题目地址：

https://www.luogu.com.cn/problem/CF1700D

题面翻译：
有$n$个容器，第$i$个容器容量为$v_i$升，可以容纳$[0,v_i]$升的水。

满出去的水会将从容器$i$转移到容器$i+1$，如果$i+1$也满了会转移得更远。满出最后一个容器的水会倒到河中。

现在要将所有容器填满。你可以选择一些容器注水，让这些容器每秒进入一升水。$q$次询问，问最初所有容器都是空的，最少选择多少个容器注水使得$t_i$秒内能填满所有容器。

$1\le n,q\le 2\times 10^5$，$1\le v_i,t_i\le 10^9$。

题目描述：
Recently in Divanovo, a huge river locks system was built. There are now $n$ locks, the $i$ -th of them has the volume of $v_i$ liters, so that it can contain any amount of water between $0$ and $v_i$ liters. Each lock has a pipe attached to it. When the pipe is open, $1$ liter of water enters the lock every second.

The locks system is built in a way to immediately transfer all water exceeding the volume of the lock $i$ to the lock $i+1$ . If the lock $i+1$ is also full, water will be transferred further. Water exceeding the volume of the last lock pours out to the river.
The picture illustrates $5$ locks with two open pipes at locks $1$ and $3$ . Because locks $1$ , $3$ , and $4$ are already filled, effectively the water goes to locks $2$ and $5$ .To make all locks work, you need to completely fill each one of them. The mayor of Divanovo is interested in $q$ independent queries. For each query, suppose that initially all locks are empty and all pipes are closed. Then, some pipes are opened simultaneously. For the $j$ -th query the mayor asks you to calculate the minimum number of pipes to open so that all locks are filled no later than after $t_j$ seconds.

Please help the mayor to solve this tricky problem and answer his queries.

输入格式：
The first lines contains one integer $n$ ( $1\le n\le 200000$ ) — the number of locks.

The second lines contains $n$ integers $v_1,v_2,\dots ,v_n$ ( $1\le v_i\le 10^9$ )) — volumes of the locks.

The third line contains one integer $q$ ( $1\le q\le 200000$ ) — the number of queries.

Each of the next $q$ lines contains one integer $t_j$ ( $1\le t_j\le 10^9$ ) — the number of seconds you have to fill all the locks in the query $j$ .

输出格式：
Print $q$ integers. The $j$ -th of them should be equal to the minimum number of pipes to turn on so that after $t_j$ seconds all of the locks are filled. If it is impossible to fill all of the locks in given time, print $-1$ .

注意题目中说，每个水槽满了之后水都会漫到其右边那个水槽里，所以如果能装满的话，优先开最左边的若干个水龙头，这样使得水漫出右边界的情况尽可能少的发生。如果只考虑第$1$个水槽，则至少需要$v_1$单位时间才能装满；如果只考虑前$2$个水槽，那么如果两个水龙头一起开，如果不溢出的话，花的时间最少是$\lceil \frac{v_1+v_2}{2}\rceil$，但是如果有溢出，意味着第$1$个水槽还没装满，但是第$2$个水槽已经漏水了，这个时候花的时间是$v_1$，所以综合两种情况，时间最少是 max ⁡ { v 1 , ⌈ v 1 + v 2 2 ⌉ } \max\{v_1,\lceil \frac{v_1+v_2}{2} \rceil\} max{v1​,⌈2v1​+v2​​⌉}，以此类推，要装满所有水槽，时间最少是$t_0={\max }_k\frac{{\sum }_{i=1}^k{v}_i}{k}$，先求出这个$t_0$，如果给出的询问$t<t_0$直接输出$-1$。

如果$t\ge t_0$，那么至少是有解的，设解为$k$，则$k\ge \lceil \frac{{\sum }_{i=1}^n{v}_i}{t}\rceil =k_0$，但是$k_0$个水龙头事实上已经足够了，因为水只会向右边漫，只要开前$k_0$个水龙头，过了$t$时间后，显然前$k_0$个水槽是满的，溢出前$k_0$个水槽的水刚好保证后面的水槽装满（可能有溢出，因为是上取整）。代码如下：

#include <iostream>
using namespace std;
using LL = long long;
 
const int N = 2e5 + 10;
int n, q;
LL a[N], low;
 
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i] += a[i - 1];
  }
 
  for (int i = 1; i <= n; i++) low = max(low, (a[i] + i - 1) / i);
 
  scanf("%d", &q);
  while (q--) {
    long t;
    scanf("%lld", &t);
    if (t < low) puts("-1");
    else printf("%lld\n", (a[n] + t - 1) / t);
  }
}
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预处理时间复杂度$O(n)$，每次询问$O(1)$，空间$O(1)$，