Problem:
Score : 300 points

Problem Statement

There are N monsters, numbered 1, 2, ..., N.

Initially, the health of Monster i is A_i.

Below, a monster with at least 1 health is called alive.

Until there is only one alive monster, the following is repeated:

A random alive monster attacks another random alive monster.

As a result, the health of the monster attacked is reduced by the amount equal to the current health of the monster attacking.

Find the minimum possible final health of the last monster alive.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible final health of the last monster alive.

Sample Input 1

4
2 10 8 40

Sample Output 1

2

When only the first monster keeps on attacking, the final health of the last monster will be 2, which is minimum.

Sample Input 2

4
5 13 8 1000000000

Sample Output 2

1

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

1000000000

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.stream.*;
import java.util.function.*;

public class Main {
    public static void main(String[] args) throws Exception {
        System.out.println(new Main().solve());
    }
    
    // パラメタ渡しもちゃんとクラス作るのも面倒なので手抜き
    // 必要な値はここで全部取ってしまう
    Scanner sc = new Scanner(System.in);
    int n = Integer.parseInt(sc.next());
    {sc.nextLine();}
    // int[] a = IntStream.range(0, n).map(i -> Integer.parseInt(sc.next())).toArray();
    // // ソートしておく
    List<Integer> a = IntStream.range(0, n)
                                .map(i -> Integer.parseInt(sc.next()))
                                .sorted()
                                .boxed().collect(Collectors.toList());
    
    int solve() {
        // 最小のan以下にはなる
        
        // 最大でもmin(a) これは他から他を引いた余りがmin(a)以下の時
        
        // なるべく小さい差を作ってそれで余りを取る？
        
        // 初期状態で最小 (これより大きい答えにはならない)
        int min = Collections.min(a);
        
        // 均す方法が必要
        // 499 501 1001 なら1になる
        // 499 501 1502 でも1になる
        // 各数を任意の回数繰り返す組み合わせ？
        // 基本的にはmodを取るのだろうが
        Set<Integer> modSet = new TreeSet<>();
        for(int i: a) {
            for(int j: a) {
                int mod = i % j;
                if(mod != 0) {
                    modSet.add(mod);
                }
            }
        }
        
        if(modSet.isEmpty()) {
            return min;
        }
        
        List<Integer> modList = new ArrayList<>(modSet);
        if(modList.get(0) == 1) {
            return 1;
        }
        int prev = 0;
        int diff = Integer.MAX_VALUE;
        
        // これを再帰とかも必要な気がする
        // mod の最小差？ 嘘くさいがこれで一回試す
        for(int mod: modList) {
            diff = Math.min(diff, mod - prev);
            prev = mod;
        }
        
        for(int i: a) {
            for(int mod: modList) {
                int mm = i % mod;
                if(mm == 0) {
                    continue;
                }
                if(mm == 1) {
                    return 1;
                }
                diff = Math.min(diff, mm);
            }
        }
        
        // 嘘くさい
        return Math.min(diff, min);
    }
    
}