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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
 
int main() {
  int X,N;
  cin >> X >> N;
  vector<int> hairetu(N);
  for(int i=0;i<N;i++){
    cin >> hairetu.at(i);
  }
  
  bitset<100> hai;
  for(int j=0;j<N;j++){
    hai.set(hairetu.at(j)-1,1);
  }
  
  int kekka = 1000;
  int count1=0;
  int count2=0;
  
  for(int k=X;k<101;k++){
    if(hai.test(k-1) == 0){
      kekka = k;
      break;
    }
    count1++;
  }
  
  for(int l=X;l>=0;l--){
    if(hai.test(l-1) == 0){
      if(count1<count2){
        break;
      }
      else if(count1>=count2){
        kekka = l;
        break;
      }
    }
    count2++;
  }
 
  if(kekka==1000){
    if(X>=51){
      cout << 101 <<endl;
    }
    else{
      cout << 0 <<endl;
    }
  }
  else{
    cout << kekka <<endl;
  }

}