Guys this is wrong. Lucky number is basically a prime number if you observe
the definition carefully, and this approach of determining primality
obviously won't work.
On Sun, Jan 4, 2009 at 7:12 PM, Channa Bankapur <channabanka...@gmail.com>wrote:
> Here is the C-version of the same. It tells you whether given n is lucky or
> not.
> void luckyOrNot(long int n){
>
> long int n2 = n;
> int i= 2;
> while(n2>=i){
>
> if(n2%i == 0){
>
> printf("%ld got unlucky when i was %d\n", n, i);
> break;
>
> }
> n2 = n2 - (n2/i);
> i++;
>
> }
> if(n2<i){
> printf("%ld is lucky\n", n);
> }
>
> }
>
> Cheers,
> Channa
>
> On Sun, Jan 4, 2009 at 10:34 AM, Sathya Narayanan <
> sathya.phoe...@gmail.com> wrote:
> > first n will be the nth number in the sequence.. after each iteration
> > (intially i=2) the new position of n is calculated as n=n- (n/i) and if
> n%i
> > then u stop.. this goes on until n<i . if n becomes less than i , then n
> is
> > lucky.. this is the approach used in tht perl prog i guess
> >
> >
> > 2009/1/4 daizi sheng <daizish...@gmail.com>
> >>
> >> check this perl program, hope it explains the algorithm itself
> >>
> >> >>>>>>START OF algo.pl<<<<<<<<<<<<
> >> use warnings;
> >> use strict;
> >>
> >> sub is_lucky_number
> >> {
> >> my ($cur, $n, $last_n, $next_n, $old_n);
> >>
> >> $cur = 1; # to remove numbers at $cur, $cur*2, ...
> >> $n = shift @_; # current index of the input number before removing
> >> $old_n = $n;
> >>
> >> die unless $n > 0;
> >> $last_n = -1; # index of the input number when after removing
> >> every $cur-1 number
> >>
> >> while($last_n != $n)
> >> {
> >> $cur++;
> >> $next_n = $n - int($n / $cur);
> >> if($n % $cur == 0)
> >> {
> >> #print "$old_n is removed at $cur-th step\n";
> >> return 0; # input number is not lucky because it will be
> >> removed at $cur-th step
> >> }
> >> $last_n = $n;
> >> $n = $next_n;
> >> }
> >>
> >> return 1;
> >> }
> >>
> >> foreach my $input (1..100)
> >> {
> >> print "$input is lucky\n" if &is_lucky_number($input);
> >> }
> >>
> >> print "Press return to exit\n";
> >> <>;
> >>
> >>
> >>
> >> On Sun, Jan 4, 2009 at 2:08 AM, kannan <kven...@gmail.com> wrote:
> >> >
> >> > All the numbers from 1 to infinity are written in sequence:
> >> >
> >> > 1,2,3,4,5,6,7,8,9,10,11,12,13,14.....
> >> > in the first iteration, every second number is removed, leaving:
> >> > 1,3,5,7,9,11,13....
> >> > in the second iteration, every third number in the above sequence is
> >> > removed, thus leaving: 1,3,7,9,13....
> >> > in the third iteration, every fourth number is removed, leaving:
> >> > 1,3,7,13....
> >> > like this the process goes on indefinitely.
> >> > the numbers which are, thus, left are called lucky numbers.
> >> >
> >> > Given a number n(can be as big as 18 digits) you must tell if n is a
> >> > lucky number or not.. how do i proceed?? all that i can think of is a
> >> > naive implementation which obviously wont work because n can be as big
> >> > as 18 digits...
> >> > could you help me or give suggestions on how to proceed....
> >> > (This question was asked in a practice contest which is over now)
> >> >
> >> > >
> >> >
> >>
> >>
> >
> >
> >
>
> >
>
--
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