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391 - Mark-up .... help !
Posted: Fri Jul 12, 2002 1:29 pm
by Dominik Michniewski
Could anyone give me any special cases for this problem ?
I use finite state machine to solve this question, but stil got WA ....
Is any other trick than I have to print "34" where I found "\s1.234" ??
And what should I print where I found "\S1.123" ??
Posted: Tue Aug 13, 2002 9:58 pm
by L33T Magus
Did you remember that if the input text ends with a '\' it needs to be printed?
Posted: Tue Aug 13, 2002 11:02 pm
by Dominik Michniewski
Yes, I remember it .....
But if input looks like
A\
B\
should I print
A\
B\
or
A
B\
?????????????????????????
Posted: Sat Aug 24, 2002 7:39 am
by arc16
Code: Select all
Is any other trick than I have to print "34" where I found "\s1.234" ??
And what should I print where I found "\S1.123" ??
output should be nothing for \s1.234 and and "S1.123"
some tricky case i know is :
Code: Select all
input:
\s3.4a\sb\s.5c\s100.d\s12e
\**\\***\\*\*\*
output:
abcde
*\**\*
also don't forget that you don't have to reset the control/mark-up flag on each line (this is where i failed before i got AC)
Posted: Tue Aug 27, 2002 9:18 am
by Dominik Michniewski
I don't know what's wrong with this code ....
I got the same result as you, arc16, but still WA
could anyone help me ?
Dominik
------------------------------
Code: Select all
I cut off the code, because it was a solution after some modifications ;-)
Posted: Tue Aug 27, 2002 9:55 am
by arc16
Dominik Michniewski wrote:I don't know what's wrong with this code ....
I got the same result as you, arc16, but still WA
could anyone help me ?
haven't take a deep look at your program, but my guess is that
your program cannot process either a sequencing '\' and '*' characters correctly. Check out the following test case:
Code: Select all
input:
\**\\***\\*\*\*
\*
\\
\*
\\ \a\b\c\d\e\f\g\\h\\\*
your output:
*\**\*
\\
\ \a\c\d\e\f\g\h\
correct output:
*\**\*
\\
\ acdefg\h\
see? in the last line, the mark-up control in your program is curreny off, while it should be on. Try to do some more testing with something like \**\\\\\\\\\\\\\\\\\***********\*\*\\*\***\\****\\\*\*\*\*\a\c
good luck
Posted: Mon Sep 23, 2002 10:30 am
by Dominik Michniewski
Thanks all for helping me
I got Acceoted now
arc16: thank you very much for very usefull test cases and hints
)
Dominik
Posted: Fri Feb 28, 2003 11:14 pm
by yahoo
I cant understan how the output for \\ be \\. From the problem description it is clear that \\ will be replaced by \.
from problem:
the mark-up \\ can be used to print a single \.
will anybody please help. I am getting wrong answer all the time.
Posted: Sat Mar 01, 2003 4:00 am
by Red Scorpion
Consider This sample:
Code: Select all
Sample Input
LSKD AODP.
\\ akdf \\
\* asifj \\ \\\
\\\\
\\ askf asf\afa\
\* \\ \bisjaf
Output:
LSKD AODP.
\ akdf \
asifj \\ \\\
\\\\
\\ askf asf\afa\
\ isjaf
Did You see the difference?
when the input is "\\" You should print the "\" when the toggle processing of mark-ups
on;
otherwise you should print exactly as they appear in the input.
Note:
At the start, processing of mark-ups should be on.
RS.
Posted: Sun Mar 02, 2003 10:30 pm
by yahoo
Thanks for your reply. My program matches with all the input and output of yours. Will anybody kindly see my program and tell me where i am wrong. Here is my code:
Code: Select all
#include <stdio.h>
#include <string.h>
main()
{
char a[10000];
int flag1,flag2,flag3,flag4,i,j,l1;
while(1)
{
if(gets(a)==NULL) break;
l1=strlen(a);
flag1=flag2=flag3=0;
for(i=0;i<l1;i++)
{
if(a[i]=='\\')
{
if(a[i+1]=='b'&&flag4!=0) {i++;flag1=1;}
else if(a[i+1]=='s'&&flag4!=0)
{
flag2=1;
for(j=i+2;;j++)
if(!((a[j]>='0' && a[j]<='9')||a[j]=='.')) break; ;
i=j-1;
}
else if(a[i+1]=='*')
{
i++;
if(flag4==0) flag4=1;
else flag4=0;
}
else if(a[i+1]=='\\'&&flag4!=0)
{
i++;
printf("\\");
}
else if (flag4==0) printf("%c",a[i]);
}
else printf("%c",a[i]);
}
printf("\n");
}
return 0;
}
Thanks in advance.
Posted: Mon Mar 03, 2003 7:41 am
by anupam
if there be space charecter after the mark up off,
will i print it or not..
i found the problem harder then i guessed before..
-- if the mark up command is off then all the charecters get printed or not..
if yes then the code of yahoo has no mistakes i think...
--if no please reply why?
--anupam
Posted: Tue May 24, 2005 3:17 pm
by N|N0
I've no idea why this does not work
Code: Select all
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define ON 1
#define OFF 0
int main(void) {
char vrsta[100000];
char *p;
int process = ON, pika, prva = 1;
while (!feof(stdin)) {
gets(vrsta);
p = vrsta;
if (!prva) putchar('\n'); else prva = 0;
while (*p != '\0') {
if (*p == '\\') {
if (*(p+1) == 'b' && process) {
p+=2;
} else if (*(p+1) == 'i' && process) {
p+=2;
} else if (*(p+1) == '*') {
process = (process==ON)?OFF:ON;
p+=2;
} else if (*(p+1) == 's' && process) {
p+=2;
pika = 0;
while (isdigit(*p) || (*p == '.' && !pika)) {
if (*p == '.') pika++;
p++;
}
} else {
if (!process) {
putchar(*p);
p++;
} else {
if (*(p+1) != '\0') {
p++;
putchar(*p);
p++;
} else {
p++;
putchar('\\');
}
}
}
} else {
putchar(*p);
p++;
}
}
}
return(0);
}
Any suggestion?
391 - Mark-up
Posted: Wed Jul 27, 2005 1:10 pm
by Navid666
remove after accept!
Posted: Mon Aug 01, 2005 1:55 pm
by Navid666
is any body there???....please help me this is very easy problem and i dont know why i got WA??????????????????
:((:(
May this helps you !
Posted: Sun Aug 07, 2005 2:40 pm
by hyperion
Hi Navid,
Test this input to your program, it should generate the output
below which my Accepted Code generated.
Hope you figure out your problem with this test cases.
Input :
\s3.4a\sb\s.5c\s100.d\s12e
\*\b no translation should occur *\\ \b*\s3.*\*
/i/b/s\bb\ii\ss
\**\\***\\*\*\*
\*
\\
\*
\\ \n\o\ \p\r\ob\l\e\m\*s\*
\* close literal text by end-of-file\
Desired Output :
abcde
\b no translation should occur *\\ \b*\s3.*
/i/b/sbis
*\**\*
\\
\ no problems
close literal text by end-of-file\