二分lower_bound()与upper_bound()的运用

<span style="color:#6633ff;">/*
G - 二分
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit
 
Status
Description
Given n points (1 dimensional) and q segments, you have to find the number of points that lie in each of the segments. A point pi will lie in a segment A B if A ≤ pi ≤ B.

For example if the points are 1, 4, 6, 8, 10. And the segment is 0 to 5. Then there are 2 points that lie in the segment.

Input
Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers denoting the points in ascending order. All the integers are distinct and each of them range in [0, 108].

Each of the next q lines contains two integers Ak Bk (0 ≤ Ak ≤ Bk ≤ 108) denoting a segment.

Output
For each case, print the case number in a single line. Then for each segment, print the number of points that lie in that segment.

Sample Input
1

5 3

1 4 6 8 10

0 5

6 10

7 100000

Sample Output
Case 1:

2

3

2

Hint
Dataset is huge, use faster I/O methods.
By Grant Yuan
2014.7.16
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
int t;
long long low,high,mid,ans;
int n,m;
int a[100005];
long long x,y;
int ct=1;
int main()
{int l,r;
    scanf("%d",&t);
    while(t--){
    scanf("%d%d",&n,&m);

    for(int i=0;i<n;i++)
      {
      scanf("%d",&a[i]);
        }
         printf("Case %d:
",ct++);
    while(m--){
         scanf("%d%d",&x,&y);
         if(x<a[0]) l=0;
         else {l=lower_bound(a,a+n,x)-a;
         }
         if(y>a[n-1]) r=n;
         else {r=upper_bound(a,a+n,y)-a;
         }
         printf("%d
",r-l);}}
         return 0;
}
</span>