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<bits/stdc++.h>
using namespace std;
typedef long long int ll;

int main()
{

int x,n;
cin>>x>>n;

bool exist[101];
for(int index = 0 ;index<100;index++)
exist[index]= false;

for(int index = 0 ;index<n;index ++)
{
    int in;
    cin>>in;
    exist[in] = true;
}
int left = x,right = x;

while((left>=0||right<=100)&&exist[left]&&exist[right])
{
    if(left>0)left--;
    if(right<100)
right++;
}
if(!exist[left])
cout<<left<<endl;
else
cout<<right<<endl;

    return 0;
}