Problem:
Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.

Initially, the hand of every clock stands still, pointing directly upward.

Now, Dolphin starts all the clocks simultaneously.

In how many seconds will the hand of every clock point directly upward again?

Constraints

1≤N≤100

1≤T_i≤10^{18}

All input values are integers.

The correct answer is at most 10^{18} seconds.

Input

Input is given from Standard Input in the following format:

N
T_1
:
T_N

Output

Print the number of seconds after which the hand of every clock point directly upward again.

Sample Input 1

2
2
3

Sample Output 1

6

We have two clocks. The time when the hand of each clock points upward is as follows:

Clock 1: 2, 4, 6, ... seconds after the beginning

Clock 2: 3, 6, 9, ... seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

Sample Input 2

5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

1000000000000000000

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            //[summary]C - Multiple Clocks
            int n = int.Parse(Console.ReadLine());
            var t = ReadLines(n);

            long lcm = t[0];
            var math = new MyMath();

            for(int i = 1; i < n; i++)
            {
                long newLcm = GetLcm(lcm, t[i]);

                if (newLcm == 0)
                {
                    lcm = math.GetLcm(lcm, t[i]);
                }
                else
                {
                    lcm = newLcm;
                }
            }

            Console.WriteLine(lcm);
        }

        static long GetLcm(long a, long b)
        {
            if (a >= b)
            {
                if (a % b == 0)
                {
                    return a;
                }
            }
            else
            {
                if (b % a == 0)
                {
                    return b;
                }
            }

            return 0;
        }

        static List<long> ReadLines(long count = 0)
        {
            //[summary]コンソールの値を読み込む
            var lines = new List<long>();
            string s;

            //読み込む行数が未指定のときは、1行目の値を"count"とみなす
            if (count == 0)
            {
                s = Console.ReadLine().TrimEnd();
                count = long.Parse(s);
            }

            for (long i = 0; i < count; i++)
            {
                s = Console.ReadLine().TrimEnd();
                lines.Add(long.Parse(s));
            }

            return lines;
        }
    }

    class MyMath
    {
        public long GetGcd(long a, long b)
        {
            //[summary]a,b2つの数の最大公約数を求める
            //以下リンクを参考に実装
            //https://qiita.com/gushwell/items/e9614b4ac2bea3fc6486

            if (a < b)
            {
                //引数を入れ替えて自分を呼び出す
                return GetGcd(b, a);
            }
            else
            {
                while (b > 0)
                {
                    long remainder = a % b;

                    a = b;
                    b = remainder;
                }

                return a;
            }
        }

        public long GetLcm(long a, long b)
        {
            //[summary]a,b2つの数の最小公倍数を求める
            //Lcm: least Common Multiple
            //以下リンクを参考に実装
            //https://qiita.com/gushwell/items/f08d0e71fa0480dbb396

            long gcd = GetGcd(a, b);
            return a * b / gcd;
        }
    }
}