清华学堂 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

这道题目，好早以前就接触过，那时候数据结构刚学，什么也不懂，今年又来刷这套题了，感觉还好

题目求 [a, b] 内所有的元素个数，二分搜索，求下界，对a求它严格的下界， 对b求它不严格的下界，搞定。

这道题目不用二分，只能拿一部分分数。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAX_SIZE = 5E5 + 10;
int num[MAX_SIZE];
int a[MAX_SIZE];

void quick_sort(int s[], int l, int r)
{
    if(l < r)
    {
        int i = l, j = r, x = s[l];
        while(i < j)
        {
            while(i < j && s[j] >= x)
                j--;
            if(i < j)
                s[i++] = s[j];

            while(i < j && s[i] < x)
                i++;
            if(i < j)
                s[j--] = s[i];
        }
        s[i] = x;
        quick_sort(s, l, i-1);
        quick_sort(s, i+1, r);
    }
}

///二分查下届
int BSearchLowerBound(int arry[], int low, int high, int target)
{
    if(high < low || target <= arry[low])
        return -1;
    int mid = (low + high + 1) / 2;
    while(low < high)
    {
        if(arry[mid] < target)
            low = mid;
        else
            high = mid - 1;
        mid = (low + high + 1)/2;
    }
    return mid;
}

int BSearchLowerBound_1(int arry[], int low, int high, int target)
{
    if(high < low || target < arry[low])
        return -1;
    int mid = (low + high + 1) / 2;
    while(low < high)
    {
        if(arry[mid] <= target)
            low = mid;
        else
            high = mid - 1;
        mid = (low + high + 1)/2;
    }
    return mid;
}

int main()
{
    int n, m;
    int x, y, ans;
    while(~scanf("%d %d", &n, &m))
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d", num+i);
        }
        //quick_sort 下标从 0 开始
        quick_sort(num, 0, n-1);

        for(int i = 0; i < m; i ++)
        {
            scanf("%d %d", &x, &y);

            ans = BSearchLowerBound_1(num, 0, n-1, y) - BSearchLowerBound(num, 0, n-1, x);

            printf("%d
", ans);
        }
    }
    return 0;
}