[algogeeks] Re: program for evaluation of remainders
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" 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 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 > > can u generalize the problem using congruences if so then please post > > it > > thnanx in advance > > > On Dec 9, 2:13 am, Dave 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 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 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<=100 > > > > > 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.
RE: [algogeeks] Re: program for evaluation of remainders
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 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 > can u generalize the problem using congruences if so then please post > it > thnanx in advance > > On Dec 9, 2:13 am, Dave 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 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 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<=100 > > > > 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.
[algogeeks] Re: find the number.
@Naresh: The sequence of numbers generated by this rule for any given
starting number is called a Collatz Sequence. Try googling it.
Here is a list of the number of iterations required for n between 1
and 10,000: http://oeis.org/A006577/b006577.txt. Maybe that will help.
Dave
On Dec 11, 7:20 am, Naresh A wrote:
> Given range of numbers between A and B (A<= B)
> Find the number within given range which has more number of iterations as
> per the following
>
> n { stop ; return iteration number } if n=1;
> n = 3n+1 if n is odd
> n = n/2 if n is even
>
> for eg :
>
> n=3 odd
>
> n=10;
> n=5;
> n=16;
> n=8;
> n=4;
> n=2;
> n=1;
>
> iterations : 7
>
> --
> *
> Time complexity= (
> *
> *NARESH ,A*
> **
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[algogeeks] find the number.
Given range of numbers between A and B (A<= B)
Find the number within given range which has more number of iterations as
per the following
n { stop ; return iteration number } if n=1;
n = 3n+1 if n is odd
n = n/2 if n is even
for eg :
n=3 odd
n=10;
n=5;
n=16;
n=8;
n=4;
n=2;
n=1;
iterations : 7
--
*
Time complexity= (http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Longest interval with maximum sum
@amir can u explain a bit more... On Tue, Dec 7, 2010 at 10:09 PM, Amir hossein Shahriari < amir.hossein.shahri...@gmail.com> wrote: > @jai : since sum of all values in C is between -n and n the last step can > be done in O(n) time and O(n) space > > > On Sun, Dec 5, 2010 at 12:44 PM, jai gupta wrote: > >> @fenghuang: the last step will take O(n logn ) . Or there is some better >> way? >> >> -- >> 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. > -- Regards Aditya Kumar B-tech 3rd year Computer Science & Engg. MNNIT, Allahabad. -- 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.
Re: [algogeeks]
Is it any puzzle On 11 Dec 2010 16:07, "parth panchal" wrote: how are you -- 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.
[algogeeks]
how are you -- 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.
