Using this idea makes my solution into x = 0; for( i = (n < N ? n : N) ; i > 0 ; --i ) x = (i * x + i) % n;
Dave On Dec 8, 7:27 am, Ashim Kapoor <ashimkap...@gmail.com> wrote: > Let me try. Any thing involving n would leave no remainder. > > so (1 + 2 ! + ... + n ! + .... + N !) mod n = (1 + 2 ! + ... + (n-1)! ) mod > n > > This should be computed from a loop. I don't know how to reduce it further. > > Ashim. > > > > On Wed, Dec 8, 2010 at 6:49 PM, ankit sablok <ankit4...@gmail.com> wrote: > > Q) can anyboy find me the solution to this problem > > > Given an integer N and an another integer n we have to write a program > > to find the remainder of the following problems > > (1! + 2! + 3! + 4! + ..................... + N!)mod(n) > > > N<=1000000 > > n<=1000; > > > please help me write a program for this problem > > thanx in advance > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algoge...@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en.- 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 algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.