1.6.6 解释器 Interpreter

思路&题解

PC/UVA 110106/10033
本题。。模拟即可。

//author: CHC
//First Edit Time:	2014-01-12 22:00
//Last Edit Time:	2014-01-13 11:30
//Filename:1.cpp
#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int n,ord[2000],sum,re[10];
char s[2000];
int main()
{
    scanf("%d",&n);
    getchar();
    getchar();
    //fgets(s,sizeof(s),stdin);
    while(n--)
    {
        int ns=0;
        memset(ord,0,sizeof(ord));
        memset(re,0,sizeof(re));
        //while(fgets(s,sizeof(s),stdin)!=NULL&&s[0]!='\n')
        while(gets(s)!=NULL)
        {
            if(strcmp(s,"")==0)break;
            sscanf(s,"%d",&ord[ns]);
            ++ns;
        }
        sum=0;
        for(int i=0;;i++)
        {
            ++sum;
            /*
            for(int j=0;j<10;j++)
                printf("%d:%d\t",j,re[j]);
            puts("");
            printf("%d\n",ord[i]);
            puts("");
            */
            int t=ord[i]/100,tg=ord[i]%10,ts=(ord[i]%100)/10;
            ord[i]%=1000;
            if(t==0&&re[tg]) i=re[ts]-1;
            else if(t==1) break;
            else if(t==2) re[ts]=tg;
            else if(t==3) re[ts]=(re[ts]+tg)%1000;
            else if(t==4) re[ts]=(re[ts]*tg)%1000;
            else if(t==5) re[ts]=re[tg];
            else if(t==6) re[ts]=(re[ts]+re[tg])%1000;
            else if(t==7) re[ts]=(re[ts]*re[tg])%1000;
            else if(t==8) re[ts]=ord[re[tg]]%1000;
            else if(t==9) ord[re[tg]]=re[ts]%1000;
            /*
            for(int j=0;j<10;j++)
                printf("%d:%d\t",j,re[j]);
            puts("");
            system("pause");
            */
        }
        printf("%d\n",sum);
        if(n)puts("");
    }
    return 0;
}

Interpreter

A certain computer has ten registers and 1,000 words of RAM. Each register or RAM location holds a three-digit integer between 0 and 999. Instructions are encoded as three-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 registers
6ds
means add the value of register s to registerd
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 1,000.

Input

The input begins with a single positive integer on a line by itself indicating the number of cases, each described as below. This is followed by a blank line, and there will be a blank line between each two consecutive inputs.
Each input case consists of up to 1,000 three-digit unsigned integers, representing the contents of consecutive RAM locations starting at 0. Unspecified RAM locations are initialized to 000.

Output

The output of each test case is a single integer: the number of instructions executed up to and including the halt instruction. You may assume that the program does halt. Separate the output of two consecutive cases by a blank line.

Sample Input

1  
299  
492  
495  
399  
492  
495  
399  
283  
279  
689  
078  
100  
000  
000  
000  

Sample Output

16