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<stdlib.h>
int main(){
  int i,x,n;
  int p,q,r;
  int min,max;
  int tmp;
  int ff=0;
  int a[109]={999};
  int b[109] = {999};
  scanf("%d %d",&x,&n);
  
  if(n==0){
    printf("%d",x);
  	return 0;
  }
  
  for(i=0;i<=n;i++){
    scanf("%d ",&a[i]);
  }
  
  b[0]=x;
  for(i=1;i<=n;i+=2){
   b[i]= x-((1+i)/2);
   b[i+1]=x+((1+i)/2);
  }
  
  for(i=0;i<101;i++){
  	for(p=0;p<n;p++){
      if(b[i]==a[p]){
       ff=1; 
        p=n;
      }
    }
    if(ff==0){
      printf("%d",b[i]);
      return 0;
    }
    ff=0;
  }
  
 return 0; 
}