题目描述
Farmer John has just arranged his NN haybales (1 leq N leq 100,0001≤N≤100,000) at various points along the one-dimensional road running across his farm. To make sure they are spaced out appropriately, please help him answer QQ queries (1 leq Q leq 100,0001≤Q≤100,000), each asking for the number of haybales within a specific interval along the road.
农夫John在一条穿过他的农场的路上（直线）放了N个干草垛（1<=N<=100,000），每个干草垛都在不同的点上。为了让每个干草垛之间都隔着一定的距离，请你回答农夫John的Q个问题（1<=Q<=100,000），每个问题都会给出一个范围，询问在这个范围内有多少个干草垛。
（其实就是有一条数轴上有N个不重复的点，再问Q个问题，每个问题是给出一个范围，问此范围内有多少个点？）
（在给出范围的边界上也算）
输入格式
The first line contains N and Q.
The next line contains N distinct integers, each in the range 0 ldots 1,000,000,000, indicating that there is a haybale at each of those locations.
Each of the next Q lines contains two integers A and B(0 leq A leq B leq 1,000,000,000) giving a query for the number of haybales between A and B, inclusive.
第一行包括N和Q
第二行有N个数字，每个数字的范围在0~1,000,000,000，表示此位置有一个干草垛。
接下来的Q行，每行包括两个数字，A和B（0<=A<=B<=1,000,000,000）表示每个询问的范围
输出格式
You should write Q lines of output. For each query, output the number of haybales in its respective interval.
总共Q行，每行输出此范围内的干草垛数量
挺简单的
就不做说明了
水水博客
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e6+10;
int a[N];
signed main(){
int n,q;
cin>>n>>q;
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
sort(a+1,a+1+n);
int A,B;
while(q--){
scanf("%d%d",&A,&B);
int op1=lower_bound(a+1,a+1+n,A)-a;
int op2=upper_bound(a+1,a+1+n,B)-a;
printf("%d
",op2-op1);
}
}