3101 Astronomy Java大数，分数的最小公倍数

最新推荐文章于 2019-09-19 17:31:29 发布

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本文介绍了一种计算星系中行星连续两次排列成直线事件间隔时间的方法。通过输入行星的数量及各自绕恒星运行的周期，利用算法找到这一特定天文现象发生的最简分数表示周期。

Astronomy

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 2704		Accepted: 553

Description

There are n planets in the planetary system of star X. They orbit star X in circular orbits located in the same plane. Their tangent velocities are constant. Directions of orbiting of all planets are the same.

Sometimes the event happens in this planetary system which is called planet parade. It is the moment when all planets and star X are located on the same straight line.

Your task is to find the length of the time interval between two consecutive planet parades.

Input

The first line of the input file contains n — the number of planets (2 ≤ n ≤ 1 000).

Second line contains n integer numbers t_i — the orbiting periods of planets (1 ≤ t_i ≤ 10 000). Not all of t_i are the same.

Output

Output the answer as a common irreducible fraction, separate numerator and denominator by a space.

Sample Input

3
6 2 3

Sample Output

3 1

Hint

Source

Northeastern Europe 2005, Northern Subregion

import java.util.*;
import java.math.*;
public class Main
{
public static void main(String args[])throws Exception
{
   Scanner cin=new Scanner(System.in);
   int n = cin.nextInt();
   int num[]=new int[n];
   int den[]=new int[n];
   int per[]=new int[n];
   for(int i = 0; i < n; i++)
    per[i]=cin.nextInt();
   Arrays.sort(per);
   int cnt=1;
   for(int i = 1; i < n; i++)
    if(per[i]!=per[cnt-1])per[cnt++]=per[i];
   for(int i = 0; i < cnt-1; i++)
   {
  num[i]=per[cnt-1]*per[i];
  den[i]=2*(per[cnt-1] - per[i]);
  int g=gcd(num[i],den[i]);
  num[i]/=g;den[i]/=g;//important
  //int c = gcd(num[i],den[i]);
  //num[i]=num[i]/c;
  //den[i]=den[i]/c;
   }
      BigInteger ansn= new BigInteger(num[0]+"");
      BigInteger ansd= new BigInteger(den[0]+"");
   for(int i = 1; i < cnt-1; i++)
   {
    BigInteger tempn = new BigInteger(num[i]+"");
    BigInteger tempd = new BigInteger(den[i]+"");
    /*ansn=ansn.multiply(tempd);
    tempn=tempn.multiply(ansd);
    ansn=lcm(ansn,tempn);
    ansd=ansd.multiply(tempd);
    BigInteger g = ansn.gcd(ansd);
    ansn=ansn.divide(g);//ansn = ansn/g;
    ansd=ansd.divide(g);//ansd = ansd/g;*/
    ansn=lcm(ansn,tempn);
    ansd=ansd.gcd(tempd);
   }
   BigInteger g = ansn.gcd(ansd);
   ansn=ansn.divide(g);
   ansd=ansd.divide(g);
   System.out.println(ansn+" "+ansd);
}
static int gcd(int x,int y)
{
   if (x == 0 || y == 0) if(x > y) return x; else return y;
      for (int t; (t = x % y) != 0; x = y, y = t);
      return y;
}
static BigInteger lcm(BigInteger a,BigInteger b)
{
  BigInteger c = a.gcd(b);
     return a.multiply(b).divide(c);
}
}