OJ Board

Posted: **Fri Mar 28, 2003 5:01 pm**

Help, I try all sample I/O in that board, I pass all, but I always got WA too.
Can you give me more test to check my program, or you give me hint??

Posted: **Tue Apr 01, 2003 1:15 pm**

may there are negatif input for this problem?

like this:
input = -1+2
output= 1-2+

Posted: **Tue Apr 01, 2003 2:35 pm**

Hello Hisoka, ... There will be no negative number. Each line in one test-case will only have 1 char.

Did you notice that this problem has multiple-input format ?

Try these test-cases:

Input

Code: Select all

7

1
+
2
-
3

1
+
2
*
3

(
1
+
2
)
/
(
3
-
4
)

(
(
1
+
2
)
*
(
3
-
4
)
+
5
)
/
(
6
+
7
)

(
1
*
(
2
+
3
+
(
4
*
5
)
)
+
1
)

(
(
(
1
)
)
)

1

Output

Code: Select all

12+3-

123*+

12+34-/

12+34-*5+67+/

123+45*+*1+

1

1

Good luck,

-turuthok-

Posted: **Tue Apr 01, 2003 7:18 pm**

Hello turuthok, I pass for all your sample I/O with my program when I got WA. And my program is multiple input too.

Posted: **Wed Apr 02, 2003 8:20 am**

I got the same problem with this question. I use simple expression's parser to create postfix notation.
I pass tests from you, turothok, and also pass tests given in other threads by Ivan Golubev
Maybe I miss something ? Please help, because problem seems to be easy ...

Dominik Michniewski

Posted: **Wed Apr 02, 2003 12:34 pm**

I got AC now, thanx turuthok

to dominik:
I try your program from your previous post, I got this.

Code: Select all

input:
1
+
2
*
3

Code: Select all

output:
+23*

you always got Wrong answer when your first input. but you get true output for your next output. so I think your mistake only from your I/O, not your algo.

Posted: **Fri Apr 18, 2003 8:25 pm**

Well, I'm getting sick & tired of this problem. I struggled with it 9 months ago, and I'm stil struggling with it now.
Can anybody, once and for all, put into plain english what is wrong with the input to this problem and how we are supposed to deal with it. I'm shure I handle all the cases in all the threads properly, but WA is all I get.

Why don't the judges correct a problem that is so long known to exist?

HELP ME, please...

Posted: **Sat Apr 19, 2003 6:19 am**

Hello, everyone!

In fact I might be a lucky person as I got AC on the problem after my first submission.

Firstly I want to say that I didn't handle cases all you talking about :

Code: Select all

1(2+3)4 
or
(1*2*3*4*5)6+2-5*9+1

I guess there is no expressions where two operands have no operator between.

To get the postfix notation I use simple stack method and it works.

Anyway I'm not sure that the judge data is correct but my AC says me that there can be two cases:
- Judge is right.
- Judge and me are equally wrong.

I guess the first is more possible than the second.

Good luck!
Andrey.

Posted: **Sat Apr 19, 2003 6:44 am**

No,
I have submitted my test program again, and find that the judge input isn't corrected yet.

Actuallly it is possible to get accepted w/o specially handled the incorrect input, if you are lucky enough that your program can output the same thing as the sample output.

Posted: **Sat Apr 19, 2003 10:24 am**

I hate it, but I'll add my code to the long list of published sources on this matter. It's a plain vanilla implementation of a RDP without any syntax checking. It glues together all expressions in one case, like Ivan Golubjev suggested, so it can handle cases like

Code: Select all

(1+2)3 -> 12+3
12 -> 12
1(7*8)6 -> 178*6
etc.

Please give me some feedback.

[pascal]program p727;

var
infix,postfix:ansistring;
parsepos,writepos:integer;

procedure expression;forward;

procedure factor;
begin
case infix[parsepos] of
'0'..'9':begin
inc(writepos);
postfix[writepos]:=infix[parsepos];
inc(parsepos);
end;
'(':begin
inc(parsepos);
expression;
inc(parsepos); //eat ')'
end;
end;
end;

{
ALL BUT THE CRITICAL SECTION REMOVED,
SINCE IT'S A SPOILER

}

[/pascal]

Posted: **Sat Apr 19, 2003 11:18 am**

Here this a part from my code:[c]void getexpression(void);

void getfact(void)
{
if (*st == '(') {
st++;
getexpression();
assert(*st == ')');
st++;
if (isdigit(*st)) {
printf("%c", *st);
st++;
}
return;
}
while (isdigit(*st)) {
printf("%c", *st);
st++;
}
}[/c]
The fix itself here is:[c] if (isdigit(*st)) {
printf("%c", *st);
st++;
}[/c]
With this modification I've got accepted.

Posted: **Sat Apr 19, 2003 11:55 am**

Thanks man!!!

That did it. Finaly got AC.

I'm not shure, but I think reading more than one digit in getfact() could be essential too.

Hope to help you back someday,

-little joey

Posted: **Sat Apr 19, 2003 5:58 pm**

I know that there is an input here like:

1(2+4)3 or 1*2(1+2)1+2

I have checked it many times
I guess that it means that there is not only one input, but more than one...
I think the answers are:

124+3 and 12*12+12+

I ask ones who solved this problem and got AC.
Please explain it if you can

Posted: **Sun Apr 20, 2003 10:23 pm**

The answer is in the previous thread; especialy the answer Ivan Golubev gave to my question. I added two 'features' to my code:

1. Read all occurring digits and print them, even if you expect only one;
2. Read any digit that follows a closing bracket and print it.

This means that for the input
12+3
I print
123+
and for the input
(1+2)3
I print
12+3
and finally got AC.

Posted: **Thu Jan 29, 2004 10:59 am**

I know and I also hate to post my code here.
BUT I don't know how to get rid of Runtime ERROR(SISIGIV)

The code is deleted.

OJ Board

727 - Equation

727

727

727