HDU 5083 Instruction --模拟

题意：给出汇编指令，解释出编码或者给出编码，解释出汇编指令。

解法：简单模拟，按照给出的规则一步一步来就好了，主要是注意“SET”的情况，还有要输出的东西最好放到最后一起输出，中间如果一旦不对就可以及时跳出去。

其他也没什么了，只要细心点，多测几组样例就好了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;
#define N 100007

string op[] = {"ADD","SUB","DIV","MUL","MOVE","SET"};
string bop[] = {"","","","","",""};
string bina[],itostr[];

void init()
{
    for(int i=;i<;i++)
    {
        string now = "";
        int tmp = i;
        for(int j=;j<;j++)
        {
            if(tmp%) now += "";
            else      now += "";
            tmp /= ;
        }
        reverse(now.begin(),now.end());
        bina[i] = now;         //i的二进制形式

        tmp = i;
        string no = "";
        while(tmp)
        {
            no += tmp%+'';
            tmp/=;
        }
        reverse(no.begin(),no.end());
        itostr[i] = no;        //i的十进制形式
    }
}

int main()
{
    string num,A;
    char ss[],R1,R2;
    int sign,i,j,a,b;
    init();
    while(scanf("%d",&sign)!=EOF)
    {
        if(sign == )
        {
            cin>>A;
            getchar();
            if(A != "SET")
                scanf("%c%d,%c%d",&R1,&a,&R2,&b);
            else
                scanf("%c%d",&R1,&a);
            for(i=;i<;i++)
            {
                if(A == op[i])
                    break;
            }
            if(i == ) { puts("Error!"); continue; }
            if(A != "SET")
            {
                if(R1 != 'R' || R2 != 'R' || a <=  || a >=  || b <=  || b >=)
                {
                    puts("Error!");
                    continue;
                }
                cout<<bop[i]<<bina[a]<<bina[b]<<endl;
            }
            else
            {
                if(R1 != 'R'|| a <=  || a >= )
                {
                    puts("Error!");
                    continue;
                }
                cout<<bop[i]<<bina[a]<<""<<endl;
            }
        }
        else
        {
            cin>>num;
            string A,B,C;
            string oA,oB,oC;
            A = num.substr(,);
            B = num.substr(,);
            C = num.substr(,);
            for(i=;i<;i++)
            {
                if(A == bop[i])
                    break;
            }
            if(i == ) { puts("Error!"); continue; }
            oA = op[i];
            if(op[i] != "SET")
            {
                for(i=;i<;i++)
                {
                    if(B == bina[i])
                        break;
                }
                if(i == ) { puts("Error!"); continue; }
                oB = "R"+itostr[i];
                for(i=;i<;i++)
                {
                    if(C == bina[i])
                        break;
                }
                if(i == ) { puts("Error!"); continue; }
                oC = "R"+itostr[i];
                cout<<oA<<" "<<oB<<","<<oC<<endl;
            }
            else
            {
                if(C != "") { puts("Error!"); continue;}
                for(i=;i<;i++)
                {
                    if(B == bina[i])
                        break;
                }
                if(i == ) { puts("Error!"); continue; }
                oB = "R"+itostr[i];
                cout<<oA<<" "<<oB<<endl;
            }
        }
    }
    return ;
}