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 <math.h>
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define PI acos(-1)
typedef unsigned long long int ull;
typedef long long int ll;
typedef long double ld;

int compare_base(const void *b, const void *f) {
  return (*(int *)b - *(int *)f);
}

ll find(ll n, ll *num) {
  ll result;
  result = 1;
  for (ll i = 0; i < n; i++) {
    if (num[i] == n) {
      result *= 0;
    } else {
      result *= 1;
    }
  }
  if (result) {
    return (0);
  } else {
    return (1);
  }
}

int main(void) {
  /* 標準入力フラグ */
  int   scan;

  /* 数入力 */
  ll    answer;
  ll    x;
  ll    n; 
  ll    *p;
  ll    *diff;

  answer = 0;
  scan = scanf("%lld", &x);
  scan = scanf("%lld", &n);
  p = (ll *)malloc(sizeof(ll) * (n + 1));
  for (ll i = 0; i < n; i++) {
    scan = scanf("%lld", &p[i]);
  }

  /* プログラムを書く */
  qsort(p, n, sizeof(ll), compare_base);
    
  if (n == 0) {
    answer = x;
    printf("%lld", answer);
    return (0);
  }

  diff = (ll *)malloc(sizeof(ll) * (n + 1));
  for (ll i = 0; i < n; i++) {
    if (x - p[i] > 0) {
      diff[i] = x - p[i];
    } else {
      diff[i] = (-1) * (x - p[i]);
    }
  }
  qsort(diff, n, sizeof(ll), compare_base);
  
  if (diff[0] != 0) {
    answer = x;
    printf("%lld", answer);
    return (0);
  }

  if (n == 1 && x == p[0]) {
    answer = x - 1;
    printf("%lld", answer);
    return (0);
  }


  // -1 と +1 から順番に見て p にないやつを出力
  for (ll i = 0; i < n; i++) {
    answer = x - i;
    if (!find(answer, p)) {
      printf("%lld", answer);
      return (0);
    } 
    answer = x + i;
    if (!find(answer, p)) {
      printf("%lld", answer);
      return (0);
    } 
  }

  return (0);
}