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[Can you send me the answer to me too??? Please!!!
Send on andrew_baggio_10@yahoo.com. I really need your help!

Thank You

Please give me the answer for problem number 325 and 335 in C Language! I really need it for my collegge assignment

Thank You

Dear Andrew Baggio,

Why do you want just the answer ?
And after you get the answer, you can just submit it to Online Judge and get Accepted ?
Isn't that makes you feel satisfied (get Accepted with other person's answer) ?
Have you tried to make it by yourself ?
If you face any problems on 325, why don't you ask something about this problem on board, like deddy one and I will help you as far as I can, but not the answer, of course.
If you want, I can help you to send my AC exe file to you, and you can re-check your output with mine. Ok ?

I think ACM is the place where we can study together to improve our skill in programming, but not just a place to submit problems as many as possible. Isn't that right ?

Dear deddy one,

You're right, e/E can only be used once like dot.
Have you got AC on this problem? If not, I'll give you another tips:
1. Using adjacent + or - means illegal (e.g: 6.5e--4 is illegal, +-1.34e9 is illegal but +3.6e13 is legal)
2. No space between real constant (e.g: 5 e 6 is illegal but 5e6 is legal)
3. Dot can't be placed before or after the real constant (e.g: .5e6 is illegal, 5e6. is illegal)

Good luck!

thx very much angga

I've found my tricky input
for future solver consider this input


123.e123123123123123123+123123123123123123123

is illegal


ps :
this problem must be faced with clear headed situation
don't try to code it under pressure like me, it can be a stress
full job

with almost the same logic thinking actually you can
try to solve 392 too (I got it AC on second submission)
in 325 I got more than 10 WA

well, actually not that same logic thinking , but in both
of my code for 325 and 392 need a lot of if's
(that's the similarity)

Your input is off course illegal, and it's easy to deduce from input statement .... ) this problem isn't involved with parsers of expressions, but it must check properly of constant ))

So if whole input line cannot be constant - it's not constant

Regards
Dominik

Have you tried to solve the problem by yourself ???

Have you tried to solve the problem by yourself ???
Tried it first, if you have any question, just put it on board.

Regards.

Dominik Michniewski wrote:Try this method:

only same states are allowed, also create biggest figure of good constant, it is: x[.(x)+][{E|e}(x)+]
where (x)+ means one or more digit
[x] means optional part of expression
{x|y} means one and exactly one of specified in brackets values

I've got AC by using Dominik's method

But " x[.(x)+][{E|e}(x)+] " need to change into these segments :
" [+ or - or empty] [(x)+.(x)+] [E or e] [+ or - or empty] [(x)+] "
I succeeded at the second try.

Good luck

Of course - I forget about signs in expression.... Sorry

Best regards
DM

I don`t know why I get WA.
Here some inputs for this problem. Maybe I have incorrect output?!
Help me PLEASE!!!
INPUT OUTPUT

Hello! I solved this problem. So maybe I can help you, but I need to see your source. Your output seems to be correct.

Hello Sanya! Thank you for reply.
Here my code for this problem:
[pascal]program acm325;
var n: integer;
s, l, i: string;
f: boolean;
begin
{ assign(input, 'input.txt');
reset(input);
assign(output, 'output.txt');
rewrite(output);}
l:=' is legal.';
i:=' is illegal.';
while not eof do begin
readln(s);
while s[1]=' ' do delete(s, 1, 1);
while s[length(s)]=' ' do delete(s, length(s), 1);
if s='*' then break;
f:=false;
write(s);
if (s[1]='+') or (s[1]='-') then delete(s, 1, 1);
while (s[1]>='0') and (s[1]<='9') and (s<>'') do begin delete(s, 1, 1); f:=true; end;
if (not f) or (s='') then begin writeln(i); continue; end;
f:=false;
if s[1]='.' then begin
delete(s, 1, 1);
if (s[1]<'0') or (s[1]>'9') then begin writeln(i); continue; end;
while (s[1]>='0') and (s[1]<='9') and (s<>'') do begin delete(s, 1, 1); f:=true; end;
end;

if s='' then begin writeln(l); continue; end;
if (s[1]='e') or (s[1]='E') then delete(s, 1, 1); f:=true;
if (not f) or (s='') then begin writeln(i); continue; end;

f:=true;
if (s[1]='-') or (s[1]='+') then delete(s, 1, 1);
for n:=1 to length(s) do if not ((s[n]>='0') and (s[n]<='9')) then f:=false;
if f then writeln(l) else writeln(i);
end;
end.
[/pascal]
So how you can see, I look into string from begin and to end according to orders.

You don't have to set "flag:=true" in "if s[1]='.' then" section.

Consider this input: 123.456+789
Your prog reaches '.', removes digits 456 and has it's flag happily set to true. It skips the following 3 "ifs", sees the + sign, removes 789 and thinks everything's OK, while it is not.

PS: Sorry, I missed the fact that there's no "begin-end" when you check if s[1] is in ['e', 'E'], so ignore the above stated. Still, your prog fails on 123.456+789.

Try to replace line
[pascal]
if (s[1]='e') or (s[1]='E') then delete(s, 1, 1); f:=true;
[/pascal]
with following:
[pascal]
if (s[1]='e') or (s[1]='E') then begin delete(s, 1, 1); f:=true; end
else begin writeln(i); continue; end;
[/pascal]