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: Accepted)

#include <iostream>
#include <string>
#include <math.h>
#include <iomanip>
#include <stdio.h>
#include <algorithm>
#include <ctime>
#include <vector>
#include <set>
#include <queue>
using namespace std;
typedef long long ll;

#define PI 3.14159265358979323846264338327950L

int res;
int f(int x)
{
    if (0 < x)
    {
        return -(x + 1);
    }
    else if (x < 0)
    {
        return -x;
    }
    else
    {
        return x = -1;
    }
}

int main()
{
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int X, N;
    cin >> X >> N;
    vector<int> P(N);
    for (int i = 0; i < N; i++)
    {
        cin >> P[i];
    }
    if (N == 0)
    {
        res = X;
        cout << X << endl;
        return 0;
    }
    int min = 100000;
    int number = 0;
    int cnt = 0;
    bool flag = true;
    while (flag)
    {
        int data = abs(X + number);
        bool isOK = true;
        for (int j = 0; j < N; j++)
        {

            if (data == P[j])
            {
                isOK = false;
            }
        }
        if (isOK)
        {
            res = X + number;
            flag = false;
        }
        //cout << number << endl;
        number = f(number);
    }
    cout << res << endl;
    return 0;
}