(Relax 数论 1.17)POJ 3101 Astronomy(分数的最小公倍数)

2个星球周期为a,b。则相差半周的长度为a*b/(2*abs(a-b)),对于n个只需求这n个

分数的最小公倍数即可！

公式：

分数的最小公倍数 = 分子的最小公倍数/分母的最大公约数

由于涉及到大数所以用java写的方便！

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class POJ_3101 {

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

		int n = scanner.nextInt();
		int an[] = new int[n];
		int a[] = new int[n];
		int b[] = new int[n];

		int i;
		for (i = 0; i < n; ++i) {
			an[i] = scanner.nextInt();
		}

		Arrays.sort(an);

		int j;
		for (i = 1, j = 1; i < n; ++i) {//去重
			if (an[i] != a[i - 1]) {
				an[j++] = an[i];
			}
		}

		int k;
		for (i = 1, k = 0; i < j; ++i) {//化简
			a[k] = (an[i] - an[i - 1]) * 2;
			b[k] = (an[i] * an[i - 1]);

			int t = gcd(a[k], b[k]);

			a[k] /= t;
			b[k++] /= t;
		}

		BigInteger ans1 = BigInteger.valueOf(a[0]);//****??
		BigInteger ans2 = BigInteger.valueOf(b[0]);
		BigInteger ans;
		for (i = 1; i < k; ++i) {
			ans1 = ans1.gcd(BigInteger.valueOf(a[i]));
			ans = ans2.multiply(BigInteger.valueOf(b[i]));
			ans2 = ans.divide(ans2.gcd(BigInteger.valueOf(b[i])));
		}

		ans = ans1.gcd(ans2);//化简输出...
		System.out.println(ans2.divide(ans) + " " + ans1.divide(ans));

	}

	public static int gcd(int a, int b) {
		int t;
		if (a < b) {
			t = a;
			a = b;
			b = t;
		}
		while (true) {
			if (b == 0)
				break;
			t = a;
			a = b;
			b = t % b;
		}
		return a;
	}
}