Bakery
Contest: 2020-2021 ICPC, NERC, Southern and Volga Russian Regional Contest (Online Mirror, ICPC Rules)
Difficulty rating: 2900
Tags: data structures, dsu

Monocarp would like to open a bakery in his local area. But, at first, he should figure out whether he can compete with other shops.

Monocarp plans that the bakery will work for $$$n$$$ days. On the $$$i$$$-th day, $$$a_i$$$ loaves of bread will be baked in the morning before the opening. At the end of the $$$n$$$-th day, Monocarp will sell all the remaining bread that wasn't sold earlier with a huge discount.

Because of how bread is stored, the bakery seller sells the bread in the following order: firstly, he sells the loaves that were baked that morning; secondly, he sells the loaves that were baked the day before and weren't sold yet; then the loaves that were baked two days before and weren't sold yet, and so on. That's why some customers may buy a rather stale bread and will definitely spread negative rumors.

Let's define loaf spoilage as the difference between the day it was baked and the day it was sold. Then the unattractiveness of the bakery will be equal to the maximum spoilage among all loaves of bread baked at the bakery.

Suppose Monocarp's local area has consumer demand equal to $$$k$$$, it means that each day $$$k$$$ customers will come to the bakery and each of them will ask for one loaf of bread (the loaves are sold according to the aforementioned order). If there is no bread left, then the person just doesn't buy anything. During the last day sale, all the remaining loaves will be sold (and they will still count in the calculation of the unattractiveness).

Monocarp analyzed his competitors' data and came up with $$$m$$$ possible consumer demand values $$$k_1, k_2, \dots, k_m$$$, and now he'd like to calculate the unattractiveness of the bakery for each value of demand. Can you help him?

Input format:
The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le m \le 2 \cdot 10^5$$$) — the number of days the bakery is open and the number of possible values of consumer demand.

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the number of bread loaves that will be baked each day.

The third line contains $$$m$$$ integers $$$k_1, k_2, \dots, k_m$$$ ($$$1 \le k_1 < k_2 < \dots < k_m \le 10^9$$$) — the possible consumer demand values in the ascending order.

Output format:
Print $$$m$$$ integers: for each consumer demand, print the unattractiveness of the bakery.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
In the first example, let's describe what happens for couple consumer demands:

If consumer demand is equal to $$$1$$$:

- at day $$$1$$$: $$$5$$$ loaves are baked and only $$$1$$$ is sold with spoilage equal to $$$1 - 1 = 0$$$;
- at day $$$2$$$: $$$4$$$ loaves are left and $$$2$$$ more are baked. Only $$$1$$$ loaf was sold and it was the loaf baked today with spoilage $$$2 - 2 = 0$$$;
- at day $$$3$$$: $$$4$$$ loaves from the first day and $$$1$$$ loaf from the second day left. One more loaf was baked and was sold this day with spoilage $$$3 - 3 = 0$$$;
- at day $$$4$$$: $$$4$$$ loaves from the first day and $$$1$$$ loaf from the second day left. $$$3$$$ more loaves were baked and one of them was sold this day with spoilage $$$4 - 4 = 0$$$;
- at day $$$5$$$: $$$4$$$ loaves from the first day, $$$1$$$ loaf from the second day and $$$2$$$ loaves from the fourth day left. $$$7$$$ more loaves were baked and, since it's the last day, all $$$14$$$ loaves were sold. $$$4$$$ loaves from the first day have the maximum spoilage equal to $$$5 - 1 = 4$$$.

If consumer demand is equal to $$$10$$$ then all baked bread will be sold in the day it was baked and will have spoilage equal to $$$0$$$.

Example input:
5 4
5 2 1 3 7
1 3 4 10

Example output:
4 2 1 0

Example input:
8 9
3 1 4 1 5 9 2 6
1 2 3 4 5 6 7 8 9

Example output:
7 5 3 3 2 1 1 1 0

Editorial:
Let's look at all $$$n$$$ days with some fixed $$$k$$$. Obviously, the seller works like a stack, so let's divide all days into segments in such a way, that the stack is emptied between consecutive segments. We can note that after we've got these segments — the answer is the maximum length among all segments. Why? Because the stalest bread on the bottom of the stack and we can't sell it until we empty the stack.
Now, let's set $$$k = 10^9$$$ and look at what happens when we gradually lower $$$k$$$. With $$$k=10^9$$$, we sell all bread that we baked on the same day, so all segments consist of one day $$$[i, i + 1)$$$. Now, if we start lowering $$$k$$$ then at some moment segments will start to merge (and only merge), since $$$k$$$ is not enough to sell all bread in this interval. Since no segment will split up, there will be only $$$n - 1$$$ merging.
So we can look only at moments $$$k$$$ when either several segments merge or when we should answer the query. With what $$$k$$$ the segment $$$[p_1, p_1)$$$ will start to merge with the next segment $$$[p_1, p_2)$$$? The answer is when $$$(p_2 - p_1) \cdot k < \sum\limits_{p_1 \le i < p_2}{a_i}$$$ or $$$k = \left\lfloor \frac{(\sum{a_i}) - 1}{p_2 - p_1} \right\rfloor$$$.
So, we can for each segment $$$[p_i, p_{i + 1})$$$ maintain its value $$$k_i$$$ when it will start merging with next segment in set. And if we want to calculate the answer for a given $$$k$$$ from the queries, we should merge all segments with $$$k_i \ge k$$$, while updating the current maximum length among all segments.
Since merging two segments require two operations with the set then the total time complexity is $$$O(m + n \log{n})$$$.