Editorial for DMOPC '15 Contest 2 P3 - Origami

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Author: cheesecake

For every cut we make, let's greedily cut the maximum amount of paper we can. Until we have more pieces of paper than $K$ ~K~, we will always double the amount that we currently have with each cut. Once we have more than $K$ ~K~, we can only obtain $K$ ~K~ pieces more with each cut, so we can use division to find the remaining cuts needed.

Time Complexity: $\mathcal{O}(\log \min(N, K))$ ~\mathcal{O}(\log \min(N, K))~

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