Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int x, n;
    cin >> x >> n;
    if(n == 0){
        cout << x << endl;  
    }else {
        vector<int> p(n);
        for(int i = 0; i < n; i++) {
            cin >> p.at(i);
        }
         vector<int> num(201);
        for(int i = 0; i <= 200; i++) {
            num.at(i) = i-100;
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= 200; j++) {
                if(p.at(i) == num.at(j)) {
                    num.at(j) = 1000;
                }
            }
        }
        /*for(int i = 0; i <= 200; i++) {
            cout << num.at(i) << endl;    
        }*/
        int ans = max(x-num.at(0), num.at(0)-x);
        cout << ans << endl;
        for(int i = 1; i <= 200; i++) {
              ans = min(ans, max(x-num.at(i), num.at(i)-x));
              cout << ans << endl;
        }
        for(int i = 0; i <= 200; i++) {
              if((x-ans) == num.at(i)) {
                 cout << x-ans << endl;
                 return 0;
              }
        }
        for(int i = 0; i <= 200; i++) {
              if((x+ans) == num.at(i)) {
                 cout << x+ans << endl;
                 return 0;;
              }
        }
        
    }
}