both will work fine
in first case in each step what is done is ans = (ans*(n+1-i))/i;
what happens is
first iteration ans = n, i =1 so i divides ans
sec. iteration ans = product of 2 consecutive nos at least one even so
divisible by 2
in third, ans is prod of 3 consecutive no.s so divisible by 3 and so on
On Tue, Jun 21, 2011 at 9:15 PM, Wladimir Tavares <wladimir...@gmail.com>
wrote:
>
> I got a doubt in the following code:
>
> if (k> n / 2)
> k = n-k;
> for (i = 1; i <= k, i + +)
> ans = ans * (n +1- i) / i;
>
> If (n +1- i) is not divisible by i this code should be wrong?
>
> I thought I needed to keep the numerator and denominator and go by
dividing the numerator and denominator gcd
>
> num = 1
> den = 1
>
> for (i = 1; i <= k, i + +) {
> num *= n +1- i;
> den *= i;
> d = gcd (num, den);
> num /= d;
> den / = d;
> }
>
> printf ("% d", num);
>
> Wladimir Araujo Tavares
> Federal University of Ceará
>
>
>
>
>
>
> On Tue, Jun 21, 2011 at 12:34 PM, PRAMENDRA RATHi rathi <
prathi...@gmail.com> wrote:
>>
>> i have also got AC...i am only asking for best algo.....
>>
>>
>>
>>
>>
>>
>> PRAMENDRA RATHI
>> NIT ALLAHABAD
>>
>>
>>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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