Me and my friends were working for upcoming ICPC regional. We got stuck at these two problems.
1-(This one is solved in comment sectiom) You are given an array with n elements, you can increase one element's value by one with one cost. Your goal is to make xor of all elements 0 with minimum cost. n <= 1000, ai <= 109
Here is the link for the problem: http://algotester.com/en/ArchiveProblem/Display/40442
2-You will be given n intervals like [li, ri]. You have to divide these intervals into two subsets A and B such that every interval belongs to exactly one subset and the intersection of A and B maximized and print this subset. The intersection of two sets of intervals is sum of every pair of intervals' (one from A, one from B) intersection length. n <= 100000, 0 <= li <= ri <= 109
Link to this problem: http://algotester.com/en/ArchiveProblem/Display/40384
Thank you for your help (:
The first one is an awesome problem as you need many observations in order to solve it. I am going to describe the solution very briefly.
Let us fix an optimal solution (that is a choice of b[i] ≥ a[i] such that their xor is 0 and is minimized). Then we make the following observations:
With this observations (that I leave to prove to the reader) it is easy to create a solution. Indeed, once we have fixed the largest value of x[i], the previous observations are enough to uniquely define an optimal strategy (fixing greedily all bits from the largest to the smallest).
In some sense the described algo is almost greedy, indeed, after a single choice, a greedy algorithm works.
"Each positive value of x[i], apart from the largest one, is assumed at most one time. The largest one can be assumed at most twice."
What do you mean here by saying "assumed"?
"V is assumed at most one time" = "there exists at most one i such that x[i] = V".
oh now i think i got it, thank you!
I think the second problem is this problem: http://mirror.codeforces.com/contest/429/problem/E. I have a comment about it here: http://mirror.codeforces.com/blog/entry/12265?#comment-169310