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;
  int con=1000,ans=0;
  scanf("%d %d", &x,&n);
  if(n==0){
    printf("%d", x);
  }else{
    int p[n];
    for(int k=0;k<n;k++){
      scanf("%d", &p[k]);
    }
    int count=0,count1=0;
    for(int i=0;i<=101;i++){
      for(int j=0;j<n;j++){
        if((x-i)!=p[j]){
          //printf("%d %d\n", p[j], x-i);
          count++;
        }
        if((x+i)!=p[j]){
          count1++;
        }
      }
      //printf("%d %d\n", count, count1);
      if(count==n||count1==n){
        if(count==n&&count1==n){
          ans = x-i;
          break;
        }
        else if(count==n){
          ans = x-i;
          break;
        }else{
          ans = x + i;
          break;
        }
      }
      count = 0;count1 = 0;
    }
    printf("%d", ans);
  }
  return 0;
}