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<bits/stdc++.h>
#define lln long long int
#define llu unsigned lln
#define sc(n) scanf("%d",&n);
#define scl(n) scanf("%lld",&n);
#define scd(n) scanf("%lf",&n);
#define pf(res) printf("%d\n",res);
#define pfl(res) printf("%lld\n",res);
#define pfd(res) printf("%lf\n",res);
#define maxii 100005
using namespace std;
typedef pair<int,int> pii;
typedef pair<lln,lln> pll;
typedef vector<int> vi;
typedef vector<lln> vl;
typedef vector<pii> vii;
typedef vector<pll> vll;

int arr[105];
int arr2[105];

int main()
{
    int x,n;
    sc(x)
    sc(n)
    if(n==0)
    {
        cout<<x;
        return 0;
    }
    for(int i=0;i<n;i++)
    {
        sc(arr[i])
        arr2[arr[i]]++;
    }

   sort(arr,arr+n);
   int res1,r,l,res2;
   if(arr2[x]==0)
   {
       cout<<x;
       return 0;
   }
   for(int i=x+1;i<=101;i++)
   {
       if(arr2[i]==0)
       {
           r=i;
           res1=i-x;
           break;
       }
   }
   for(int i=x-1;i>=0;i--)
   {
       if(arr2[i]==0)
       {
           l=i;
           res2=x-i;
           break;
       }
   }
   //cout<<res1<<" "<< res2<<endl;;
   if(res1>=res2)
   {
       cout<<min(l,r);
   }
   else if(res1<res2)
   {
       cout<<r;
   }
   //else cout<<l;
}