OJ Board

Posted: **Sun Mar 04, 2007 6:18 pm**

no idea why WA on such an easy problem

cut...

Posted: **Sun Mar 04, 2007 6:25 pm**

your program fails on

1 abcde

the output should be:

Code: Select all

edcba

have fun

Posted: **Sun Mar 04, 2007 6:29 pm**

Posted: **Tue Jun 26, 2007 6:05 pm**

HI, everybody i solve this problem but i got WA.
I don't know the problem of my code.

I want more sample input/output.
Is this sample input/output only for the special case:G=1;

Ples help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Posted: **Wed Jun 27, 2007 10:48 pm**

Try the cases.

Input:

Code: Select all

1 ABCDEFGHIJKLMNOPQRSTUVWX
2 ABCDEFGHIJKLMNOPQRSTUVWX
3 ABCDEFGHIJKLMNOPQRSTUVWX
4 ABCDEFGHIJKLMNOPQRSTUVWX
6 ABCDEFGHIJKLMNOPQRSTUVWX
8 ABCDEFGHIJKLMNOPQRSTUVWX
0

Output:

Code: Select all

XWVUTSRQPONMLKJIHGFEDCBA
LKJIHGFEDCBAXWVUTSRQPONM
HGFEDCBAPONMLKJIXWVUTSRQ
FEDCBALKJIHGRQPONMXWVUTS
DCBAHGFELKJIPONMTSRQXWVU
CBAFEDIHGLKJONMRQPUTSXWV

Hope these help.

Posted: **Thu Jun 28, 2007 8:43 pm**

thanks GURU, i got AC
THANKS A LOT

Posted: **Mon May 26, 2008 10:58 am**

It seemed very easy to me.
I just calculated the length and then divided the length by G to find the group length.
then in next step run a loop to length. and when it's divisible then reverse.

Posted: **Thu Jun 19, 2008 12:21 pm**

thank u shiplu vai
i hv got ac

Posted: **Fri Jun 20, 2008 11:46 am**

input

Code: Select all

1 S

your code output a null after S
write the output in a file for this input and see.

Good luck.

Posted: **Tue Jan 31, 2012 3:54 pm**

Hello, i solved the problem but i always got Wrong Answer, i can't identify what's wrong in the code.
Could you please give me a help to check what's wrong?

Thanks in advance.

Posted: **Tue Jan 31, 2012 4:02 pm**

Hello, i solved the problem and tested with some sample inputs, the outputs are always right, but when submit i got Wrong Answer, i can't identify what's wrong in the code.
Could you please give me a help to check what's wrong?

Code: Select all

int main()
{
    unsigned char nbrGroups=48, lines=0, offset = 0x00, out_offset=0,
    outstr[100], chars, count, groupSize, i;
    string str;

    //number of groups valid between 0 and 9
    while((nbrGroups >= 48) && (nbrGroups <= 57) && (++lines < 11))
    {
        cin >> nbrGroups;
        cin >> str;
        offset=0;
        out_offset=0;

        if(nbrGroups != 48)
        {
            //calculate the group size
            for(count=0;str[count] != 0x00; count++);

            //check if the total length is multiple of nbrGroups
            if(count%(nbrGroups-48) == 0)
            {
                groupSize = count/(nbrGroups-48);

                //go throught the groups
                for(count=0;count<(nbrGroups-48);count++)
                {
                    //invert the group
                    for(i=offset+groupSize;i != offset ;i--)
                        outstr[out_offset++] = str[i-1];

                    //set the position to the next group
                    offset+=(groupSize);
                }

                outstr[out_offset] = 0x00;
                cout << outstr << endl;
            }
        }
    }

    return 0;
}

Thanks in advance.

Posted: **Fri Dec 14, 2012 9:06 pm**

Please can any one tell why i am getting compiler Error????I m really fed up with this things...

#include<stdio.h>
#include<string.h>
int main()
{
int n,i,j,k,l,t;
char a[1000],b[1000];
while(scanf("%d",&n)!=EOF)
{
if(n==0)break;
else
{
scanf("%s",&a);
l=strlen(a)/n;
for(i=0,j=0;i<n;i++)
{
for(k=0,t=0;t<l;k++,t++)
{
b[k]=a[j];
j++;
}
b[k]='\0';
strrev(b);
printf("%s",b);
}
printf("\n");
}
}
return 0;
}

Posted: **Sat Dec 15, 2012 6:19 am**

You can see the reason for your Compiler Error by clicking My Submissions.

In this case it's because strrev() isn't included in the judge's compiler. Write your own strrev().

Posted: **Mon Dec 17, 2012 6:42 pm**

Thanks got ac....@brianfry713

Posted: **Tue Oct 29, 2013 12:15 pm**

Code: Select all

Remove after AC

OJ Board

11192 - Group Reverse

11192 - Group Reverse

11192,WA

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse

11192 - Group Reverse

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse

Re: 11192 - Group Reverse