UVA10033 Interpreter【模拟】

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. Thefirst 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

299

492

495

399

492

495

399

283

279

689

078

100

000

Sample Output

问题链接：UVA10033 Interpreter

问题简述：（略）

问题分析：

这是一个模拟题，模拟指令的执行过程，内容包括指令、寄存器和存储器。

程序说明：

需要注意输入格式的处理。

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA10033 Interpreter */#include <bits/stdc++.h>using namespace std;const int N1 = 1000;
const int N2 = 10;
int ram[N1], reg[N2];int main()
{int t;cin >> t;cin.get();cin.get();while(t--) {memset(reg, 0, sizeof reg);memset(ram, 0, sizeof ram);int add = 0, ans;string inst;while(getline(cin, inst)) {if(inst.empty())break;for(auto c : inst)ram[add] = ram[add] * 10 + c - '0';add++;}ans = add = 0;while(true) {ans++;int cmd = ram[add++];if(cmd == 100)break;int op = cmd / 100, operand1 = cmd % 100 / 10, operand2 = cmd % 10;switch(op) {case 0:if(reg[operand2]) add = reg[operand1];break;case 2:reg[operand1] = operand2;break;case 3:(reg[operand1] += operand2) %= 1000;break;case 4:(reg[operand1] *= operand2) %= 1000;break;case 5:reg[operand1] = reg[operand2];break;case 6:(reg[operand1] += reg[operand2]) %= 1000;break;case 7:(reg[operand1] *= reg[operand2]) %= 1000;break;case 8:reg[operand1] = ram[reg[operand2]];break;case 9:ram[reg[operand2]] = reg[operand1];break;}}cout << ans << '\n';if(t)cout << '\n';}return 0;
}

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