@vaibhav :acc. to u Its giving 9 for "BGBBGGGBBBGBGB" bt it should be 8
....how?
my code:
#include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
        while(t--)
        {
        int i=0,s=0,count=0,c=0,flag=1;
        char a[100005],ch;
        scanf("%s",a);
        while(a[i]!='\0')
        {       if(flag) while(a[i]=='B') {i++; flag=0;}  //except heading
'B'
                while(a[i]=='G')
                        {count++; i++;}
                        s+=(count-1);
                        count=0;
                while(a[i]=='B')
                        {c++; i++;}
                        s+=c;
                        c=0;

        }
        printf("%d\n",s);
        }
return 0;
}

On Tue, Jun 28, 2011 at 12:50 AM, vaibhav agarwal <
vibhu.bitspil...@gmail.com> wrote:

> @kartik also consider the case of  GBGGGG the answer is 1 BGGGGG. ie
> trailing G's left.
> BBBBBGB leading B's left hence only one G followed by B therefore only one
> iteration.
>
> try out some cases u will find hw it wrks.
>
>
> On Tue, Jun 28, 2011 at 12:42 AM, vaibhav agarwal <
> vibhu.bitspil...@gmail.com> wrote:
>
>> @kartik
>>
>> yup frgt to mention the last case 1 followed by zero's in that case number
>> of iterations is the no. of trailing zeroes.
>> GBGBBB
>> will have four iterations
>> 101000 = 1(one zero b/w two ones) + 3(last 1 followed by 3 zero's)
>>
>> BGBGBB,BBGBGB,BBBGBG,BBBBGG
>>
>> well logic is how hw mny jump of 1's u need to do to get it to the end of
>> the sequence.
>> GBG requires 1 iteration ie BGG
>>
>>
>> On Mon, Jun 27, 2011 at 11:27 PM, kartik sachan 
>> <kartik.sac...@gmail.com>wrote:
>>
>>> HEY DUDE I AM NOT GETTING UR LOGIC AT ALL I THINK HOW U WILL SATISFY THIS
>>> CASE GBGBBB
>>>
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-- 
Dileep Kumar
Computer Science & Engineering
NIT, Allahabad

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