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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.PrintWriter
import java.lang.StrictMath.*
import kotlin.comparisons.compareBy

var out: PrintWriter = PrintWriter(System.out)  // これを使って出力すると早い
fun main(args: Array<String>) {
    try {
        val n = ReadLib.readInt()
        var currentTime = ReadLib.readLong()
        for (i in 1 until n) {
            currentTime = MyMath.lcm(currentTime, ReadLib.readLong())
        }
        out.println(currentTime)
    } finally {
        out.flush()   // 消さない
    }
}

/**
 * 読み取り
 */
class ReadLib {
    companion object {
        private var reader: BufferedReader = System.`in`.bufferedReader()

        fun reset() {
            reader = System.`in`.bufferedReader()
        }

        fun readInt(): Int {
            return Integer.parseInt(reader.readLine())
        }

        // Int
        fun readIntArray(delimiters: String = " "): IntArray {
            return reader
                .readLine()
                .split(delimiters)
                .map { s -> Integer.parseInt(s) }
                .toIntArray()
        }

        fun readNTimesInt(n: Int): IntArray {
            return (1..n).map { readInt() }.toIntArray()
        }

        fun readNTimesIntArray(n: Int): Array<IntArray> {
            return (1..n).map { readIntArray() }.toTypedArray()
        }

        // long
        // TODO 別の方法を考える
        fun readLong(): Long {
            return reader.readLine().toLong()
        }

        fun readLongArray(delimiters: String = " "): LongArray {
            return reader.readLine()
                .split(delimiters)
                .map { s -> s.toLong() }
                .toLongArray()
        }

        fun readNTimesLong(n: Int): LongArray {
            return (1..n).map { readLong() }.toLongArray()
        }

        fun readString(): String {
            return reader.readLine()
        }

        fun readStringArray(delimiters: String = " "): Array<String> {
            return readString().split(delimiters).toTypedArray()
        }

        fun readNTimesCharArray(n: Int): Array<CharArray> {
            return (1..n).map { readString().toCharArray() }.toTypedArray()
        }

        fun readNTimesString(n: Int): Array<String> {
            return (1..n).map { readString() }.toTypedArray()
        }

    }
}

/**
 * 要素とその個数を出力
 */
fun <T> Array<T>.ofCountMap(): Map<T, Int> {
    return this.groupBy { it }.mapValues { it.value.count() }
}

/**
 * 文字とその個数を出力
 */
fun String.ofCountMap(): Map<Char, Int> {
    return this.toCharArray().ofCountMap()
}

/**
 * 文字とその個数を出力
 */
fun CharArray.ofCountMap(): Map<Char, Int> {
    return this.groupBy { it }.mapValues { it.value.count() }
}

/**
 * Combination, nCk mod p を求める
 * https://ja.wikipedia.org/wiki/%E3%83%A2%E3%82%B8%E3%83%A5%E3%83%A9%E9%80%86%E6%95%B0
 * http://drken1215.hatenablog.com/entry/2018/06/08/210000
 * @param maxN 最大のn
 * @param mod 素数p
 */
class ModCombination(private val maxN: Int, private val mod: Int = 1000000007) {
    private val fac: IntArray = IntArray(maxN)
    private val finv: IntArray = IntArray(maxN)

    init {
        val inv: IntArray = IntArray(maxN)
        fac[0] = 1
        fac[1] = 1

        inv[0] = 1
        inv[1] = 1
        for (i in 2..maxN) {
            fac[i] = fac[i - 1] * i % mod   // 累積
            // mod % i < i
            inv[i] = mod - ((inv[mod % i].toLong()) * (mod / i) % mod).toInt()
            finv[i] = finv[i - 1] * inv[i] % mod
        }
    }

    /**
     * nCk mod p
     */
    fun Calc(n: Int, k: Int): Int {
        if (n < k || n < 0 || k < 0) {
            return 0
        }

        return ((fac[n].toLong() * finv[k] % mod) * finv[n - k] % mod).toInt()
    }
}

class MyMath {
    companion object {
        // 最大公約数
        fun gcd(m: Long, n: Long): Long {
            var M = m
            var N = n
            while (M % N > 0) {
                val pN = N
                N = M % N
                M = pN
            }

            return N
        }

        fun lcm(m: Long, n: Long): Long {
            return (m / gcd(m, n)) * n
        }

        // 素因数分解（試し割り）
        // http://sucrose.hatenablog.com/entry/2014/10/10/235805
        fun primeFactorization(n: Long): Map<Long, Long> {
            val primes = enumeratePrime(ceil(sqrt(n.toDouble())).toInt())
                .map { it.toLong() }
                .toLongArray()
            primes.sort()

            val counter = mutableMapOf<Long, Long>()
            var cn = n
            for (p in primes) {
                if (cn < p) {
                    break
                }

                while (cn % p == 0L) {
                    counter[p] = counter.getOrElse(p, { 0 }) + 1
                    cn /= p
                }
            }

            if (cn > 1) {
                counter[cn] = 1
            }

            return counter
        }

        // Nまでの素数列挙
        fun enumeratePrime(n: Int): IntArray {
            val isPrime = BooleanArray(n + 1) { true }
            val primes = mutableListOf<Int>()
            for (i in 2..n) {
                if (!isPrime[i]) {
                    continue
                }

                primes.add(i)
                for (j in i..n step i) {
                    isPrime[i] = false
                }
            }

            return primes.toIntArray()
        }
    }
}