Simple Computers
Time Limit: 1 Second
Memory Limit: 32768 KB
| 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
题意简要说明：一个简单的8位计算机，32个字节的内存单元，8位累加器，5位PC（即可寻址能力为32 bytes）。每一个存储器中的byte的高三位是指令码，低五位是存储器地址。其指令含义如上面题目中描述。现在给出存储器的32个字节的初始状态，pc（程序计数器）和accu（累加器）初始值为0，要求输出计算机对上面的代码段运行结束时的累加器值（按照二进制方式输出）。
题目很基本，因此基本解题步骤是：解析输入，即把类似“11110000”这样的字符串换算成相应的数字，然后模拟运行，然后在输出累加器值，即再把值换算成字符串。代码如下：
Code_1098
/* ZOL - 1098 Simple Computers */
#include <stdio.h>
#include <string.h>
/*计算机的32bytes内存*/
unsigned char memory[32];
/*pc: program counter, 程序计数器，即程序运行时的地址指针 */
unsigned char pc;
/*accu: accumulator, 即累加器*/
unsigned char accu;
/*每次读取的一行（byte）*/
char line[9];
/*解析读入的一行字节的值*/
unsigned char ParseFromLine(char* s)
{
unsigned char result = 0;
int base = 128;
while(*s)
{
result += (*s-'0')*base;
base = (base>>1);
s++;
}
return result;
}
/*把值解析成二进制表达的字符串，准备打印*/
void ParseToLine(unsigned char number, char* buf)
{
int i=7;
memset(buf, '0', 8);
buf[8]=0;/* NULL terminated */
while(number)
{
buf[i--] = (number % 2 ) + '0';
number = (number>>1);
}
}
/*运行，返回累加器的值*/
unsigned char Run()
{
/*分别是当前指令，指令码（高三位），地址（低五位）*/
unsigned char instruction, code, addr;
/*初始化*/
pc=0;
accu=0;
/*运行循环*/
while(1)
{
/*取指*/
instruction = memory[pc++];
/*一旦pc越过内存界限，就回退到0。没有这个语句会WA，这个语句导致AC！*/
if(pc>=32) pc=0;
code = (instruction >> 5);
addr = (instruction & 0x1f);
switch(code)
{
/* 000xxxxx STA x store the value of the accu into memory byte x */
case 0:
memory[addr] = accu;
break;
/* 001xxxxx LDA x load the value of memory byte x into the accu */
case 1:
accu = memory[addr];
break;
/* 010xxxxx BEQ x if the value of the accu is 0 load the value x into the pc */
case 2:
if(accu == 0) pc=addr;
break;
/* 011----- NOP no operation */
case 3:
break;
/* 100----- DEC subtract 1 from the accu */
case 4:
accu--;
break;
/* 101----- INC add 1 to the accu */
case 5:
accu++;
break;
/* 110xxxxx JMP x load the value x into the pc */
case 6:
pc=addr;
break;
/* 111----- HLT terminate program */
case 7:
return accu;
break;
}
}
return accu;
}
int main()
{
int i=0;
unsigned char result;
while(scanf("%s",line)!=EOF)
{
memory[0]=ParseFromLine(line);
for(i=1;i<32;i++)
{
scanf("%s",line);
memory[i]=ParseFromLine(line);
}
result = Run();
ParseToLine(result, line);
printf("%s\n", line);
}
return 0;
}
