可以明显代入找到通项公式 x = (s+n) / 2^(n+1) - 1
逐一用BigDecimal去计算，除数会出现无穷小数，所以要用整除方法，然后加1 即可，因为答案不能小于这个小数
而且答案是整数，所以+1

import java.util.*;
import java.math.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String str = sc.next();
        BigDecimal s = new BigDecimal(str);
        s = s.add(new BigDecimal(n));

        BigDecimal devide = new BigDecimal(1);

        while(n-- != -1){
            devide = devide.multiply(new BigDecimal(2)); 
        }
        devide = devide.subtract(new BigDecimal(1));

        s = s.divideToIntegralValue(devide).add(new BigDecimal(0.5)).setScale(0, BigDecimal.ROUND_HALF_UP);
        System.out.println(s);

    }
}

简短优化后

import java.util.*;
import java.math.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        BigDecimal s = new BigDecimal(sc.next()).add(new BigDecimal(n));
        BigDecimal devide = new BigDecimal(1);
        while(n-- != -1)
            devide = devide.multiply(new BigDecimal(2));
        System.out.println(s.divideToIntegralValue(devide.subtract(new BigDecimal(1))).add(new BigDecimal(1)));

    }
}