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

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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;}}