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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
 

long long euclidGcd(long long inputNum1, long long inputNum2)
{
	long long minNumGcd = 0;
	long long maxNumGcd = 0;
	long long euclideanAlgorithmGcd = 1;	//ユークリッドの互除法によりGcd(greatest common divisor)最大公約数

	//--
	//数値の大小関係(--> if,else)
	//ユークリッドの互除法により最大公約数を求める。euclideanAlgrithimGcdを0にしないように、while実行されない(--> for(交換))
	//最大公約数を返す(--> return)
	if (inputNum1 > inputNum2)
	{
		maxNumGcd = inputNum1;	
		minNumGcd = inputNum2;
	}
	else
	{
		maxNumGcd = inputNum2;
		minNumGcd = inputNum1;
	}


	while (euclideanAlgorithmGcd != 0)
	{
		//見にくい
		euclideanAlgorithmGcd = maxNumGcd % minNumGcd;	// aaa = bbb % ccc;
		maxNumGcd = minNumGcd;							// bbb = ccc;
		minNumGcd = euclideanAlgorithmGcd;				// ccc = aaa;
	}

	
	euclideanAlgorithmGcd = maxNumGcd;
	return euclideanAlgorithmGcd;
	//--
}

long long euclidLcm(long long inputNum1, long long inputNum2, long long inputNumGcd)
{
	long long euclideanAlgorithmLcm = 0;	

	//--
	//最小公倍数の公式　＝ inputNum1 * inputNum2 /euclideanAlgorithmGcd(最大公約数);(--> 公式
	euclideanAlgorithmLcm = inputNum1 * (inputNum2 / inputNumGcd);
	//--
	return euclideanAlgorithmLcm;

}


int main(void)
{
	int input = 0;
	long long answer = 0;
	long long euclideanAlgorithmGcd = 0;

	long long t[100] = {};

	//数字を受け取る(--> scanf N個入力される予定)
	//数値の受け取り(--> for(scanf))
	//ユークリッドの互除法により、最小公倍数を計算(--> for(euclidGcd,euclidLcm))
	//答えの表示(--> printf)
	//-
	scanf("%d", &input);

	for (int i = 0; i < input;++i)
	{
		scanf("%lld",&t[i]);
	}

	answer = t[0];

	for (int i = 0; i < input - 1; i++) {
		euclideanAlgorithmGcd = euclidGcd(answer, t[i + 1]);
		answer = euclidLcm(answer, t[i + 1], euclideanAlgorithmGcd);
	}
	
	printf("%lld", answer);
	//--

	return 0;

}