组合数（2019湖南acm第二题）

思路：高精度来求

代码：

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

public class Main {
    public static void main(String[] args) {
        Scanner scan=new Scanner(System.in);
        int n,k;
        while(scan.hasNext()){//求最小值min
            n=scan.nextInt();
            k=scan.nextInt();
            if(n-k<=k) k=n-k;
            BigInteger ans=new BigInteger("1");
            BigInteger max=new BigInteger("1000000000000000000");
            BigInteger tmp=new BigInteger("1");
            int m=n;
            for (int i=1;i<=k;i++){
                tmp=BigInteger.valueOf(i);
                ans=ans.multiply(BigInteger.valueOf(m));
                ans=ans.divide(tmp);
                if (ans.compareTo(max)>=0)
                    break;
                m--;
            }
            if (ans.compareTo(max)>=0) System.out.println(max);
            else System.out.println(ans);
        }
    }
}