不需要考虑mid+1、mid-1的二分查找模板

一种更便于记忆的二分法

1. l = -1, r = n;
2. l = mid, r = mid;
3. while(l + 1 != r);

模板

int L=-1,R=n;
while(L+1!=R)
{
	int mid=L+R>>1;
	if(check()) L=mid;
	else R=mid;
	//最后根据你所分左右两边区间的结果
	//选取L或者R作为结果
}

AcWing 789. 数的范围

#include<iostream>
using namespace std;
const int N = 1e6 + 10;

int n, q, k, a[N];
int main(){
    cin >> n >> q;
    for(int i = 0; i < n; i++)scanf("%d", &a[i]);
    while(q--){
        scanf("%d", &k);
        int l = -1, r = n;
        while(l + 1 != r){
            int mid = l + r >> 1;
            if(a[mid] < k) l = mid;
            else r = mid;
        }
        if(a[r] != k){
            cout << -1  << " " << -1 << endl;
            continue;
        }
        int ll = -1, rr = n;
        while(ll + 1 != rr){
            int midd = ll + rr >> 1;
            if(a[midd] <= k) ll = midd;
            else rr = midd;
        }

        cout << r << " " << ll << endl;
    }
}

https://blog.csdn.net/WJPnb1/article/details/126360962?spm=1001.2014.3001.5502