nice prakash. Algorithm is definitely better than brute force. But
here is brute force anyway, just go from 2 to square root of n ;P
I initially misread it and thought you were asking for ANY {c,d} set
in which c,d are factors of n (but not necessarily c*d = n) . That
wouldve been all of the factors (12) , choose 2, unless I am
mistaken. 12 nCr 2 = 66.
#!/usr/bin/ruby -w
#how many unique sets {a,b} can be formed if a,b are factors of N and
axb=N
# N = 24*33
n=792
#factors = [2,2,2,3,3,11]
#results
res=['1x792']
2.upto(Math.sqrt(n)) {|c|
next unless (c&1).zero? || c%3==0 || c%11==0
res<<c.to_s+'x'+(n/c).to_s if n%c==0
}
p res, res.size
# output $ ./set_factors.rb
#["1x792", "2x396", "3x264", "4x198", "6x132", "8x99", "9x88",
"11x72", "12x66", "18x44", "22x36", "24x33"]
#12
On Aug 23, 2:39 pm, Nikhil Gupta <nikgp...@iitr.ernet.in> wrote:
> sorry I wrote them in different order:
> if (a,b) and (b,a) are considered same then answer is 12
> and if they are considered different it is 24.
>
> --
> Nikhil Gupta
> Indian Institute of Technology Roorkee,
> Roorkee, Uttarakhand,
> India , 247667
> Phone: +91 9634990161
> email: nikgp...@iitr.ernet.in
> nikhilg.8...@gmail.com
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