Oh. That is supposed to be t. So the line actually could read
t /= 10000;
Sorry for the typo.
Dave
On Feb 4, 11:58 am, Rahul Verma <[email protected]> wrote:
> @Dave
> In the line t=p/1000000;
> what is the value of p here or what it means?
>
> On Feb 4, 9:57 pm, Dave <[email protected]> wrote:
>
>
>
> > @Rahul: Given that 100! < 100^100 = 10^200, we know that 100! has less
> > than 200 digits. Assuming 32-bit integers, we might choose to store
> > anywhere from 1 to 7 digits in an integer, realizing that we have to
> > multiply the integer by a number up to 100 and still want the result
> > to be less than 2^31. I'll use 5 digits in an integer. Then the
> > computation can look something like this, assuming that you have
> > already read n, the number whose factorial you want:
>
> > int a[40], i, j, l, t;
> > a[0] = 1; // rightmost "digit" of number = 0!
> > l = 1; // the active number of elements of a[] used.
> > for( i = 2 ; i < n ; ++i )
> > {
> > t = 0; // the carry into the next product
> > for( j = 0 ; j < l ; ++j )
> > {
> > t += i * a[j]; // multiply digit by i and add carry
> > a[j] = t % 10000; // extract 5 digits
> > t = p / 100000; // calculate next carry
> > }
> > if( t > 0 ) // does the number need to add a digit?
> > a[l++] = t;
> > }
> > printf("%i",a[l-1]);
> > for( i = l-2 ; i >= 0 ; --i )
> > printf("%.5i",a[i]);
> > printf("\n");
>
> > Dave
>
> > On Feb 4, 10:19 am, Rahul Verma <[email protected]> wrote:
>
> > > Hi Group,
>
> > > Let me help to solve this problem of
> > > SPOJhttps://www.spoj.pl/problems/FCTRL2/
>
> > > The approach to solve the problem is very easy and I'm confused that
> > > how to store such a large no. like 100! in a variable using C++
> > > Language.
>
> > > Thanx & Regards,
> > > Rahul Verma- Hide quoted text -
>
> - Show quoted text -
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
