作者：未知 来源：月光软件站 加入时间：2005-2-28　月光软件站

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 Specification
The 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 Specification
For each test case, output on a line the value of the accumulator on termination in binary representation, again highest bits first. 
Sample Input
00111110
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 Output
10000111

#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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