UVA 10033 - Interpreter

A certain computer has 10 registers and 1000 words of RAM. Each register or RAM location holds a

3-digit integer between 0 and 999. Instructions are encoded as 3-digit integers and stored in RAM. The

encodings are as follows:

100 means halt

2dn means set register d to n (between 0 and 9)

3dn means add n to register d

4dn means multiply register d by n

5ds means set register d to the value of register s

6ds means add the value of register s to register d

7ds means multiply register d by the value of register s

8da means set register d to the value in RAM whose address is in register a

9sa means set the value in RAM whose address is in register a to the value of register s

0ds means goto the location in register d unless register s contains 0

All registers initially contain 000. The initial content of the RAM is read from standard input. The

first instruction to be executed is at RAM address 0. All results are reduced modulo 1000.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input to your program consists of up to 1000 3-digit unsigned integers, representing the contents of consecutive RAM locations starting at 0. Unspecified RAM locations are initialized to 000.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

The output from your program is a single integer: the number of instructions executed up to and including the halt instruction. You may assume that the program does halt.

Sample Input

1

299

492

495

399

492

495

399

283

279

689

078

100

000

000

000

Sample Output

16

题意：给出一系列的指令，求总共执行的指令数；

Instructions	Meaning
100	停止执行
2dn	设置寄存器d的值为n(0-9)
3dn	寄存器d的值加上n
4dn	寄存器d的值乘上n
5ds	设置寄存器d的值为寄存器s的值
6ds	寄存器d的值加上寄存器s的值
7ds	寄存器d的值乘上寄存器s的值
8da	设置寄存器d的值为RAM中地址为寄存器a的值
9sa	设置RAM中地址为寄存器a的值为寄存器s的值
0ds	跳转到RAM中地址为寄存器d的值处，除非此时寄存器s的值为0

思路：按照所给的指令进行模拟即可，要注意的是在输入第一个测试数的时候后面要跟一个空行，并且每两个测试用例之间要有一个空行；再有一个需要注意的是，虽然题目告诉我们RAM中的指令数只有1000条，但是不代表输入的指令数不超过1000条，所以数组RAM开的大一点，否则会出现Runtime Error；最后一点，在计算总共执行的指令数时，要加上停止指令的执行；

#include <cstdio>
#include <cstring>
#include <memory>

int RAM[10000];
int reg[10+5];

#define op1(num) (num/100)
#define op2(num) (num/10%10)
#define op3(num) (num%10)

int execute(){
int cur = 0;   //the position of RAM
int cnt = 0+1; //the number of instructions executed
int op;        //current instruction

while(1){
op = RAM[cur++];
switch(op1(op)){
case 0:
if(reg[op3(op)])
cur = reg[op2(op)];
break;
case 1:
if(!op2(op) && !op3(op))
return cnt;
break;
case 2:
reg[op2(op)] = op3(op);
break;
case 3:
reg[op2(op)] = (reg[op2(op)] + op3(op))%1000;
break;
case 4:
reg[op2(op)] = (reg[op2(op)] * op3(op))%1000;
break;
case 5:
reg[op2(op)] = reg[op3(op)];
break;
case 6:
reg[op2(op)] = (reg[op2(op)] + reg[op3(op)])%1000;
break;
case 7:
reg[op2(op)] = (reg[op2(op)] * reg[op3(op)])%1000;
break;
case 8:
reg[op2(op)] = RAM[reg[op3(op)]];
break;
case 9:
RAM[reg[op3(op)]] = reg[op2(op)];
break;
default:
break;
}
cnt++;
}
}

int main(){
int N;
scanf("%d\n\n", &N);

while(N--){
char buff[10];
int cnt = 0;
memset(RAM, 0, sizeof(RAM));
memset(reg, 0, sizeof(reg));
while(gets(buff)){
if(!strcmp(buff, ""))
break;
sscanf(buff, "%d", RAM+cnt);
cnt++;
}
printf("%d\n", execute());
if(N)
printf("\n");
}

return 0;
}