Instruction (hdu 5083)
Instruction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 343 Accepted Submission(s): 99
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.
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.
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.
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.
line.
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!
pid=5085" target="_blank" style="color:rgb(26,92,200); text-decoration:none">5085
5084 5081pid=5080" target="_blank" style="color:rgb(26,92,200); text-decoration:none">5080
5079。代码比較乱啊啊啊啊啊
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std; char ope[6][5]={{"ADD"},{"SUB"},{"DIV"},{"MUL"},{"MOVE"},{"SET"}};
char code[6][7]={{"000001"},{"000010"},{"000011"},{"000100"},{"000101"},{"000110"}};
char co[31][6]={ {"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"}};
char s1[5],s2[10]; int OP(char str[])
{
for (int i=0;i<6;i++)
{
if (strcmp(str,ope[i])==0)
return i;
}
} int main()
{
int n;
while (~scanf("%d",&n))
{
if (n==1)
{
scanf("%s%s",s1,s2);
int x=OP(s1);
int dd=x;
printf("%s",code[x]);
x=s2[1]-'0';
int ok=0;
if (isdigit(s2[2]))
{
ok=1;
x=x*10;
x=x+s2[2]-'0';
}
x--;
printf("%s",co[x]);
if (dd==5)
{
printf("00000\n");
continue;
}
if (ok)
{
x=s2[5]-'0';
if (isdigit(s2[6]))
{
x=x*10;
x=x+s2[6]-'0';
}
}
else
{
x=s2[4]-'0';
if (isdigit(s2[5]))
{
x=x*10;
x=x+s2[5]-'0';
}
}
x--;
printf("%s\n",co[x]);
}
else if (n==0)
{
scanf("%s",s1);
int len=strlen(s1);
int t=1;
int sum=0;
for (int i=5;i>=0;i--)
{
int x=s1[i]-'0';
sum+=t*x;
t=t*2;
}
if (sum>6||sum==0)
{
printf("Error!\n");
continue;
}
sum--;
int o=sum;
int flag=sum;
// printf("%s",ope[sum]);
t=1;
sum=0;
for (int i=10;i>=6;i--)
{
int x=s1[i]-'0';
sum+=t*x;
t=t*2;
}
if (sum>31||sum==0)
{
printf("Error!\n");
continue;
}
int aa=sum;
// printf(" R%d",sum);
if (flag==5)
{
t=1;
sum=0;
for (int i=15;i>=11;i--)
{
int x=s1[i]-'0';
// printf("x=%d\n",x);
sum+=t*x;
t=t*2;
}
if (sum==0)
{
printf("%s",ope[o]);
printf(" R%d",aa);
printf("\n");
continue;
}
else
{
printf("Error!\n");
continue;
}
}
t=1;
sum=0;
for (int i=15;i>=11;i--)
{
int x=s1[i]-'0';
// printf("x=%d\n",x);
sum+=t*x;
t=t*2;
}
if (sum>31||sum==0)
{
printf("Error!\n");
continue;
}
printf("%s",ope[o]);
printf(" R%d",aa);
printf(",R%d\n",sum);
}
}
return 0;
}
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