acwing789.数的范围(二分法)

#include <iostream>
using namespace std;
const int N=1e6+10;
int q[N];
int n,m;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++) scanf("%d",&q[i]);
    while(m--){
        int x;
        scanf("%d",&x);
        int l=0,r=n-1;
        while(l<r){
            int mid = (l+r)>>1;
            if(q[mid]>=x) r=mid;//让端点右边的数都≥x；因此是寻找左端点
            else l=mid+1;
        }
        if(q[l]!=x) cout<<"-1 -1"<<endl;//异常情况，没有x
        else{
            cout<<l<<" ";
            int l=0,r=n-1;
            while(l<r){
                int mid = (l+r+1)>>1;
                if(q[mid]<=x) l=mid;//让端点左边的数都≤x；因此是寻找右端点
                else r=mid-1;
            }
            cout<<l<<endl;
        }
    }
}