmp, x[], *y;

    y=(unsigned int *)Input;

    for(int i=;i<;i++) x[i]=y[i];

    for(int i=;i<;i+=){

        for(int j=;j<;j++){

            x[j]^=L1((unsigned int)SboxTrans(x[(j+)%]^x[(j+)%]^x[(j+)%]^rk[i+j]));

            xx[i+j]=x[j];

        }

    }

    y=(unsigned int *)Output;

    for(int i=;i<;i++) y[i]=x[-i];

}

unsigned int key[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int miwen[]={0x595298c7,0xc6fd271f,0x0402f804,0xc33d3f66};

unsigned int mingwen[]={0x01234567,0x89abcdef,0xfedcba98,0x76543210};

unsigned int rk[];

unsigned int output[]={};

void solve1(){

    printf("题目1:\n");

    SM4KeyExt(key,rk,);

    printf("rk数组:\n");

    for(int i=;i<;i++) printf("%08x\n",rk[i]);

    printf("\n");

    printf("x数组：\n");

    SM4Crypt(mingwen,output,rk);

    for(int i=;i<;i++) printf("%08x\n",xx[i]);

    printf("结果：\n");

    for(int i=;i<;i++) printf("%08x ",output[i]);

    printf("\n");

}

void solve2(){

    printf("题目2：\n");

    SM4KeyExt(key,rk,);

    for(int i=;i<;i++){

        SM4Crypt(mingwen,mingwen,rk);

    }

    for(int i=;i<;i++) cout<<hex<<mingwen[i]<<" ";

    cout<<"\n";

}

int main(){

    //freopen("out.txt","w",stdout);

    solve1();

    solve2();

    return ;

}

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