Use two loops:
num_triangles = 0
for i:1 to (N-1)/2
for j:i+1 to i+j<N
num_triangles += N-(i+j)
This gives values in increasing order such that c>b>a (c,b,a being lengths
of sides of the triangle) and sum of any two sides is greater than the
third.
Consider : 1,2,3,4,5,6
Possible side combinations are:
1,2,4
1,2,5
1,2,6
1,3,5
1,3,6
1,4,6
2,3,6
On Wednesday, 13 June 2012 22:18:01 UTC+5:30, payel roy wrote:
>
>
> Let's say there are N sides are given. Length of them are like
> 1,2,3,4,5,....N.
>
> How do you determine how many tri-angles can be made out of these N sides?
>
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