Number of upward triangles of side length k where biggest triangle size is n (n>=k) is 1 + 2 + ... + (n-k+1) = (n-k)(n-k+1)/2
Number of downward triangles of side length k where biggest triangle size is n (n>=2k) is 1 + 2 + ... + (n-2k+1) = (n-2k)(n-2k+1)/2 Therefore answer is [(sigma, 1, n) (n-k)(n-k+1)/2] + [(sigma, 1, (int)n/2) (n-2k)(n-2k+1)/2 ] Using formulas for sigma of 1 to n of k and k^2, you can easily derive the formula now -Bharath --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/google-code?hl=en -~----------~----~----~----~------~----~------~--~---
