AcWing 789. 数的范围

找区间段的左/右端点

1、找左端点，则认为自己在分界点右侧，如果m[mid]>=a,则往左侧去，mid本身可能是答案，故保留r=mid，否则右侧

2、找右端点，认为自己在左侧，如果m[mid]<=a,则向右侧缩小区间,mid本身可能是答案保留l=mid，否则左侧

然后套模板，细究细节边界

#include<iostream>
#include<cstdio>

using namespace std;
const int N = 100010;
int n, q;
int m[N];
int main() {
    cin >> n >> q;
    for(int i = 0; i < n; i++) scanf("%d",&m[i]);
    int a;
    for(int i = 0; i < q; i++) {
        scanf("%d", &a);
        int l = 0, r = n-1, mid;
        while(l < r) {
            mid = l + r >> 1;
            if(m[mid] >= a) r = mid;
            else l = mid + 1;
        }
        if(m[r] == a) {
            cout << r << " ";
            r = n-1;
            while(l < r) {
                mid = l + r + 1>> 1;
                if(m[mid] <= a) l = mid;
                else r = mid - 1;
            }
            cout << r << endl;
        } else {
            cout << "-1 -1" << endl;
        }
    }

    return 0;
}