The Chosen One（规律题，java大数乘法）

Welcome to the 2017 ACM-ICPC Asia Nanning Regional Contest.
Here is a breaking news. Now you have a chance to meet alone with the Asia Director through a game.
All boys and girls who chase their dreams have to stand in a line. They are given the numbers in the order in which they stand starting from 1.
The host then removes all boys and girls that are standing at an odd position with several rounds.
For example if there are n=8 boys and girls in total. Initially standing people have numbers 1,2,3,4,5,6,7 and 8. After the first round, people left are 2,4,6 and 8. After the second round, only two people, 4 and 8, are still there.
The one who stays until the end is the chosen one.
I know you want to become the chosen one to meet alone with your idol. Given the number of boys and girls in total, can you find the best place to stand in the line so that you would become the chosen one?
Input
First line of the input contains the number of test cases t(1≤t≤1000).Each of the next t lines contains the integer n which is the number of boys and girls in total, where 2≤n≤10^50.
Output
The output displays t lines, each containing a single integer which is the place where you would stand to win the chance.
找规律

答案 ；个数 ；；；；范围：

可知，找到<=人数的2的倍数即是答案，但因10^50太大，改学Java,代码来源

import java.math.BigInteger;
import java.util.*;
public class Main 
{
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        int t = cin.nextInt();
        while (t-- != 0)
        {
            BigInteger n = cin.nextBigInteger();
            BigInteger ans = new BigInteger("2");
            System.out.println(ans = ans.pow(n.bitLength() - 1));
        }
    }
}