简简单单读懂题后就能明白，这就是一道查找题了。

然后看一下数据量，1e6那基本就是明着告诉你暴力遍历会超时了。

根据以上两点，不难想到二分查找，而本题就是考察的二分查找的左边界与右边界，找到目标编号的左边界与右边界然后除以2就得出答案了。

参考代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int nums[maxn];
int key,n;


int left_bound() {
    if (n == 0) return -1;
    int left = 0;
    int right = n - 1; 

    while (left <= right) { 
        int mid = left + ((right - left) >> 1);
        if (nums[mid] == key) {
            right = mid - 1 ;
        } else if (nums[mid] < key) {
            left = mid + 1;
        } else if (nums[mid] > key) {
            right = mid - 1; 
        }
    }
    if (left == n || nums[left] != key) return -1;
    return left;
    }

int right_bound() {
    if (n == 0) return -1;
    int left = 0;
    int right = n - 1; 

    while (left <= right) { 
        int mid = left + ((right - left) >> 1);

        if (nums[mid] == key) {
            left = mid + 1 ;
        } else if (nums[mid] < key) {
            left = mid + 1;
        } else if (nums[mid] > key) {
            right = mid - 1; 
        }
    }
    if (left == n || nums[left - 1] != key) return -1;
    return left - 1;
    }
    
    
int main() {
	cin >> n >> key;
    for (int i = 0; i < n; i++) cin >> nums[i]; 
    sort(nums,nums + n);
    if (left_bound() == -1) {
    cout <<  "Not Found";
	return 0;	
	}
    int ans = (left_bound() +  right_bound()) / 2 + 1;
    cout << ans;
    return 0;
}