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acm.jlu.edu.cn-1099-Simple Computers

王朝other·作者佚名  2006-01-09
窄屏简体版  字體: |||超大  

You are to write an interpreter for a simple computer. This computer uses a processor with a small number of machine instructions. Furthermore, it is equipped with 32 byte of memory, one 8-bit accumulator (accu) and a 5-bit program counter (pc). The memory contains data as well as code, which is the usual von Neumann architecture. The program counter holds the address of the instruction to be executed next. Each instruction has a length of 1 byte - the highest 3 bits define the type of instruction and the lowest 5 bits define an optional operand which is always a memory address (xxxxx). For instructions that don't need an operand the lowest 5 bits have no meaning (-----). Here is a list of the machine instructions and their semantics: 000xxxxx STA x store the value of the accu into memory byte x

001xxxxx LDA x load the value of memory byte x into the accu

010xxxxx BEQ x if the value of the accu is 0 load the value x into the pc

011----- NOP no operation

100----- DEC subtract 1 from the accu

101----- INC add 1 to the accu

110xxxxx JMP x load the value x into the pc

111----- HLT terminate program

In the beginning, program counter and accumulator are set to 0. After fetching an instruction but before its execution, the program counter is incremented. You can assume that programs will terminate. Input SpecificationThe input file contains several test cases. Each test case specifies the contents of the memory prior to execution of the program. Byte 0 through 31 are given on separate lines in binary representation. A byte is denoted by its highest-to-lowest bits. Input is terminated by EOF. Output SpecificationFor each test case, output on a line the value of the accumulator on termination in binary representation, again highest bits first.

Sample Input00111110

10100000

01010000

11100000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00111111

10000000

00000010

11000010

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

00000000

11111111

10001001

Sample Output10000111

#include<iostream>

using namespace std;

int mem[32];

int ins;

int ope;

int pc;

int ac;

int loadmem(char* s)

{

int d=0;

for(int i=0;i<8;i++)

d=d*2+s[i]-'0';

return d;

}

void loadins()

{

ins=(mem[pc]&224)>>5;

}

void loadope()

{

ope=(mem[pc]&31);

}

int main()

{

char s[9];

while(cin>>s)

{

mem[0]=loadmem(s);

for(int i=1;i<32;i++)

{

cin>>s;

mem[i]=loadmem(s);

}

pc=0;ac=0;

loadins();

loadope();

pc++;

while(!(ins==7))

{

switch(ins)

{

case 0:

mem[ope]=ac;

break;

case 1:

ac=mem[ope];

break;

case 2:

if(ac==0)

pc=ope;

break;

case 3:

break;

case 4:

ac--;

if(ac==-1)

ac=255;

break;

case 5:

ac++;

if(ac==256)

ac=0;

break;

case 6:

pc=ope;

break;

case 7:

goto L1;

default:

;

}

loadins();

loadope();

pc++;

if(pc==32)

pc=0;

}

L1:int r=128;

for(int i=0;i<8;i++)

{

if(ac>=r)

{

cout<<1;

ac-=r;

}

else

cout<<0;

r/=2;

}

cout<<endl;

}

return 0;

}

 
 
 
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