10033

tgoulart · Post by **tgoulart** » Wed Jan 17, 2007 6:10 am

Actually, my program stops when 125 is reached. It is the halt instruction, so the program finishes there, and the subsequent lines should be ignored. This way we have 5 instructions executed, including the 125.

Bandrosh · Post by **Bandrosh** » Wed Jan 17, 2007 11:46 pm

u have accepted? in ur question?
hehe thnks man, i

Bandrosh · Post by **Bandrosh** » Thu Jan 18, 2007 1:01 am

Bandrosh wrote:u have accepted? in ur question?
hehe thnks man, now i have win one Sig Error heheh Ipm staying more crazy

tgoulart · Post by **tgoulart** » Thu Jan 18, 2007 5:09 am

Yes, I got accepted and I hardly doubt that cases like this one would be in the judge's input, unless it's the last case, because of the lines that you have to ignore.

Try to recode your program, you might have done some mistakes, and don't forget that all results are mod 1000!

Bandrosh · Post by **Bandrosh** » Thu Jan 18, 2007 11:48 pm

tgoulart · Post by **tgoulart** » Sat Jan 20, 2007 3:46 pm

I hardly doubt that cases like this one would be in the judge's input

Sorry, I didn't noticed that I store all the data before calculating, so It's not true...

Q.Qnick · Post by **Q.Qnick** » Tue Mar 20, 2007 3:23 pm

here is my code
how can i solve the problem of time limit

#include<stdio.h>

int s,i,j;
int RAM[20][3],REG[10],ans;
char save[6];
main(){
scanf("%d\n",&s);
j=0;
for(i=0;i<s;i++){
j=0;
while(gets(save)){
if(!save[0])break;
if(save[0]=='0'&& save[1]=='0' &&save[2]=='0')
break;
RAM[j][0]=save[0]-'0';
RAM[j][1]=save[1]-'0';
RAM[j][2]=save[2]-'0';
j++;

}
for(;j<20;j++){
RAM[j][0]=0;
RAM[j][1]=0;
RAM[j][2]=0;
}
for(ans=0,j=0;;){
if(RAM[j][0]==1)
{
ans++;
break;
}
else if(RAM[j][0]==2)
{
REG[ RAM[j][1] ] = RAM[j][2];
}
else if(RAM[j][0]==3)
{
REG[ RAM[j][1] ] += RAM[j][2];
REG[ RAM[j][1] ]%=1000;
}
else if(RAM[j][0]==4)
{
REG[ RAM[j][1] ] *= RAM[j][2];
REG[ RAM[j][1] ]%=1000;
}
else if(RAM[j][0]==5)
{
REG[ RAM[j][1] ] = REG[ RAM[j][2] ];
}
else if(RAM[0]==6)
{
REG[ RAM[j][1] ] += REG[ RAM[j][2] ];
REG[ RAM[j][1] ]%=1000;
}
else if(RAM[j][0]==7)
{
REG[ RAM[j][1] ] *= REG[ RAM[j][2] ];
REG[ RAM[j][1] ]%=1000;
}
else if(RAM[j][0]==8)
{
REG[ RAM[j][1] ] = RAM[j][ REG[ RAM[j][2] ] ];
}
else if(RAM[j][0]==9)
{
RAM[j][ REG[ RAM[j][2] ] ] = REG[ RAM[j][1] ];
}
else if(RAM[j][0]==0 && REG[ RAM[j][2] ]!=0)
{
j = REG[ RAM[j][1] ];
}
ans++;
j++;
}
printf("%d\n",ans);
if(i!=s-1){
printf("\n");
}
}
}

Q.Qnick · Post by **Q.Qnick** » Tue Mar 20, 2007 4:35 pm

some body help me　Ｔ.Ｔ

Q.Qnick · Post by **Q.Qnick** » Wed Mar 21, 2007 2:42 pm

i do some change , but it isn't work .
can some one let me know where is the plb.
thx a lot .

#include<stdio.h>

int s,i,j,k;
int RAM[1000][3],REG[10]={0},ans;
char save[6];
main(int argc,char *argv[]){
/*FILE *fp;
fp=fopen(argv[1],"r");*/
fscanf(stdin,"%d\n",&s);
/*printf("%d\n",s);*/
j=0;
for(i=0;i<s;i++){
j=0;
while(fgets(save,6,stdin)!=NULL){
if(save[0]=='\n')
{
break;
}
else
{
RAM[j][0]=save[0]-'0';
RAM[j][1]=save[1]-'0';
RAM[j][2]=save[2]-'0';
j++;
}
}
for(k=j;k<1000;k++){
RAM[k][0]=0;
RAM[k][1]=0;
RAM[k][2]=0;
}
/*for(k=0;k<j;k++){
printf("%d %d %d %d\n",i,RAM[k][0],RAM[k][1],RAM[k][2]);
}
printf("\n==\n");*/

for(k=0;k<10;k++){
REG[k]=0;
}

ans=0;
for(j=0;;){
if(RAM[j][0]==1)
{
ans++;
break;
}
else if(RAM[j][0]==2)
{
REG[ RAM[j][1] ] = RAM[j][2];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==3)
{
REG[ RAM[j][1] ] += RAM[j][2];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==4)
{
REG[ RAM[j][1] ] *= RAM[j][2];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==5)
{
REG[ RAM[j][1] ] = REG[ RAM[j][2] ];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==6)
{
REG[ RAM[j][1] ] += REG[ RAM[j][2] ];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==7)
{
REG[ RAM[j][1] ] *= REG[ RAM[j][2] ];
REG[ RAM[j][1] ]%=1000;
j++;
}
else if(RAM[j][0]==8)
{
REG[ RAM[j][1] ] = RAM[j][ REG[ RAM[j][2] ] ];
j++;
}
else if(RAM[j][0]==9)
{
RAM[j][ REG[ RAM[j][2] ] ] = REG[ RAM[j][1] ];
j++;
}
else if(RAM[j][0]==0)
{
if(REG[ RAM[j][2] ]!=0)
{
j = REG[ RAM[j][1] ];
}
else
{
j++;
}
}
if(j==1000){
j=0;
}
ans++;
}
printf("%d\n",ans);
if(i!=s-1){
printf("\n");
}
}
}

helloneo · Post by **helloneo** » Wed Mar 21, 2007 3:23 pm

Don't create a new thread if there is one already..
Use old threads..

http://online-judge.uva.es/board/viewtopic.php?t=13985
http://online-judge.uva.es/board/viewtopic.php?t=9667
http://online-judge.uva.es/board/viewtopic.php?t=940
http://online-judge.uva.es/board/viewtopic.php?t=8923

and so on..

pbfox · Post by **pbfox** » Wed Mar 21, 2007 4:03 pm

Help me! Input Code?

Input:

Code: Select all

pbfox · Post by **pbfox** » Wed Mar 21, 2007 4:14 pm

my code is wrong, but anyway...
help me,please!

Bandrosh · Post by **Bandrosh** » Wed May 09, 2007 7:02 pm

sim gourlat to precisando de ajuda, agora q terminou o semestre v

hardcode · Post by **hardcode** » Thu Jul 05, 2007 3:00 am

tgoulart wrote:Actually, my program stops when 125 is reached. It is the halt instruction, so the program finishes there, and the subsequent lines should be ignored. This way we have 5 instructions executed, including the 125.

Why does your program stop when 125 is reached tgoulart? Isn't 100 mean to be the halt instruction? I'm confused. Thanks for your time!

tgoulart · Post by **tgoulart** » Thu Jul 05, 2007 4:28 am

hardcode wrote:
tgoulart wrote:Actually, my program stops when 125 is reached. It is the halt instruction, so the program finishes there, and the subsequent lines should be ignored. This way we have 5 instructions executed, including the 125.
Why does your program stop when 125 is reached tgoulart? Isn't 100 mean to be the halt instruction? I'm confused. Thanks for your time!

My program stops because I verify the instruction by the first digit, and 1 is the halt operation. But if you take a look at the table, you will see that 125 isn't a valid instruction.

OJ Board

10033 - Interpreter

Help me! Input code?

Re: 10033 - Interpreter - I need a multi-testcase input