1.6.6 解释器 Interpreter

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:

100means halt2dnmeans set register d to n (between 0 and 9)3dnmeans add n to register d4dnmeans multiply register d by n5dsmeans set register d to the value of registers6dsmeans add the value of register s to registerd7dsmeans multiply register d by the value of register s8dameans set register d to the value in RAM whose address is in register a9sameans set the value in RAM whose address is in register a to the value of register s0dsmeans 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

1299492495399492495399283279689078100000000000

Sample Output

16