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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 100007string op[8] = {"ADD","SUB","DIV","MUL","MOVE","SET"};string bop[8] = {"000001","000010","000011","000100","000101","000110"};string bina[36],itostr[35];void init(){    for(int i=1;i<32;i++)    {        string now = "";        int tmp = i;        for(int j=0;j<5;j++)        {            if(tmp%2) now += "1";            else      now += "0";            tmp /= 2;        }        reverse(now.begin(),now.end());        bina[i] = now;         //i的二进制形式        tmp = i;        string no = "";        while(tmp)        {            no += tmp%10+0;            tmp/=10;        }        reverse(no.begin(),no.end());        itostr[i] = no;        //i的十进制形式    }}int main(){    string num,A;    char ss[3],R1,R2;    int sign,i,j,a,b;    init();    while(scanf("%d",&sign)!=EOF)    {        if(sign == 1)        {            cin>>A;            getchar();            if(A != "SET")                scanf("%c%d,%c%d",&R1,&a,&R2,&b);            else                scanf("%c%d",&R1,&a);            for(i=0;i<7;i++)            {                if(A == op[i])                    break;            }            if(i == 7) { puts("Error!"); continue; }            if(A != "SET")            {                if(R1 != R || R2 != R || a <= 0 || a >= 32 || b <= 0 || b >=32)                {                    puts("Error!");                    continue;                }                cout<<bop[i]<<bina[a]<<bina[b]<<endl;            }            else            {                if(R1 != R|| a <= 0 || a >= 32)                {                    puts("Error!");                    continue;                }                cout<<bop[i]<<bina[a]<<"00000"<<endl;            }        }        else        {            cin>>num;            string A,B,C;            string oA,oB,oC;            A = num.substr(0,6);            B = num.substr(6,5);            C = num.substr(11,5);            for(i=0;i<7;i++)            {                if(A == bop[i])                    break;            }            if(i == 7) { puts("Error!"); continue; }            oA = op[i];            if(op[i] != "SET")            {                for(i=0;i<32;i++)                {                    if(B == bina[i])                        break;                }                if(i == 32) { puts("Error!"); continue; }                oB = "R"+itostr[i];                for(i=0;i<32;i++)                {                    if(C == bina[i])                        break;                }                if(i == 32) { puts("Error!"); continue; }                oC = "R"+itostr[i];                cout<<oA<<" "<<oB<<","<<oC<<endl;            }            else            {                if(C != "00000") { puts("Error!"); continue;}                for(i=0;i<32;i++)                {                    if(B == bina[i])                        break;                }                if(i == 32) { puts("Error!"); continue; }                oB = "R"+itostr[i];                cout<<oA<<" "<<oB<<endl;            }        }    }    return 0;}
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HDU 5083 Instruction --模拟