igdor99's blog

By igdor99, 7 years ago, translation, In English

580A - Kefa and First Steps

Note, that if the array has two intersecting continuous non-decreasing subsequence, they can be combined into one. Therefore, you can just pass the array from left to right. If the current subsequence can be continued using the i-th element, then we do it, otherwise we start a new one. The answer is the maximum subsequence of all the found ones.

Asymptotics — O(n).


580B - Kefa and Company

At first we sort all friends in money ascending order. Now the answer is some array subsegment. Next, we use the method of two pointers for finding the required subsegment.

Asymptotics — O(n log n).


580C - Kefa and Park

Let's go down the tree from the root, supporting additional parameter k — the number of vertices in a row met with cats. If k exceeds m, then leave. Then the answer is the number of leaves, which we were able to reach.

Asymptotics — O(n).


580D - Kefa and Dishes

A two-dimensional DP will be used to solve the problem. The first dimention is the mask of already taken dishes, and the second — the number of the last taken dish. We will go through all the zero bits of the current mask for the transitions. We will try to put the one in them, and then update the answer for a new mask. The answer will consist of the answer of the old mask, a dish value, which conforms to the added bit and the rule, that can be used. The final answer is the maximum of all the values of DP, where mask contains exactly m ones.

Asymptotics — O(2n * n2).


580E - Kefa and Watch

At first, we calculate the hash for all line elements depending on their positions. That is, the hash of the number k, standing on the i-th position will be equal to gi * k, where g is the base of the hash. We construct the segment tree of sums, which support a group modification, for all hashes. Thus, we can perform queries for modification in O(log n). It remains to deal with the queries of the second type. Let us assume, that we want to process the query 2 l r d. Obviously, the substring from l to r have a d-period, if a substring from l + d to r is equal to substring from l to r - d. We can find out the sum of hashes at the subsegment with the help of the sums tree, so we can compare the two strings in O(log n).

Asymptotics — O(m log n).


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