|
|
back to board1509 Posted by svr 5 Dec 2006 11:23 Problem really simple. Use harsh criterion of koeff between max and min distances stable for rounding ang if N=7 also number of pairs with min koeff. Most difficult situation if N=2 and distance near 50*sqrt(2), if using rounding rooles here maybe very many tricky tests but authors kindly didn't use this opportunities. Re: 1509 Posted by Lomir 20 Jan 2007 00:40 Yep, problem rather simple, but i made more then 30 submission before AC, and fully rewrited my program 3 times. Also there are something strage with the precision. Re: 1509 Also there are something strage with the precision. Yes, there are strange things with precision. After normalizing max distance to 1, not all metrics worked: sum of absolute values of differences between corresponding distances -> ac sum of squares of differences between corresponding distances -> wa8 maximum of absolute values of differences between corresponding distances -> wa8 Re: 1509 problem rather simple, but i made more then 30 submission before AC I made exactly 30 submissions before AC. I think this number is standard for this problem. :-) Re: 1509 Posted by svr 21 Jan 2007 23:29 I think that float numbers troubles tipical thing. For float math lows don't satisfies and result depend on algorithm when for right math result dosn't depend on and we can use different algos. It's humor then men play with epsilon trying to ajust to tests list. |
|
|