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Views: 932 | Replies: 5
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I think not less than 2N.
Right?
Right?
B.Sc Computer Science, Helwan University
Microsoft Student Partner
Personal blog http://ramymahrous.blogspot.com/
Arabic technical blog http://fci-h-ar.blogspot.com/
English technical blog http://fci-h.blogspot.com/
Microsoft Student Partner
Personal blog http://ramymahrous.blogspot.com/
Arabic technical blog http://fci-h-ar.blogspot.com/
English technical blog http://fci-h.blogspot.com/
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Join Date: Aug 2007
Location: Adelaide, South Australia
Posts: 428
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Solved Threads: 53
a -> 1 step
For i = 1 to n -> takes n steps -> n+1 steps so far -> the 1 is negligible as n gets big so approx n steps so far
For j = 1 to i -> takes i steps for each n -> i*n so far -> approx n*n so far
For k = 1 to j -> takes j steps for each n*n -> j*n*n so far -> approx n*n*n so far
a = a+1 -> 1 step for each n*n*n -> 1*n*n*n
= n^3
For i = 1 to n -> takes n steps -> n+1 steps so far -> the 1 is negligible as n gets big so approx n steps so far
For j = 1 to i -> takes i steps for each n -> i*n so far -> approx n*n so far
For k = 1 to j -> takes j steps for each n*n -> j*n*n so far -> approx n*n*n so far
a = a+1 -> 1 step for each n*n*n -> 1*n*n*n
= n^3
yea darkgen, you're right I think your solution is eligable
B.Sc Computer Science, Helwan University
Microsoft Student Partner
Personal blog http://ramymahrous.blogspot.com/
Arabic technical blog http://fci-h-ar.blogspot.com/
English technical blog http://fci-h.blogspot.com/
Microsoft Student Partner
Personal blog http://ramymahrous.blogspot.com/
Arabic technical blog http://fci-h-ar.blogspot.com/
English technical blog http://fci-h.blogspot.com/
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