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 <stdio.h>
#include <math.h>

int main(){
    int x, n;
    scanf("%d%d", &x, &n);

    if(n == 0)
        printf("%d", x);
    else{
        int p[101],pp[101];
        int menor_num = 101;
        int maior_num = 0;
        int cont = 0;

        for(int i = 0; i < n; ++i){
            scanf("%d", &p[i]);
        }

        for(int j = 0; j < n; ++j){
            if(p[j] <= menor_num)
                menor_num = p[j];
            if(p[j] >= maior_num)
                maior_num = p[j];
        }

        for(int t = 0; t < 101; ++t){
            int c = 0;
            for(int s = 0; s < n; ++s){
                if(t == p[s])
                    c = -1;
            }
            if(c == 0){
                pp[cont] = t;
                ++cont;
            }
        }

        int resp = 101;

        for(int k = 0; k < cont; ++k){
             if(abs(pp[k] - x) <= abs(resp - x)){
                if(abs(pp[k] - x) == abs(resp - x)){
                    if(pp[k] < resp)
                        resp = pp[k];
                }
                else
                    resp = pp[k];
             }
        }
        
    
        printf("%d", resp);
    }

    return 0;
}