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 <stdio.h>
#define max(a, b) ((a) > (b) ? (a) : (b))


int main(void){
    int f, i,ii,x , n,a[105], q[105], tmp;
    scanf("%d%d", &x, &n);
    if (n == 0)
    {
        printf("%d", x);
        return (0);
    }
    for (i = 0; i < n; i++)
    {
        scanf("%d", q + i);
        a[i] =  max(q[i] - x, x - q[i]);
    }
    f = 0;
    while (f == 0)
    {
        f = 1;
        for (i = 0; i < n - 1; i++)
        {
            if (a[i] > a[i + 1])
            {
                f = 0;
                a[i] ^= a[i + 1];
                a[i + 1] ^= a[i];
                a[i] ^= a[i + 1];
            }
        }
    }
       if (a[0] > 0)
    {
        printf("%d", x);
        return (0);
    }
       
       
       f = 2;
       
     for (i = 1; i < n - 1; i++)
    {
        if (a[i + 1] != a[i])
        {
            tmp = a[i];
            f = 3;
            break;
        }
        i++;
        if (a[i + 1] > a[i] + 1)
        {
            printf("%d", x - a[i] - 1);
            return (0);
        }
    } 
       if (a[1] > 1)
    {
        printf("%d", x - 1);
        return (0);
    }
    
    if (f == 3)
    {
        for (i = 0; i < n; i++)
        {
            if (max(q[i] - x, x - q[i]) == tmp)
            {
                printf("%d", x + x - q[i]);
                return (0);
            }
        }
    }
    
    if (n % 2 == 0)
    {
        printf("%d", x - (n / 2) - 1);
        return (0);
    }
    
    for (i = 0; i < n; i++)
    {
        if (q[i] == x + 1 + (n / 2))
        {
            printf("%d", x - n / 2 - 1);
            return (0);
        }
    }
    printf("%d", x + n / 2 + 1);
    return(0);
}