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Maybe he implies that the problems are not sorted by difficulty

Держи в курсе

and also not a geometry one!

I like mathforces

who cares about you to be honest

Не жду задачу А

Seems so, coz his problems are generally based on good idea/observation + quick/short implementation but if you haven't thought of the idea clearly then implementation can go pretty lengthy and fail on system tests

I think you should block both users

The number will obviously fit.

E can be solved with segment tree + lazy propagation.

2nd query will be discarded (just calling the get function).
To do the 1st query, first update that single element without lazy propagation, then perform updating all elements id in the range [i + 1, mx] (inclusive) with , with mx as the maximum index that . mx could be found by binary search.

Tried to create a difference array from array A and array K, and tried updating the values of this diff array for query of type 1, it will turn most of the values of this diff array to 0.

Query of type 1 will increase the diff value of only first element and make all other diff value 0. So, at most 2*n values need to be updated at most.

Tried to maintain the index of non-zero values of this diff array using set,and then range update and range sum query using segment tree lazy propagation.

The answer will fit inside of a long long. max(A) starts out  ≤ 10⁹ and you can add at most 10¹¹ to it, so A[i] will at most be around 10¹¹, meaning the sum of the As will at most be around 10¹⁶. So everything should fit inside a long long without any problems.

About the solution: You can reduce the problem to all k being 0 by working on instead of A. For B we have that B[i] ≤ B[i + 1] so it must always be increasing.

To do the two types of queries I used binary search together with a lazy segment tree that allowed sum on intervals and set on intervals.

To do the add query you need to add x to B[i] and then update the rest of B so that it is increasing, to do this you can apply binary search on B to find all indices j such that j > i and B[i] > B[j]. Then use the set on interval operation to make all of those B[j] equal to B[i]. This takes per query.

To do the sum query just use the built in sum in the lazy segment tree. But this is a sum over B and not A, so you need to adjust the answer by removing the extra k you added in the beginning, which can be done by playing around with cumulative sums. This takes per query.

There is a solution that do not uses segment tree tricks, just two plain seg trees, one with range += lazy and range sum other with range = and one get one element.

You can notice that if you update a[i] += x that leads to a[i + 1] = a[i] + k[i] and so on from i to some l, than if you'll update a[i] += y again, you'll just make += on range(i, l). So this observations leads us to smth like DSU (disjoint set union). But there is one thing, when we get query update a[i] += x and i belongs to some [l, r] segment where l is strictly less than i this segment [l, r] splits to two [l, i — 1], [i, r]. So you need to do unusual DSU with segment tree (the one with range=).

When you get a[i] += x query your steps is split range to which i belongs than go to right and merge some ranges (continue while conditions from statement is true), while merging make += on right range. It is clear that we will do at most n + q mergins, because we can merge only by two reasons 1 — it was originally disjont (thats n) or 2 — it was splitted (that's q). Complexity O((n + q) * log(n))

Submission for details 51198410

You can solve it using plain segment tree with lazy updates.

Notice the fact that the values will be increasing. So, suppose that the i+1 th index is changed once due to the update on i'th index. Now, if somehow a[i] increases again, a[i+1] will increase by the same amount. Based on this, you can perform range updates. And number of these updates won't be that much.

If you can swap, do it immediately otherwise take the previous pupil to the farthest position from you, until nothing change

We will iterate from the back of the queue (person before Nastya) to the front. We will store for each person i how many persons are after this person in the queue and are favored by person i, in fav[i]. Initially we set fav[i] for each person i that favors Nastya to 1. We will also maintain a variable that stores how many persons have been removed from the queue so far. Now if we are at person i and there are x persons removed, a total of rem = n - 1 - i - x persons remain to the right of this person. If fav[Q[i]] == rem, every person that remains to the right of this person is favored by this person, and thus this person can swap with each one of these. In this case we remove this person and increase our answer by 1. Otherwise we cannot remove it and we will increase fav for all the persons that favor this person.

The complexity of this approach is O(N + M). https://mirror.codeforces.com/contest/1136/submission/51189237

I just pushed every person that can let Nastya stay before him to the end of p[] (to the place before such persons that are already at the end), and then counted number of such persons in the end of p[].

Observation: if person can not be pushed to the end, he can not ever let Nastya ahead.

51192297

Moving from natasha to first guy, if you find the guy that can move to back of natash,then move it.

Each respective diagonal in the matrix must have the same elements.

Notice that you can imitate transposition of every matrix by little transpositions of matrices 2x2. That's because transposition of matrix 2x2 swaps two nearby elements on the diagonal, and transposition of matrix nxn swaps some elements on every it's diagonal (elements that are on a diagonal remain on that diagonal). But you can imitate every permutation by some swapping nearby elements.

I noticed that transpose does not change the sum of row+column. That's why I just verify that both matrices have values with equivalent sums. I did not prove it though.

Each respective diagonal in the matrix must have the same elements.

Check in your "Compiler settings" if you'd enabled C++11 features yet.