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 Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use proconio::{fastout, input, marker::*};

#[fastout]
fn main() {
    input! {
        x: usize,
        n: usize,
        p: [usize; n],
    }
    let mut dp = vec![true; 103];
    for i in 0..n {
        dp[p[i]] = false;
    }

    for i in 0..102 {
        if i <= x && dp[x - i] {
            println!("{}", x - i);
            break;
        }

        if dp[x + i] {
            println!("{}", x + i);
            break;
        }
    }
}

// #[allow(unused_imports)]
// use proconio::{fastout, input, marker::*};

// #[fastout]
// fn main() {
//     input! {
//         x: usize,
//         n: usize,
//         p: [usize; n],
//     }
//     let mut ans = if n == 0 { x } else { 1000 };

//     for i in (0..x).rev() {
//         if 0 < p.iter().filter(|&&l| l == i).count() {
//             continue;
//         }
//         ans = i;
//         break;
//     }

//     for i in x..101+1 {
//         if 0 < p.iter().filter(|&&l| l == i).count() {
//             continue;
//         }
//         if ans <= x {
//             if x - ans == i - x {
//                 break;
//             }

//             if x - ans < i - x {
//                 break;
//             }
//         }

//         ans = i;
//         break;
//     }

//     println!("{}", ans);
// }