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 <string.h>
#include <stdio.h>
#include <stdlib.h>
#include<math.h>
#define ll long long
#define pai 3.1415926535897932384626433832795028841971693

int main(){
  int x,n,min=100001,ans=0,k;
  scanf("%d %d",&x,&n);
  int p[n],a[n];
  for(int i=0;i<n;i++){
    scanf("%d",&p[i]);
  }

  for(int i=-1;i<201;i++){
    for(int j=0;j<n;j++){
      if(i==p[j]){
        k=1;
        break;
      }
      k=0;
    }
    if(k==1){
      continue;
    }
    else{
      if(min>abs(x-i)){
        min=abs(x-i);
        ans=i;
      }
      else if(min==abs(x-i)){
        if(ans>i){
          ans=i;
        }
      }
    }
  }
  printf("%d",ans);
   return 0;
}