njuoj 递归查询字符串

 思路：每次将pos映射到对称区间的相应位置。直到达到[1,5]区间

其中需要注意的是，需要找到$的分布规律。单独处理

package com.company.dp;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class 递归查询子字符串 {
    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int case_num=Integer.parseInt(scanner.nextLine());
        for (int i = 0; i < case_num; i++) {
            long pos=Long.parseLong(scanner.nextLine());
            char res=getNum(pos);
            System.out.println(res);
        }
        scanner.close();
    }

    private static char getNum(long pos) {
        long[] s=new long[50];
        s[0]=5;
        for (int i = 1; i < s.length; i++) {
            s[i]=s[i-1]*2+i;
            
        }
        for (int i = 0; i < s.length; i++) {
            System.out.println(s[i]);
        }

        String str0="12345";

        while(pos>5){
            for (int i = 1; i < s.length; i++) {
                if(pos>=s[i-1]&&pos<=s[i]){

                    if(2*pos-s[i]>=0&&2*pos-s[i]<=i)return '$';
                    pos=s[i]-pos+1;
                    break;
                }

            }

        }
        return str0.charAt(((int) pos)-1);
        
    }

}