In a strange planet there are n races. They are completely different as well as their food habits. Each race has a food-eating period. That means the ith race eats after every xi de-sec (de-sec is the unit they use for counting time and it is used for both singular and plural). And at that particular de-sec they pass the whole day eating.
The planet declared the de-sec as ‘Eid’ in which all the races eat together.
Now given the eating period for every race you have to find the number of de-sec between two consecutive Eids.
Input
Input starts with an integer T (≤ 225), denoting the number of test cases.
Each case of input will contain an integer n (2 ≤ n ≤ 1000) in a single line. The next line will contain n integers separated by spaces. The ith integer of this line will denote the eating period for the ith race. These integers will be between 1 and 10000.
Output
For each case of input you should print a line containing the case number and the number of de-sec between two consecutive Eids. Check the sample input and output for more details. The result can be big. So, use big integer calculations.
Sample Input
2
3
2 20 10
4
5 6 30 60
Sample Output
Case 1: 20
Case 2: 60
-
题意:
n个数的最小公倍数
分析:
java大数实现即可
神奇的发现System.gc();//这个不超内存
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T=in.nextInt();
int cas=0;
while(T-->0)
{
cas++;
int n = in.nextInt();
BigInteger ans=in.nextBigInteger();
for(int i=1;i<n;i++)
{
BigInteger a=in.nextBigInteger();
ans=ans.multiply(a).divide(ans.gcd(a));
}
System.out.println("Case "+cas+": "+ans);
System.gc();//这个不超内存
}
}
}