poj 3101 Astronomy

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!

 

 1 import java.math.*;
 2 import java.util.*;
 3 
 4 public class Main {
 5     public static void main(String arg[]){
 6         Scanner cin = new Scanner(System.in);
 7         int n = cin.nextInt();
 8         int an[] = new int[n];
 9         int a[] = new int[n];
10         int b[] = new int[n];
11         int i,j,k;
12         for (i=0;i<n;i++)
13             an[i] = cin.nextInt();
14         Arrays.sort(an);
15         for (i=1,j=1;i<n;i++)
16             if (an[i]!=an[i-1])
17                 an[j++] = an[i];
18         for (i=1,k=0;i<j;i++){
19             a[k] = (an[i] - an[i-1])*2;
20             b[k] = an[i] * an[i-1];
21             int t = gcd(a[k],b[k]);
22             a[k] /= t;
23             b[k++] /= t;
24         }
25         BigInteger ans1 = BigInteger.valueOf(a[0]),ans2 = BigInteger.valueOf(b[0]),ans;
26         for (i=1;i<k;i++){
27             ans1 = ans1.gcd(BigInteger.valueOf(a[i]));
28             ans = ans2.multiply(BigInteger.valueOf(b[i]));
29             ans2 = ans.divide(ans2.gcd(BigInteger.valueOf(b[i])));
30         }
31         ans = ans2.gcd(ans1);
32         System.out.println(ans2.divide(ans)+" "+ans1.divide(ans));
33     }
34     public static int gcd(int a,int b){
35         int t;
36         if (a<b){
37             t = a;
38             a = b;
39             b = t;
40         }
41         while(true){
42             if (b==0) break;
43             t = a;
44             a = b;
45             b = t%b;
46         }
47         return a;
48     }
49 }
View Code

 

 

 

posted @ 2013-07-29 19:25  _随心所欲_  阅读(240)  评论(0编辑  收藏  举报