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)

//0614 C問題
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int main(void) {

  int x, n, l, r, i=0, j=0, cl, cr;

  scanf("%d%d", &x, &n);
  if ((x<1)||(100<x)||(n<0)||(100<n)) {
    return -1;
  }
  if (n==0) {
    printf("\n%d\n", x);
    return 0;
  }
  int p[n];

  while (i<n) {
    if (scanf("%d", &p[i])==0) {
      printf("%d\n", x);
      return 0;
    }
    if ((p[i]<1)||(100<p[i])) {
      return -1;
    }
    i++;
  }

  while (1) {
    cl = 0;
    cr = 0;
    l = x-j;
    r = x+j;
    for (i = 0; i < n; i++) {
      if (l==p[i]) {
        cl += 1;
      }
      if (r==p[i]) {
        cr += 1;
      }
    }
    if (cl==0) {
      printf("%d\n", l);
      return 0;
    }
    if (cr==0) {
      printf("%d\n", r);
      return 0;
    }
    j++;
  }

  return 0;
}