清華OJ——資料結構與演算法實驗（中國石油大學）

範圍查詢(Range)

Descriptioin

Let S be a set of n integral points on the x-axis. For each given interval [a, b], you are asked to count the points lying inside.

Input

The first line contains two integers: n (size of S) and m (the number of queries).

The second line enumerates all the n points in S.

Each of the following m lines consists of two integers a and b and defines an query interval [a, b].

Output

The number of points in S lying inside each of the m query intervals.

Example

Input

5 2
1 3 7 9 11
4 6
7 12

Output

0
3

Restrictions

0 <= n, m <= 5 * 10^5

For each query interval [a, b], it is guaranteed that a <= b.

Points in S are distinct from each other.

Coordinates of each point as well as the query interval boundaries a and b are non-negative integers not greater than 10^7.

Time: 2 sec

Memory: 256 MB

描述

數軸上有n個點，對於任一閉區間 [a, b]，試計算落在其內的點數。

輸入

第一行包括兩個整數：點的總數n，查詢的次數m。

第二行包含n個數，為各個點的座標。

以下m行，各包含兩個整數：查詢區間的左、右邊界a和b。

輸出

對每次查詢，輸出落在閉區間[a, b]內點的個數。

樣例

見英文題面

限制

0 ≤ n, m ≤ 5×105

對於每次查詢的區間[a, b]，都有a ≤ b

各點的座標互異

各點的座標、查詢區間的邊界a、b，均為不超過10^7的非負整數

時間：2 sec

記憶體：256 MB

 1 #include<cstdio>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=1e7+10;
 5 int a[N];
 6 int main()
 7 {
 8     int n,m,now;
 9     int maxx=0;
10     scanf("%d%d",&n,&m);
11     for(int i=0;i<n;i++)
12     {
13         scanf("%d",&now);
14         a[now]=1;
15     }
16     for(int i=1;i<N;i++)
17     {
18         a[i]=a[i-1]+a[i];
19     }
20     for(int i=0;i<m;i++)
21     {
22         int x,y;
23         scanf("%d%d",&x,&y);
24         printf("%d\n",a[y]-a[x-1]);
25     }
26     return 0;
27 }