## 题目

### Problem Statement

N hotels are located on a straight line. The coordinate of the i-th hotel $(1≤i≤N)$ is $x_i$.

Tak the traveler has the following two personal principles:

He never travels a distance of more than L in a single day.
He never sleeps in the open. That is, he must stay at a hotel at the end of a day.
You are given Q queries. The $j$-th $(1≤j≤Q)$ query is described by two distinct integers aj and bj. For each query, find the minimum number of days that Tak needs to travel from the aj-th hotel to the bj-th hotel following his principles. It is guaranteed that he can always travel from the aj-th hotel to the bj-th hotel, in any given input.

### Constraints

• $2≤N≤105$
• $1≤L≤109$
• $1≤Q≤105$
• $1≤x_i<x_2<…<x_N≤109$
• $x_i+1−x_i≤L$
• $1≤a_j,b_j≤N$
• $a_j≠b_j$
• $N, L, Q, x_i, a_j, b_j$ are integers.

### Partial Score

$200$ points will be awarded for passing the test set satisfying $N≤10^3$ and $Q≤10^3$.

### Input

The input is given from Standard Input in the following format:

$N$
$x_1$ $x_2$ $…$ $x_N$
$L$
$Q$
$a_1$ $b_1$
$a_2$ $b_2$
$:$
$a_Q$ $b_Q$

### Output

Print Q lines. The j-th line $(1≤j≤Q)$ should contain the minimum number of days that Tak needs to travel from the $a_j$-th hotel to the $b_j$-th hotel.

## 题意

有n个点，第i个点的坐标为x[i]，有一个人每天最多走L米，他每天必须从某个点出发，然后再终止于某个点，有Q个询问，代表他的起点和终点，问最少需要多少天可以到达。

## 思路

预处理一下，倍增的过程中统计答案即可，路径不唯一，但答案一定是唯一的。

## 实现

#include <bits/stdc++.h>
const int maxn = int(1e5) + 7, inf = 0x3f3f3f3f;
int n, x[maxn], up[maxn], l, q, u, v, e = {1}, ans, tmp;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", x + i);
scanf("%d%d", &l, &q);
memset(up, 0x3f, sizeof(up));
int cur = 1;
for (int i = 1; i <= n; i++) while (x[i] > x[cur] + l) up[cur++] = i - 1;
for (int j = 1; j < 20; j++) {
e[j] = e[j - 1] << 1;
for (int i = 1; i <= n; i++) if (up[i][j - 1] < inf)
up[i][j] = up[up[i][j - 1]][j - 1];
}
while (ans = 0, q--) {
scanf("%d%d", &u, &v);
if (u > v) std::swap(u, v);
for (int i = 19; i >= 0; i--) if (up[u][i] < v) u = up[u][i], ans += e[i];
printf("%d\n", ans + 1);
}
return 0;
}