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: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class Solve{
    static public int mod = 1000000007;
    public static void Main(){
        // 方針
        // 
        int n = rint();
        long lcm = rlong();
        for(int i=0;i<n-1;i++){
            long temp = rlong();
            if(temp > 1000000000000000000/2){
                lcm = temp;
            }else{
                lcm = LCM(lcm,temp);
            }
        }
        WriteLine(lcm);
        
        
        
        
    }
    static int binarysearchd(int[] a,int b){
        int leng = a.Length;
        int min = 0;
        int max = leng-1;
        while(max - min > 1){
            int mid = (min+max)/2;
            if(a[mid] < b){
                min = mid;
            }else{
                max = mid;
            }
        }
        if(a[0] >= b){
            return 0;
        }else if(a[leng-1] < b){
            return leng;
        }else{
            if(a[max] < b){
                return max+1;
            }
            return min+1;
        }
    }
    static int binarysearchu(int[] a,int b){
        int leng = a.Length;
        int min = 0;
        int max = leng-1;
        while(max - min > 1){
            int mid = (min+max)/2;
            if(a[mid] < b){
                min = mid;
            }else{
                max = mid;
            }
        }
        if(a[0] > b){
            return leng;
        }else if(a[leng-1] <= b){
            return 0;
        }else{
            if(a[max] <= b){
                return leng-(max+1);
            }
            return leng-(min+1);
        }
    }
    // 1からaまでの数字並べ替え全通り返すやつ
    static List<string> permutation(int a){var list = new List<string>();if(a == 1){list.Add("1");return list;}else{foreach(var k in permutation(a-1)){var ch = k.ToCharArray();for(int i=0;i<=ch.Length;i++){string intemp = "";if(i == 0){intemp += a.ToString();}for(int j=0;j<ch.Length;j++){intemp += ch[j].ToString();if(i == j+1){intemp += a.ToString();}}list.Add(intemp);}}}return list;}
    public static void swap(ref int a,ref int b){int temp = a;a= b;b = temp;}
    static void charswap(ref char a,ref char b){char temp = a;a= b;b = temp;}
    static int ncr(int n,int r){r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    static int square2(int a,int b){long output = 1;var list = new List<long>();int sh = 1;long n = a;list.Add(a);while(sh < b){sh *= 2;n = n*n%mod;list.Add(n);}for(int i=list.Count-1;i>=0;i--){if(b > sh){b -= sh;sh /= 2;output = output*list[i]%mod;}}return (int)output;}
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static char[] chara(){ string[] a=stra();string b="";for(int i=0;i<a.Length;i++){b+=a[i];}return b.ToCharArray();}
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long divideup(long divided,long divid){ long temp_divide = divided/divid; if(divided % divid > 0){ temp_divide++; } return temp_divide; }
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){
        long gcd = GCD(a,b);
        if(a%gcd == 0){
            return a/gcd * b;
        }else if(b%gcd == 0){
            return b/gcd * a;
        }else{
            return a * b / GCD(a,b); 
        }
    }
}