11221 - Magic square palindromes.

[pipo]

hi all...

I got WA

In my thought, this problem is easy... but I got WA

I don't know where is wrong...

Could you give me some sample inputs and outputs ??

thanks in advance...

[mmonish]

I also think it's an easy problem.
Try this cases

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5
sdfg ?!@# dfdsfg.,./?><45344354)(*&^%^%%     gfsddfgfds
abcdbfgc,    ,cgfbdcba*(][
a
a,b.a
aaa!!!!aaa

My AC output

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Case #1:
No magic :(
Case #2:
4
Case #3:
1
Case #4:
No magic :(
Case #5:
No magic :(

Hope this helps.

[pipo]

hm... mmonish...

Is the 2 case right ??

The sentence of the second way in problem statement is not equal to original sentence.

[mmonish]

Thanks pipo u r right. Probably there is no test case like this in the judge data.
Now i edit my post.

[chinmoy kanti dhar]

this program gives run time error(signal 11).help me plz

code:
cut

[tanvir]

I think ur array(for array d) index is bouncing for the line:

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 if(n==6){k++;n=1;} 

check it...
tanvir

[chinmoy kanti dhar]

thanks tanvir for ur reply.but i dont understanad why it will cause pb.my program's pb is invalid memory reference.it is necessary to increase array size?

[tanvir]

look a bit more carefully?
what was the reason to write this line:

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if(n==6){k++;n=1;} 

why u r comparing n with 6??????
hope it helps
tanvir

[chinmoy kanti dhar]

thanks tanvir.now i have accepted the pb.thanks for the help.

[ferng1021]

I passed sample I/O and the above test data, but still got WA...
can any body help me?

thank you!

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#include<stdio.h>
#include<string.h>

int csN, csk, L, K;
char S[10002];

inline bool Equal(char a, char b, char c, char d)
{
	return a == b && a == c && a == d;
}

int main()
{
	int i, j, t;
	char ok;
	scanf("%d\n", &csN);
	for(csk = 1; csk <= csN; ++csk)
	{
		fgets(S, 10001, stdin);
		L = strlen(S);
		for(i = t = 0; i < L; ++i)
		{
			if(S[i] < 'a' || S[i] > 'z') ++t;
			else S[i-t] = S[i];
		}
		L -= t;
		S[L] = '\0';
		for(K = 1; K*K < L; ++K) ;
		if(K*K != L) ok = 0;
		else
		{
			for(i = 0; i < K; ++i)
			{
				for(j = 0; j < K; ++j)
					if(!Equal(S[i*K+j], S[(K-i-1)*K+(K-j-1)], S[j*K+i], S[(K-j-1)*K+(K-i-1)])) break;
				if(j < K) break;
			}
			ok = (i==K);
		}
		if(ok) printf("Case #%d:\n%d\n", csk, K);
		else printf("Case #%d:\nNo magic :(\n", csk);
	}
}

[rhsumon]

Hi, I have tried it wth a simple thinking it is:
1. I count the letters.
2. Chack this is a sqr number or not
3. than check the str palindrom or not

Oh a silly mistake AC now!!!!!!!!!!!!

[Fuad Hassan EWU]

why WA? is mmonish's last 2 output is right?how?

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AC

[Jan]

Use str[1001][1001]. And check the case below.

Input:

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1
abcdef bptuse cqzxrd drxzqc esutpb fedcba

Output:

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Case #1:
No magic :(

Hope it helps.

[Fuad Hassan EWU]

jan vai my code is giving its output 6. if i arrange the input it has the figure--

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abcdef
bptuse
cqzxrd
drxzqc
esutpb
fedcba

is not this magic sqare palindrome? if not then why? plz explain.

[Jan]

Read the description again...

A magic square palindrome is a sentence whose characters can be divided in a K x K square table with the property that the original sentence can be read from the table in four different ways:

1. Start from the (1,1) cell, move right until the end of the line and than proceed to the next line.
2. Start from the (1,1) cell, move down until the end of the column and then proceed to the next column.
3. Start from the (K,K) cell, move left until the beginning of the line and then proceed to the previous line.
4. Start from the (K,K) cell, move up until the beginning of the column and then proceed to the previous column.

If we read them from the square we get...

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1. abcdefbptusecqzxrddrxzqcesutpbfedcba
2. abcdefbpqrsectzxudduxztcesrqpbfedcba
3. abcdefbptusecqzxrddrxzqcesutpbfedcba
4. abcdefbpqrsectzxudduxztcesrqpbfedcba

And you can see that all of them are not equal. Hope it helps.