1295. Simple Computers

单点时限: 2.0 sec

内存限制: 256 MB

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.

输入格式

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.

输出格式

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

样例

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
Output
10000111

3 人解决,5 人已尝试。

3 份提交通过,共有 15 份提交。

8.5 EMB 奖励。

创建: 14 年,1 月前.

修改: 4 年前.

最后提交: 6 月,2 周前.

来源: Ulm Local 2000

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