Lanqiao Cup Test Questions-Basic Practice Hexadecimal to Octal

Resource limit
Time limit: 1.0s
Memory limit: 512.0MB

Problem description 　　
given positive integer n hexadecimal, octal output thereof corresponds.

Input format 　　
input acts a first positive integer n (1 <= n <= 10). In the next n lines, each line contains a string consisting of 0 _{9 and uppercase letters A} F, which represents the hexadecimal positive integer to be converted, and the length of each hexadecimal number does not exceed 100000.
Output format 　　
output n lines, each act input corresponding octal positive integer.

[Note] The entered hexadecimal number will not have leading 0, such as 012A.　
The output octal number cannot have a leading 0 either.

Sample input 　　
2 39 123ABC

Sample output
71 4435274

[Prompt] 　　
First convert the hexadecimal number into a certain hexadecimal number, and then convert the certain hexadecimal number into octal.

import java.util.Scanner;
 
public class Main {
    
    
	
	public static void main(String[] args) {
    
    
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		for(int cnt=0;cnt<n;cnt++){
    
    
			String str = sc.next();
			StringBuilder str2 = To_binary(str);
			To_Octal(str2);
		}
	}
	
	public static StringBuilder To_binary(String str){
    
         //16进制转换为2进制
		StringBuilder str2 = new StringBuilder();      //StringBuilder用来申明可变字符串
		for(int i=0;i<str.length();i++){
    
    
			char c = str.charAt(i);
			switch(c){
    
    
			case '0':
				str2.append("0000");break;
			case '1':
				str2.append("0001");break;
			case '2':
				str2.append("0010");break;
			case '3':
				str2.append("0011");break;
			case '4':
				str2.append("0100");break;
			case '5':
				str2.append("0101");break;
			case '6':
				str2.append("0110");break;
			case '7':
				str2.append("0111");break;
			case '8':
				str2.append("1000");break;
			case '9':
				str2.append("1001");break;
			case 'A':
				str2.append("1010");break;
			case 'B':
				str2.append("1011");break;
			case 'C':
				str2.append("1100");break;
			case 'D':
				str2.append("1101");break;
			case 'E':
				str2.append("1110");break;
			case 'F':
				str2.append("1111");break;
			default:break;
			}
		}
		return str2;
	}
	
	public static void To_Octal(StringBuilder str){
    
      //转为8进制，转1位输出1位
		int len = str.length();
		int ans=0,s = 0;
		if(len % 3==1){
    
                              //最高位的补零模拟
			ans = str.charAt(0)-'0';
			s+=1;
		}else if(len % 3==2){
    
    
			ans = (str.charAt(0)-'0')*2 + (str.charAt(1)-'0');
			s+=2;
		}
		if(ans>0)
			System.out.print(ans);
		boolean flog = true;
		for(int i=s;i<str.length();i+=3){
    
    
			ans =(str.charAt(i)-'0')*4+ (str.charAt(i+1)-'0')*2 + (str.charAt(i+2)-'0');
			if(flog){
    
    
				if(ans==0){
    
    
					continue;
				}else{
    
    
					flog = false;
				}
			}
			System.out.print(ans);
		}
		System.out.println();
	}
}