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By xzyxzy, history, 2 years ago, , I'm sorry to ask a foolish problem about my submission.So I share a way to solve 808E by ternary searching.

As we can see,W[i] = 1, 2, 3. We sort the Souvenirs which have the same W[i] by their C[i]. Then we know we will just choose a prefix for each W[i].

If we get X Souvenirs that W[i] = 3,we remain (m - 3X) to get Souvenirs that W[i] = 1, 2. Let's define F[i][x] the prefix sum of Souvenirs that W[i] = x. If we get Y Souvenirs that W[i] = 2,our cost is F[X] + F[Y] + F[m - 3X - 2Y] (m-3X-2Y>=0).

We can try every X and use ternary searching to find the best Y.The complexity is O(nlogn).

My submission:http://codeforces.com/contest/808/submission/27174539 Comments (5)
 » 2 years ago, # | ← Rev. 3 →   But we know that as X increases,the optimal Y is decreasing,so you can use two-pointers to solve it.And if you use radix sort,you can get an algorithm which complexity is O(n).
•  » » That's right!Thanks for your better solution.
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 » I'm trying to learn how to prove when a function (like this one) is strictly increasing-then-decreasing (or viceversa) so to apply ternary search. How can i formally show this function holds this property ??? Other problems are welcome too