hdu 5083(模拟)

时间：2016-07-10 14:05:33 阅读：188 评论：0 收藏：0 [点我收藏+]

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Instruction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 991 Accepted Submission(s): 244

Problem Description

Nowadays, Jim Green has produced a kind of computer called JG. In his computer, the instruction is represented by binary code. However when we code in this computer, we use some mnemonic symbols. For example, ADD R1, R2 means to add the number in register R1 and R2, then store the result to R1. But this instruction cannot be execute directly by computer, before this instruction is executed, it must be changed to binary code which can be executed by computer. Each instruction corresponds to a 16-bit binary code. The higher 6 bits indicates the operation code, the middle 5 bits indicates the destination operator, and the lower 5 bits indicates the source operator. You can see Form 1 for more details.

15

instruction ADD Ra,Rb SUB Ra,Rb DIV Ra,Rb MUL Ra,Rb MOVE Ra,Rb SET Ra

Operation

Input

Multi test cases (about 50000), every case contains two lines.
First line contains a type sign, ‘0’ or ‘1’.
‘1’ means you should transfer an instruction into a 16-bit binary code;
‘0’ means you should transfer a 16-bit binary code into an instruction.
For the second line.
If the type sign is ‘1’, an instruction will appear in the standard form which will be given in technical specification;
Otherwise, a 16-bit binary code will appear instead.
Please process to the end of file.

[Technical Specification]
The standard form of instructions is
ADD Ra,Rb
SUB Ra,Rb
DIV Ra,Rb
MUL Ra,Rb
MOVE Ra,Rb
SET Ra
which are also listed in the Form 2.

Output

For type ‘0’，if the 16-bit binary code cannot be transferred into a instruction according to the description output “Error!” (without quote), otherwise transfer the 16-bit binary code into instruction and output the instruction in the standard form in a single line.
For type ‘1’, transfer the instruction into 16-bit binary code and output it in a single line.

Sample Input

1
ADD R1,R2
0
0000010000100010
0
1111111111111111

Sample Output

0000010000100010
ADD R1,R2
Error!

Source

BestCoder Round #15

直接模拟.

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
char opr0[][10] = {"0","ADD","SUB","DIV","MUL","MOVE","SET"};
char opr1[][10] = {"0","000001","000010","000011","000100","000101","000110"};
char opr2[][10] = {"0","00001","00010","00011","00100","00101","00110","00111","01000",
                   "01001","01010","01011","01100","01101","01110","01111","10000","10001","10010","10011",
                   "10100","10101","10110","10111","11000","11001","11010","11011","11100","11101","11110","11111"
                  };
int main()
{
    char s[20],s1[20],s2[100];
    int opr;
    while(scanf("%d",&opr)!=EOF)
    {
        if(opr==1)
        {
            scanf("%s %s",s,s1);
            if(strcmp(s,"ADD")==0) printf("%s",opr1[1]);
            if(strcmp(s,"SUB")==0) printf("%s",opr1[2]);
            if(strcmp(s,"DIV")==0) printf("%s",opr1[3]);
            if(strcmp(s,"MUL")==0) printf("%s",opr1[4]);
            if(strcmp(s,"MOVE")==0) printf("%s",opr1[5]);
            if(strcmp(s,"SET")==0) printf("%s",opr1[6]);
            int len = strlen(s1);
            int sum = 0;
            if(strcmp(s,"SET")==0)
            {
                for(int i=1; i<len; i++) sum = sum*10+s1[i]-‘0‘;
                printf("%s",opr2[sum]);
                printf("00000");
            }
            else
            {
                int i;
                for(i=1; i<len&&s1[i]!=‘,‘; i++) sum = sum*10+s1[i]-‘0‘;
                printf("%s",opr2[sum]);
                sum = 0;
                i+=2;
                for(; i<len; i++) sum = sum*10+s1[i]-‘0‘;
                printf("%s",opr2[sum]);
            }
            printf("\n");
        }
        else
        {
            scanf("%s",s2);
            char s3[100],s4[100],s5[100];
            int len = strlen(s2);
            if(strlen(s2)!=16)
            {
                printf("Error!\n");
                continue;
            }
            int cnt = 0,cnt1=0,cnt2=0;
            for(int i=0; i<6&&i<len; i++) s3[cnt++] = s2[i];
            s3[cnt]=‘\0‘;
            for(int i=6; i<11&&i<len; i++) s4[cnt1++] = s2[i];
            s4[cnt1]=‘\0‘;
            for(int i=11; i<=15&&i<len; i++) s5[cnt2++] = s2[i];
            s5[cnt2]=‘\0‘;
            bool flag = false;
            int t1=7;
            for(int i=1; i<=6; i++)
            {
                if(strcmp(s3,opr1[i])==0)
                {
                    t1 = i;
                    break;
                }
            }
            if(t1==7) flag = true;
            int t2 = 32;
            for(int i=1; i<32; i++)
            {
                if(strcmp(s4,opr2[i])==0)
                {
                    t2 = i;
                    break;
                }
            }
            if(t2==32) flag = true;
            if(t1==6)
            {
                if(strcmp(s5,"00000")!=0) flag = true;
                if(!flag) printf("%s R%d\n",opr0[t1],t2);
                else printf("Error!\n");
            }
            else
            {
                int t3 = 32;
                for(int i=1; i<32; i++)
                {
                    if(strcmp(s5,opr2[i])==0)
                    {
                        t3 = i;
                        break;
                    }
                }
                if(t3==32) flag = true;
                if(flag) printf("Error!\n");
                else printf("%s R%d,R%d\n",opr0[t1],t2,t3);
            }
        }
    }
    return 0;
}

hdu 5083(模拟)

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原文地址：http://www.cnblogs.com/liyinggang/p/5657520.html

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