Tsinghua OJ：范围查询(Range)

范围查询(Range)

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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 21 3 7 9 114 67 12

Output

03

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

数轴上有n个点，对于任一闭区间 [a, b]，试计算落在其内的点数。

输入

第一行包括两个整数：点的总数n，查询的次数m。

第二行包含n个数，为各个点的坐标。

以下m行，各包含两个整数：查询区间的左、右边界a和b。

输出

对每次查询，输出落在闭区间[a, b]内点的个数。

样例

见英文题面

限制

0 ≤ n, m ≤ 5×105

对于次查询的区间[a, b]，都有a ≤ b

各点的坐标互异

各点的坐标、查询区间的边界a、b，均为不超过10^7的非负整数

时间：2 sec

内存：256 MB

解题思路：

1、用points数组标识，将有坐标位置置1（points数组初始化0）。

2、更新数组：points[i] += points[i-1]  (i>1)。这时points[i]，即表示[0,i]之间的坐标数。

3、用points[b] - points[a] 即得结果。

注意：由于第2步中，可能出现的下标为小于0的情况。因此，第1步置1的时候，可以将其往后移一位。

具体代码如下：

#include <cstdio>  #include <cstdlib>const int SZ = 1 << 20;  //提升IO buff struct fastio{char inbuf[SZ];char outbuf[SZ];fastio(){setvbuf(stdin, inbuf, _IOFBF, SZ);setvbuf(stdout, outbuf, _IOFBF, SZ);}}io;#define N 10000001int points[N];int main(){int n, m, i, x;int a, b;scanf("%d %d", &n, &m);for (i = 0; i < N; i++)points[i] = 0;while (n--){scanf("%d", &x);points[x + 1] = 1;}for (i = 2; i <= N; ++i)points[i] += points[i - 1];while (m--){scanf("%d %d", &a, &b);printf("%d\n", points[b + 1] - points[a]);}return 0;}