So, m writing this post for some more problems of SPOJ. The problems I m going to discuss are of an average level.
BISHOPS: Its an easy problem, if we try verbally. The answer is (2*n-2) i.e. equal to the number of diagonals. But the difficulty lies in the range of the inputs. N<=10^100..!! OMG!!.. :).. Try strings..
MINCOUNT: Actually what I did.. I tried it out for certain test cases and found that the answer is.. (n*(n+1)/2)/3.. i.e. the total number of coins in the triangle divided by 3. But why its so.. its still unanswered for me.. kindly post the explanation if you get one..!! :)
HUBULLU: Same mincount type problem here too.. seeing the discussion in the forum I got some hints that who so ever plays first wins.. So, it became a very very easy question for me to crack but why it happens is yet not understood by me..
TRICOUNT: Its another mathematics problem.. try it for sometime with all yours geometrical anlysis and at one time you will definitely come out with a solution that is much like (t*(t+2)*(2*t+1))/8.. where t is the level of triangle. Try it now yourself..!!
SAMER08F: Another age old problem of how many squares are there in a given bunch of squares?? Answer lies in the sum all squared numbers till the number of squares in each side of the grid i.e. if N=8 then answer= (1*1) + (2*2) + ... + (8*8).
Seeing my last three posts of SPOJ I think now its easy for any newbie to get started. So, no more discussing algorithms of SPOJ problems from my side. Keep practicing and who knows one day you beat [Trichromatic] XilinX..!! :p
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