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
nN
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=100
n=1000;
please help me write a program for this problem
thanx in advance- Hide quoted text -
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