2008: Simple Computers

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.

 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）。

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.

For each test case, output on a line the value of the accumulator on termination in binary representation, again highest bits first.