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Instruction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 347 Accepted Submission(s): 101
15 operation code(6 bits)109destination operator code(5 bits)54source operator code(5 bits)0Form 1
In JG system there are 6 instructions which are listed in Form 2.
instructionADD Ra,RbSUB Ra,RbDIV Ra,RbMUL Ra,RbMOVE Ra,RbSET RafunctionAdd the number in register Ra and Rb, then store the result to Ra.Subtract the number in register Ra to Rb, then store the result to Ra.Divide the number in register Ra by Rb, then store the result to Ra.Mulplicate the number in register Ra and Rb, then store the result to Ra.Move the number in register Rb to Ra.Set 0 to Ra.Form 2
Operation code is generated according to Form 3.
OperationADDSUBDIVMULMOVESETOperation code000001000010000011000100000101000110Form 3
Destination operator code and source operator code is the register code of the register which is related to.
There are 31 registers in total. Their names are R1,R2,R3…,R30,R31. The register code of Ri is the last 5 bits of the number of i in the binary system. For eaxample the register code of R1 is 00001, the register code of R2 is 00010, the register code of R7 is 00111, the register code of R10 is 01010, the register code of R31 is 11111.
So we can transfer an instruction into a 16-bit binary code easyly. For example, if we want to transfer the instruction ADD R1,R2, we know the operation is ADD whose operation code is 000001, destination operator code is 00001 which is the register code of R1, and source operator code is 00010 which is the register code of R2. So we joint them to get the 16-bit binary code which is 0000010000100010.
However for the instruction SET Ra, there is no source register, so we fill the lower 5 bits with five 0s. For example, the 16-bit binary code of SET R10 is 0001100101000000
You are expected to write a program to transfer an instruction into a 16-bit binary code or vice-versa.
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.
1≤a,b≤31
There is exactly one space after operation, and exactly one comma between Ra and Rb other than the instruction SET Ra. No other character will appear in the instruction.
For type ‘1’, transfer the instruction into 16-bit binary code and output it in a single line.
1 ADD R1,R2 0 0000010000100010 0 1111111111111111
0000010000100010 ADD R1,R2 Error!
解题思路：
阅读理解模拟题。指令为“Set”时要单独且特别注意。
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <stdlib.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
using namespace std;
map<string,string>z;//操作符与二进制指令相相应
map<string,int>z2;//二进制数与十进制数相相应
void prepare()
{
z["000001"]="ADD";
z["000010"]="SUB";
z["000011"]="DIV";
z["000100"]="MUL";
z["000101"]="MOVE";
z["000110"]="SET";
z2["00001"]=1;z2["00010"]=2;
z2["00011"]=3;z2["00100"]=4;
z2["00101"]=5;z2["00110"]=6;
z2["00111"]=7;z2["01000"]=8;
z2["01001"]=9;z2["01010"]=10;
z2["01011"]=11;z2["01100"]=12;
z2["01101"]=13;z2["01110"]=14;
z2["01111"]=15;z2["10000"]=16;
z2["10001"]=17;z2["10010"]=18;
z2["10011"]=19;z2["10100"]=20;
z2["10101"]=21;z2["10110"]=22;
z2["10111"]=23;z2["11000"]=24;
z2["11001"]=25;z2["11010"]=26;
z2["11011"]=27;z2["11100"]=28;
z2["11101"]=29;z2["11110"]=30;
z2["11111"]=31;
}
string find(string op)//寻找操作符相应的二进制指令
{
map<string,string>::iterator i;
for(i=z.begin();i!=z.end();i++)
{
if(i->second==op)
return (i->first);
if(i->first==op)
return (i->second);
}
return "Error!";
}
string find2(int op)//寻找十进制相应的二进制
{
map<string,int>::iterator i;
for(i=z2.begin();i!=z2.end();i++)
{
if(i->second==op)
return (i->first);
}
return "Error!";
}
int find22(string op)//寻找二进制数相应的十进制
{
map<string,int>::iterator i;
for(i=z2.begin();i!=z2.end();i++)
{
if(i->first==op)
return (i->second);
}
return -1;
}
string op,source;//输入
string zhi;//输入
string wa="Error!";
int sigh;
int main()
{
prepare();
while(scanf("%d",&sigh)!=EOF)
{
if(sigh==1)
{
cin>>op>>source;
if(op=="SET")//单独推断
{
cout<<"000110";
int a=0;//提取出来十进制数
for(int i=1;i<source.length();i++)
a=a*10+source[i]-'0';
cout<<find2(a);
cout<<"00000"<<endl;
}
else
{
string ans=find(op);
//提取出来两个数字
int p;
int a1=0,a2=0;
bool deng=0;
for(p=0;p<source.length();p++)
{
if(source[p]==',')
deng=1;
if(p>0&&deng==0)
a1=a1*10+source[p]-'0';
if(deng==1)
break;
}
p=p+2;
for(;p<source.length();p++)
a2=a2*10+source[p]-'0';
cout<<ans;
cout<<find2(a1)<<find2(a2);
cout<<endl;
}
}
else
{
cin>>zhi;
string zhiling=zhi.substr(0,6);
string temp=find(zhiling);
if(temp==wa)
{
cout<<wa<<endl;
continue;
}
string a1=zhi.substr(6,5);
string a2=zhi.substr(11,5);
int f1=find22(a1);
int f2=find22(a2);
if(temp=="SET")
{
if(a2!="00000"||f1==-1)
{
cout<<wa<<endl;
continue;
}
cout<<temp<<" R"<<f1<<endl;
continue;
}
if(f1==-1||f2==-1)//推断两个数字是否合法
{
cout<<wa<<endl;
continue;
}
cout<<temp<<" ";
cout<<"R"<<f1<<",R"<<f2<<endl;
}
}
return 0;
}
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