HDOJ1019 Least Common Multiple（求多个数的最小公倍数）

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56425    Accepted Submission(s): 21492

Problem Description

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

 

Input

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

 

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

 

Sample Input

2
3 5 7 15
6 4 10296 936 1287 792 1

 

Sample Output

105
10296

 

求一组数的最小公倍数，就相当于求前面的两个的最小公倍数lcm，然后再拿第三个数来和lcm 求最小公倍数.......

最后求到的就是一组数的最小公倍数。

怎么求两个数的最小公倍数呢？

可以先得到两个数的最大一方，拿他的倍数（1,2,3.....）去与小的值取余，除得尽就是得到最小公倍数了。

下面AC代码：

import java.util.Scanner;

public class Main{
	private static Scanner scanner;

	public static void main(String[] args) {
		scanner = new Scanner(System.in);
		int cases = scanner.nextInt();
		while (cases-- > 0) {
			int n = scanner.nextInt();
			int arr[] = new int[n];
			for (int i = 0; i < arr.length; i++) {
				arr[i] = scanner.nextInt();
			}
			int lcm = 1;
			for (int j = 0; j < arr.length; j++) {
				lcm = findlcm(lcm, arr[j]);
			}
			System.out.println(lcm);
		}
	}

	private static int findlcm(int lcm, int i) {
		if (lcm < i) {
			int t = lcm;
			lcm = i;
			i = t;
		}
		int n = 1;
		while (lcm * n % i != 0) {
			n++;
		}
		return lcm*n;
	}
}