@Pankaj: You can do with a one-dimensional array since you can update the
array one row at a time. It would look like this:
int P = 1000000007;
c[0] = 1;
for( j = 1 ; j <= n ; ++j )
{
// at this point, c[i]. for i = 0, 1, 2, ..., j-1, holds (j-1) choose i
t = c[0];
for( i = 1 ; i < j ; ++i )
{
u = c[i] + t;
if( u > P )
u -= P;
t = c[i];
c[i] = u;
}
c[j] = t;
// at this point, c[i], for i = 0, 1, 2, ..., j, holds j choose i
}
Dave
On Saturday, March 3, 2012 6:44:11 AM UTC-6, Pankaj wrote:
> @jai-thanks!!! , it worked...,is there any method for big numbers like
> 100000,in my case 1000 so storing in array[10^3][10^3] worked but what if
> >10^5,storing will not work then??,anyway thanks a lot again!!!!
>
On Saturday, March 3, 2012 6:44:11 AM UTC-6, Pankaj wrote:
>
> @jai-thanks!!! , it worked...,is there any method for big numbers like
> 100000,in my case 1000 so storing in array[10^3][10^3] worked but what if
> >10^5,storing will not work then??,anyway thanks a lot again!!!!
>
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