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<iostream>
#include<string>
#include<vector>
#include<cmath>
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
int arr[1000];
int main() {
    int x,n;
    cin >> x >> n;
    int min = 10000000;
    int max = -1;
    int p;
    FOR(i, 0, n) {
        cin >> p;
        arr[p] = 1;
        if (p < min)
            min = p;
        if (p > max)
            max = p;
    }
    // int dist = 100000;
    // int buf = x;
    // int a;
    // FOR(i , min , max + 1) {
    //     if (arr[i] == 0) {
    //         if ((a = abs(x - i)) < dist) {
    //             dist = a;
    //             buf = i;
    //         }
    //     }
    // }
    if (n == 0) {
        cout << x << endl;
        return 0;
    }
    
    int off = 0;
    int i = x;
    while(i < min)
        i++;
    while(i >= max)
        i--;
    while (off < 100) {
        if (i - off > 0 && arr[i - off] == 0) {
            cout << i - off << endl;
            return 0;
        }
        else if (i + off <= 100 && arr[i + off] == 0) {
            cout << i + off << endl;
            return 0;
        }
        off++;
    }
    return 0;
}