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)

/**
*    created: 14.06.2020 20:53:42
**/

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
const int inf =1000000000;

int vector_finder(std::vector<int> vec, int number) {
  auto itr = std::find(vec.begin(), vec.end(), number);
  size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}

int main() {
    int x,n;
    cin >> x>>n;
    vector<int> p(n);
    rep(i, n) cin >> p[i];
    int ans = inf;
    int annns;
    for (int i = -100; i <= 100; i++)
    {
        if (!vector_finder(p, i))
        {
            if (ans > abs(i - x))
            {
                annns = i;
            }
            ans = min(ans, abs(i - x));
        }
            }
            /*
    sort(p.begin(), p.end());
    int j=0;
    for (int k = j; k < n;k++){
        for (int i = 1; i <= 100;i++){
            if(p[k]!=i){
                ans = min(ans, abs(i - x));
                cout << ans << endl;
                break;
            }
        }
    }*/
    cout << annns << endl;
}