Lowest Common Multiple Plus

2 4 6
3 2 5 7

12
70

分析：两个数的最小公倍数 lcm( x , y ) = x * y / gcd( x , y )。其中 gcd() 是这两个数的最大公约数，可以采用“辗转相除法”求解。所以这题的关键是求最大公约数。

import java.util.Scanner;

public class Main {

	// 最大公约数
	static long gcd(long a, long b) {
		long remainder = a % b;
		while (remainder != 0) {
			a = b;
			b = remainder;
			remainder = a % b;
		}
		return b;
	}

	// 最小公倍数
	static long lcm(long a, long b) {
		return a * b / gcd(a, b);
	}

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		while (scanner.hasNext()) {
			int n = scanner.nextInt();

			long temp = 1;
			for (long i = 0; i < n; i++) {
				long num = scanner.nextLong();
				temp = lcm(temp, num);
			}

			System.out.println(temp);
		}
	}
}

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[hdu 2028] Lowest Common Multiple Plus

原文：http://blog.csdn.net/u011506951/article/details/23618581