Are all of you talking about getting the result in closed form, so that no loop is involved?
Other than mine, I haven't seen an implementation of a working algorithm. Let's see your code! My algorithm avoids calculating the factorials, which overflow 32-bit integers for N > 12, and is O(min(N,n)). When you beat that, let me know. Dave On Dec 11, 10:15 am, "Shiv Shankar" <mca.shivshan...@gmail.com> wrote: > Hi, > I agree with ankit sablok. And if we get the factorial of n in 1!, 2!, 3! > Etc. Then we can find the number easily. In its complexity will be O(N) > > > > -----Original Message----- > From: algogeeks@googlegroups.com [mailto:algoge...@googlegroups.com] On > > Behalf Of Dave > Sent: Friday, December 10, 2010 8:10 PM > To: Algorithm Geeks > Subject: [algogeeks] Re: program for evaluation of remainders > > @Ankit: Why not just use the algorithm I proposed > inhttp://groups.google.com/group/algogeeks/msg/2941ab071a39517c: > > x = 0; > for( i = (n < N ? n : N) ; i > 0 ; --i ) > x = (i * x + i) % n; > > Dave > > On Dec 10, 4:23 am, ankit sablok <ankit4...@gmail.com> wrote: > > @Dave > > we will use residues then i think the property of modulus > > > 1!mod997 + 2!mod997 + 3!mod997 ...... + 997!mod997 > > > i just proposed the solution using congruences for the case > > n<N > > > can u generalize the problem using congruences if so then please post > > it > > thnanx in advance > > > On Dec 9, 2:13 am, Dave <dave_and_da...@juno.com> wrote: > > > > @Ankit: So how does that work with, e.g., N = n = 997? I.e., what is > > > the calculation? > > > > Dave > > > > On Dec 8, 11:33 am, ankit sablok <ankit4...@gmail.com> wrote: > > > > > @ all the authors thanx for the suggestions actually wt i know about > > > > the problem is i think we can solve the problem mathematically if we > > > > know about congruences > > > > > for instance > > > > if N=100 > > > > 1! + 2! + ......... + 100! > > > > and n=12 > > > > > we find that > > > > 4!mod24=0 > > > > > hence the above equation reduces to the > > > > (1!+2!+3!)mod 12 =9 > > > > hence the answer is 9 > > > > > so can anyone write a program for this logic > > > > > On Dec 8, 6:19 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- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- 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 > athttp://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.
