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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.util.*;

public class Main {

	public static void main(String[] args) {

		// 入力
		Scanner sc = new Scanner(System.in);
		Integer target = new Integer(Integer.parseInt(sc.next()));
		int n = Integer.parseInt(sc.next()); // 禁止リストの長さ

		// 禁止リストがない場合は目標がそのまま答え
		if (n == 0) {
			System.out.println(target);
			return;
		}

		// 入力（続き）
		Integer x[] = new Integer[n];
		for (int i = 0; i < n; i++) {
			x[i] = new Integer(Integer.parseInt(sc.next()));
		}

		// 目標が禁止されていない場合は目標がそのまま答え
		if (!Arrays.asList(x).contains(target)) {
			System.out.println(target);
			return;
		}

		// ソート
		Arrays.sort(x);
		// System.out.println(Arrays.toString(x));

		// 目標は禁止リストの何番目？
		int index = Arrays.asList(x).indexOf(target);

		// 左方向に答えを探していく
		int now = target.intValue();
		int minus = 0;
		while (now <= x[x.length - 1].intValue()) {

			// １歩く
			minus++;

			// そこは？
			now = target.intValue() - minus;
			// System.out.println("ここは" + now);

			// 禁止リストにない数に到達したらwhileループを抜ける
			if (!Arrays.asList(x).contains(new Integer(now))) {
				break;
			}
		}

		// 右方向に答えを探していく
		now = target.intValue();
		int plus = 0;
		while (x[0].intValue() <= now) {

			// １歩く
			plus++;

			// そこは？
			now = target.intValue() + plus;
			// System.out.println("ここは" + now);

			// 禁止リストにない数に到達したらwhileループを抜ける
			if (!Arrays.asList(x).contains(new Integer(now))) {
				break;
			}
		}

		// System.out.println(minus + " " + target + " " + plus);

		// 最も近いほうを出力
		if (minus <= plus) {
			System.out.println(target.intValue() - minus);
		} else {
			System.out.println(target.intValue() + plus);
		}

	}
}