i have already mentioned that if final answer fits in 64 bit integer then it
will not overflow
On Tue, Jun 21, 2011 at 5:34 PM, Dumanshu <duman...@gmail.com> wrote:
> @Sunny: the value can still overflow because
> Using the while loop u r dividing the res value whenever possible. so
> the while checks if the current j divides res or not. What if current
> res value is not divisible by that particular j ... it will multiply
> by i for next res value and it will keep doing this until u get a res
> value divisible by that specific current j. So in the meantime, when
> the res value is increasing just because j doesn't divide it, it can
> overflow.
>
> On Jun 21, 4:12 pm, sunny agrawal <sunny816.i...@gmail.com> wrote:
> > i am doing the same thing without using array
> >
> > long long int res = 1;
> > int j = 2;
> > for(int i = n-k+1; i <= n; i++){
> > res *= (long long int)i;
> > while(j <= k && res%j == 0){
> > res/=j;
> > j++;
> > }
> > }
> >
> > On Tue, Jun 21, 2011 at 4:25 PM, kartik sachan <kartik.sac...@gmail.com>
> > wrote:
> >
> >
> >
> >
> >
> > > @ sunny i am not getting ur apporach
> > > but i am thinking like this......
> > > taking an array from and intilize it to n to n-r
> >
> > > a[1000];
> > > int k=0;
> > > for(int i=n;i>=n-r;i--)
> > > a[k++]=i;
> >
> > > for(int y=n-r;y>1;y--)
> > > for(j=0;j<k;j++)
> > > if(a[j]%y==0)
> > > {a[j]=a[j]/y;break;}
> >
> > > for example we take 7c3
> >
> > > so first array is intilize to {7,6,5,4}
> > > then we check divisibilty by {3,2}
> > > so 6 is divisible by 3 so put 6/3 back in array 2
> > > so finally 7*2*5*2;
> >
> > > --
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> >
> > --
> > Sunny Aggrawal
> > B-Tech IV year,CSI
> > Indian Institute Of Technology,Roorkee- Hide quoted text -
> >
> > - Show quoted text -
>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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