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 in http://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 at http://groups.google.com/group/algogeeks?hl=en. -- 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.
