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Alun's code
Posted: Sat May 08, 2004 2:58 pm
by Algoritmo
Alun, you checked for mirroring and palindroming, both things at the same time. I think it's easier if you check independently:
[cpp]int Len = strlen(S);
bool IsPal = true, IsMir = true;
for (int i = 0; i < Len; i++) {
char C = S
;
char Op = S[Len-1-i];
if (C != Op) IsPal = false;
if (C != Rev(Op)) IsMir = false;
}
[/cpp]
One thing that bored me about this problem is this sentence:
"A mirrored string is a string for which when each of the elements of the string is changed to its reverse (if it has a reverse) and the string is read backwards the result is the same as the original string."
This sentence is ambuguous. What if it does not have a reverse? Should I not replace that character, should I replace it with blank space or all characters must have an reverse for that input to be mirrored?
The sample input does not answer my question, but I discovered the answer by sending mutiple codes until I got Accepted.
I replaced characters that didn't have a reverse with blank space.
401Why WA??
Posted: Sun Jun 06, 2004 4:41 am
by Jewel of DIU
Here is my code for prob 401, It causes WA.
[c]
#===========#
CUT AFTER AC
#===========#
[/c]
silly mistake
Posted: Sun Jun 06, 2004 7:59 am
by sohel
Reversing
V yields
V and not
U.
A part of your code
[c]
rev['T'] = 'T';
rev['U'] = 'U';
rev['V'] = 'U';
[/c]
Hope you can see the mistake...
Posted: Wed Jun 09, 2004 3:57 pm
by Jewel of DIU
Now I have got AC
.
Thank you Sohel Bhai.
401- Palindromes
Posted: Sun Jun 20, 2004 11:57 am
by RustB
This seems too simple to give this much trouble
Can anyone see what the problem is?
Code: Select all
#include <stdio.h>
char rev[40]="01SE0Z0080A000300HIL0JM0O0002TUVWXY5";
int ispalin(char *x, int n)
{
int i;
for(i=0;i<n;i++)
if(x[i] != x[n-i-1])
return 0;
return 1;
}
int ismirror(char *x, int n)
{
int i,flag=0;
for(i=n/2;i<n;i++)
if(isalpha(x[i]))
if(rev[x[i]-'A'+10]!=0)
{
x[i]=rev[x[i]-'A'+10];
flag=1;
}
else
if(rev[x[i]])
{
x[i]=rev[x[i]];
flag=1;
}
if(!flag) return 0;
for(i=0;i<n;i++)
if(x[i] != x[n-i-1])
return 0;
return 1;
}
int main(void)
{
char inp[25],xinp[25];
while(!feof(stdin))
{
int isp,ism,n;
scanf("%s",inp);
n=strlen(inp);
strcpy(xinp,inp);
isp=ispalin(xinp,n);
ism=ismirror(xinp,n);
if(!isp && !ism)
printf("%s -- is not a palindrome.\n\n",inp);
else if(isp && !ism)
printf("%s -- is a regular palindrome.\n\n",inp);
else if(!isp && ism)
printf("%s -- is a mirrored string.\n\n",inp);
else
printf("%s -- is a mirrored palindrome.\n\n",xinp);
}
return 0;
}
Posted: Sun Jun 20, 2004 12:31 pm
by sohel
Consider these inputs and outputs..
input
[c]
B
E
[/c]
your program outputs
[c]
0 -- is a mirrored palindrome.
3 -- is a mirrored palindrome.
[/c]
Hope you can find your mistake and rectify it.
Posted: Sun Jun 20, 2004 6:50 pm
by RustB
Okay... Thanks for that. I made some stupid assumptions about ASCII conversions :/ But I have a different set of problems now. I suppose I have to sort this out the hard way :/
Thanks anyway
Posted: Mon Jun 21, 2004 2:19 pm
by RustB
I still don't understand. This is my test data
Code: Select all
NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA
MAIAM
123ESI
123ES1
DEO3D
9339
A
B
E
M
1
3
2
4
is:
Code: Select all
NOTAPALINDROME -- is not a palindrome.
ISAPALINILAPASI -- is a regular palindrome.
2A3MEAS -- is a mirrored string.
ATOYOTA -- is a mirrored palindrome.
MAIAM -- is a mirrored palindrome.
123ESI -- is not a palindrome.
123ES1 -- is a mirrored string.
DEO3D -- is a mirrored string.
9339 -- is a regular palindrome.
A -- is a mirrored palindrome.
B -- is a regular palindrome.
E -- is a mirrored palindrome.
M -- is a mirrored palindrome.
1 -- is a mirrored palindrome.
3 -- is a mirrored palindrome.
2 -- is a mirrored palindrome.
4 -- is a regular palindrome.
Is everything correct? Why do I still get a WA? Please help me
Edit: The formatting is correct in the actual output
Posted: Mon Jun 21, 2004 2:27 pm
by sohel
By giving a quick glance I see that the output for
E
is not correct.
E is a palindrome - right.... but its not a mirrored one.
So I think the output should be
E -- is a regular palindrome.
Posted: Mon Jun 21, 2004 5:06 pm
by RustB
I changed that, but it's still not accepted
Posted: Tue Jun 22, 2004 8:17 pm
by jagadish
i lost my AC code for this one ..anyway consider these inputs
2S2
SS22
output:
222 -- is a mirror.... // original string is changed
SS22 -- is not a palindrome // must be mirrored string
Posted: Wed Jun 23, 2004 6:44 am
by RustB
This is my modified code
Code: Select all
#include <stdio.h>
char rev[40]="01SE0Z0080A000300HIL0JM0O0002TUVWXY5";
int ispalin(char *x, int n)
{
int i;
for(i=0;i<n;i++)
if(x[i] != x[n-i-1])
return 0;
return 1;
}
int ismirror(char *x, int n)
{
int i,flag=0;
if(n<2) return 0;
for(i=n/2;i<n;i++)
if(isalpha(x[i]))
{
if(rev[x[i]-'A'+10]!='0')
{
x[i]=rev[x[i]-'A'+10];
flag=1;
}
}
else
{
if(rev[x[i]-'0']!= '0')
{
x[i]=rev[x[i]-'0'];
flag=1;
}
}
if(!flag) return 0;
for(i=0;i<n;i++)
if(x[i] != x[n-i-1])
return 0;
return 1;
}
int main(void)
{
char inp[25],xinp[25];
int i;
while(1)
{
int isp,ism,n;
fgets(inp,sizeof(inp),stdin);
n=strlen(inp);
inp[n-1]='\0';
n--;
if(strcmp(inp,"")==0) break;
strcpy(xinp,inp);
isp=ispalin(xinp,n);
ism=ismirror(xinp,n);
if(!isp && !ism)
printf("%s -- is not a palindrome.\n\n",inp);
else if(isp && !ism)
printf("%s -- is a regular palindrome.\n\n",inp);
else if(!isp && ism)
printf("%s -- is a mirrored string.\n\n",inp);
else
printf("%s -- is a mirrored palindrome.\n\n",inp);
}
return 0;
}
The OJ says it gets a wrong answer.
It gives the correct answer for your inputs jagadish
Posted: Wed Jun 23, 2004 7:54 pm
by jagadish
since sohel told pointed out that program is not working for B ,E you
probably added this line [c] if(n<2) return 0; [/c] to fix it ,now it doesnt work for 8 ! ..its an easy program you need to test properly
btw i re-wrote the solution if u still get WA i can send that ..
Posted: Thu Jun 24, 2004 5:02 am
by RustB
For 8 I'm getting
8 -- is a regular palindrome
What's wrong with that?
Posted: Sun Jun 27, 2004 6:41 am
by sohel
8 is a mirrored palindrome.. I think.
