Range of numbers (dichotomous classic template topic)

Given a length in ascending order is n Array of integers for , as well as q A query .

For each query , Returns an element k Starting and ending positions of （ Location slave 0 Start counting ）.

If the element does not exist in the array , Then return to -1 -1.

Input format

The first line contains integers n and q, Represents the length of the array and the number of queries .

The second line contains n It's an integer （ Both in 1∼10000 Within the scope of ）, Represents a complete array .

Next q That's ok , Each line contains an integer k, Represents a query element .

Output format

common q That's ok , Each line contains two integers , Represents the starting and ending positions of the desired element .

If the element does not exist in the array , Then return to -1 -1.

Data range

1≤n≤100000
1≤q≤10000
1≤k≤10000

sample input ：

6 3
1 2 2 3 3 4
3
4
5

sample output ：

3 4
5 5
-1 -1

Ideas ： This topic belongs to Integer binary search A classic example of , For an ordered sequence of numbers , lookup x Start address and end address , You can find out the first ratio through dichotomy x The position of a large number a[mid]>=x, Then search from the last position to... Through two points n-1 Mid position ratio x The small first number a[mid]<=x.

  Code ：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int a[100010];
int main()
{
    int n,m;cin>>n>>m;
    for(int i=0;i<n;i++)cin>>a[i];
    while(m--)
    {
        int x;cin>>x;
        // Two points search , Defect search interval 
        int l=0,r=n-1;
        while(l<r)
        {
            int mid=l+r>>1;
            if(a[mid]>=x){    // Find the first one greater than x The location of r
                r=mid;
            }
            else l=mid+1;
        }
        int ll=r,rr=n-1;
        while(ll<rr)
        {
            int mid=ll+rr+1>>1;
            if(a[mid]<=x){    // Find the first one less than x The location of rr
                ll=mid;
            }
            else rr=mid-1;
        }
        if(a[r]==x)
        {
            cout<<r<<" "<<ll<<endl;
        }
        else cout<<"-1 -1"<<endl;
    }
    return 0;
    
}