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"。=。=
1 ////////////////////////////////////////////////////////////////////////// 2 // POJ1049 Microprocessor Simulation 3 // Memory: 164K Time: 0MS 4 // Language: C++ Result: Accepted 5 ////////////////////////////////////////////////////////////////////////// 6 7 #include <cstdio> 8 9 using namespace std; 10 11 char mem[257]; 12 char A, B; 13 int pc; 14 15 int Hex2Dec(char c) { 16 if (c >= '0' && c <= '9') return c - '0'; 17 else return c - 'A' + 10; 18 } 19 20 int DHex2Dec(char a, char b) { 21 return Hex2Dec(a) * 16 + Hex2Dec(b); 22 } 23 24 char Dec2Hex(int i) { 25 if (i >= 10) return 'A' + i - 10; 26 else return i + '0'; 27 } 28 29 int main(void) { 30 while (scanf("%s", mem) && mem[0] != '8') { 31 pc = 0; 32 char command; 33 char buf; 34 int buf1, buf2; 35 A = B = '0'; 36 while ((command = mem[pc]) != '8') { 37 switch (command) { 38 case '0'://LD 39 A = mem[DHex2Dec(mem[pc + 1], mem[pc + 2])]; 40 pc += 3; 41 break; 42 case '1'://ST 43 mem[DHex2Dec(mem[pc + 1], mem[pc + 2])] = A; 44 pc += 3; 45 break; 46 case '2'://SWP 47 buf = A; 48 A = B; 49 B = buf; 50 ++pc; 51 break; 52 case '3'://ADD 53 buf1 = Hex2Dec(A); 54 buf2 = Hex2Dec(B); 55 B = Dec2Hex((buf1 + buf2) / 16); 56 A = Dec2Hex((buf1 + buf2) % 16); 57 ++pc; 58 break; 59 case '4'://INC 60 if (A == 'F') A = '0'; 61 else if (A == '9') A = 'A'; 62 else ++A; 63 ++pc; 64 break; 65 case '5'://DEC 66 if (A == '0') A = 'F'; 67 else if (A == 'A') A = '9'; 68 else --A; 69 ++pc; 70 break; 71 case '6'://BZ 72 if (A == '0') pc = DHex2Dec(mem[pc + 1], mem[pc + 2]); 73 else pc += 3; 74 break; 75 case '7'://BR 76 pc = DHex2Dec(mem[pc + 1], mem[pc + 2]); 77 break; 78 } 79 } 80 printf("%s\n", mem); 81 } 82 return 0; 83 }
