POJ1049 Microprocessor Simulation

题目来源：http://poj.org/problem?id=1049

题目大意：

　　一种小型的微处理器有以下特性：

　　1. 每个字长4bit.

　　2. 地址用2个字进行编码。先高位字后低位字，即高位字大的地址占据内存中靠前的字。

　　3. 内存大小为256个字。

　　4. 有两个微处理器A和B，每个存储一个字。

　　5. 有9个指令编码。每条指令需要至少一个字来存储编码，其中有4条指令含参数，并需要额外的2个字。

　　每4个bit组成的字可以取值0-15(10进制)，下文中我们将用16进制来表示这些数。

　　9条指令说明如下：

Code	Words	Description
0	3	LD: Load accumulator A with the contents of memory at the specified argument.
1	3	ST: Write the contents of accumulator A to the memory location specified by the argumen
2	1	SWP: Swap the contents of accumulators A and B.
3	1	ADD: Add the contents of accumulators A and B. The low word of the sum is stored in A, and the high word in B
4	1	INC: Increment accumulator A. Overflow is allowed; that is, incrementing F yields 0.
5	1	DEC: Decrement accumulator A. Underflow is allowed; that is, decrementing 0 yields F.
6	3	BZ: If accumulator A is zero, the next command to be executed is at the location specified by the argument. If A is not zero, the argument is ignored and nothing happens.
7	3	BR: The next command to be executed is at the location specified by the argument.
8	1	STP: Stop execution of the program.

程序总是最先执行地址00处的指令，然后依次执行后面的指令直到遇到Stop指令。

下面的例子展示了一些片段程序并描述了它的作用。

Program	Description
01A8	Load accumulator A with the contents of memory location 1A (26 in decimal) and stop.
01A512F8	Load accumulator A with the contents of memory location 1A (26 in decimal), decrement it, store the result to memory location 2F, then stop.

输入：输入由若干行组成，每行256个16进制数字（1-9, A-F）.每行表示的是内存的内容，地址编码从00到FF。00地址的内存单元为Stop指令时标志着输入的结束。输入的程序保证了程序的指令不会位于地址F0到FF的内存单元中。

输出：对每个输入的内存块，模拟微处理器的执行，输出程序执行后的每个内存字，格式与输入一致。

Sample Input

0102011311321128FF0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Sample Output

0102011311321128FF1E00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

标题已经告诉我们方法：模拟。关键是细心，比如数字的INC和DEC要注意从数字到字母的跳变，还有分清楚值和址。勾起了当年做组成原理课程设计的"美好回忆"。=。=

//////////////////////////////////////////////////////////////////////////
//        POJ1049 Microprocessor Simulation
//        Memory: 164K        Time: 0MS
//        Language: C++        Result: Accepted
//////////////////////////////////////////////////////////////////////////

#include <cstdio>

using namespace std;

char mem[257];
char A, B;
int pc;

int Hex2Dec(char c) {
if (c >= '0' && c <= '9') return c - '0';
else return c - 'A' + 10;
}

int DHex2Dec(char a, char b) {
return Hex2Dec(a) * 16 + Hex2Dec(b);
}

char Dec2Hex(int i) {
if (i >= 10) return 'A' + i - 10;
else return i + '0';
}

int main(void) {
while (scanf("%s", mem) && mem[0] != '8') {
pc = 0;
char command;
char buf;
int buf1, buf2;
A = B = '0';
while ((command = mem[pc]) != '8') {
switch (command) {
case '0'://LD
A = mem[DHex2Dec(mem[pc + 1], mem[pc + 2])];
pc += 3;
break;
case '1'://ST
mem[DHex2Dec(mem[pc + 1], mem[pc + 2])] = A;
pc += 3;
break;
case '2'://SWP
buf = A;
A = B;
B = buf;
++pc;
break;
case '3'://ADD
buf1 = Hex2Dec(A);
buf2 = Hex2Dec(B);
B = Dec2Hex((buf1 + buf2) / 16);
A = Dec2Hex((buf1 + buf2) % 16);
++pc;
break;
case '4'://INC
if (A == 'F') A = '0';
else if (A == '9') A = 'A';
else ++A;
++pc;
break;
case '5'://DEC
if (A == '0') A = 'F';
else if (A == 'A') A = '9';
else --A;
++pc;
break;
case '6'://BZ
if (A == '0') pc = DHex2Dec(mem[pc + 1], mem[pc + 2]);
else pc += 3;
break;
case '7'://BR
pc = DHex2Dec(mem[pc + 1], mem[pc + 2]);
break;
}
}
printf("%s\n", mem);
}
return 0;
}

View Code