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;
    vector<int> hukumanai(n);
    vector<int> hukumu(0);
    for (int i = 0; i < n; i++){
        cin >> hukumanai.at(i);
    }
    for (int i = -100; i < 101; i++){
        bool a = 1;
        for (int j = 0; j < n; j++){
            if(hukumanai.at(j) == i)a = 0;
        }
        if(a)hukumu.push_back(i);
    }
    vector<pair<int, int>> p(hukumu.size());
    for (int i = 0; i < hukumu.size(); i++){
        p.at(i).second = hukumu.at(i);
        p.at(i).first = abs((x - hukumu.at(i)));
    }
    sort(p.begin(), p.end());
    cout << p.at(0).second << endl;
}