AES简单实现

#include <iostream>#include <stdio.h>#include <iomanip>using namespace std;typedef unsigned char byte;//加法byte add(byte a,byte b){    return a^b;}//乘法byte multiply(byte a,byte b){    //16bit    unsigned short int res, i, temp;    res=0,temp=a;    //乘    for(i=1; i<=0x80; i<<=1,temp<<=1)    {        if(i&b)        {            res^=temp;        }    }    //取余    for(i=0x8000, temp=0x8d80; i>0x80; i>>=1, temp>>=1)        if(i&res)            res^=temp;    return (byte)res;}//倍乘byte xtime(byte a){    if(a&0x80)//最高位为1    {        //先左移然后和AES要求多项式对应的二进制 0xb1 取模        return (a<<1)^0x1b;    }    else    {        return a<<1;    }}int main(){    int a,b;    int num;    while(1)    {        cout<<"please choose  1.add  2.multiply  3.xtime 4.esc"<<endl;        cin>>num;        switch(num)        {        case 1:            cout<<"please enter two Hexadecimal numbers."<<endl;            cin>>hex>>a;            cin>>hex>>b;            printf("%X\n",add(a,b));            break;        case 2:            cout<<"please enter two Hexadecimal numbers."<<endl;            cin>>hex>>a;            cin>>hex>>b;            printf("%X\n",multiply(a,b));            break;        case 3:            cout<<"please enter a Hexadecimal number."<<endl;            cin>>hex>>a;            printf("%X\n",xtime(a));            break;            //cout<<setiosflags(ios::uppercase)<<hex<<(xtime(a))<<endl;        case 4:            goto end;        default:            cout<<"enter the wrong number!"<<endl;        }    }    end:    return 0;}