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>
#include<stdlib.h>
//絶対値小さい順にソート
int c[102];
void merge(int i,int j,int *s){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d]=s[b];
        b++;
      }
      else if(j<b){
        c[d]=s[a];
        a++;
      }
      else if(abs(s[a])<abs(s[b])){
        c[d]=s[a];
        a++;
      }
      else if(abs(s[a])==abs(s[b]) && s[a]<s[b]){
        c[d]=s[a];
        a++;
      }
      else{
        c[d]=s[b];
        b++;
      }
    }
    for(d=i;d<=j;d++)
      s[d]=c[d];
  }
}

int main(){
  int x,n,i;
  scanf("%d%d",&x,&n);
  int p[n+1];
  if(n==0){
    printf("%d\n",x);
    return 0;
  }
  for(i=0;i<n;i++){
    scanf("%d",&p[i]);
    p[i]=p[i]-x;
  }
  merge(0,n-1,p);
  if(p[0]!=0){
    printf("%d\n",x);
    return 0;
  }
  else{
    for(i=1;i<n/2;i++){
      if(p[2*i-1]!=-i){
        printf("%d\n",x-i);
        return 0;
      }
      else if(p[2*i]!=i || 2*i==n){
        printf("%d\n",x+i);
        return 0;
      }
    }
  }
  printf("%d\n",x-n/2-1);
  return 0;
}