问题描述 给定n个十六进制正整数,输出它们对应的八进制数。
输入格式 输入的第一行为一个正整数n (1<=n<=10)。 接下来n行,每行一个由0~9、大写字母A~F组成的字符串,表示要转换的十六进制正整数,每个十六进制数长度不超过100000。
输出格式 输出n行,每行为输入对应的八进制正整数。
【注意】 输入的十六进制数不会有前导0,比如012A。 输出的八进制数也不能有前导0。
样例输入 2 39 123ABC
样例输出 71 4435274
【提示】 先将十六进制数转换成某进制数,再由某进制数转换成八进制。
import java.util.HashMap;
import java.util.Scanner;
public class Main {
static String[] Binary = {"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
static String[] Octal = {"000","001","010","011","100","101","110","111"};
static HashMap<String, String> map;
/*
* Binary是0-15的二进制数的数组
* Octal是0-7的二进制数的水族
* map是0-7与其二进制数的映射
*/
//16进制转2进制方法
public static String HexToBinary(String hex){
StringBuffer binary = new StringBuffer();
for ( int i = 0 ; i < hex.length() ; i++){
char ch = hex.charAt(i);
if ( ch >= '0' && ch <= '9' ){
binary.append(Binary[(int)(ch-'0')]);
}else{
binary.append(Binary[(int)(ch-'A')+10]);
}
}
return binary.toString();
}
//2进制转8进制方法
public static String BinaryToOctal(String binary){
StringBuffer octal = new StringBuffer();
int reminder = binary.length() % 3;
//sub为2进制数需要补充0的个数
int sub = 0;
if ( reminder != 0)
sub = 3 - reminder;
StringBuffer tmp = new StringBuffer();
for ( int i = 0 ; i < sub ; i++){
tmp.append("0");
}
tmp.append(binary);
for ( int i = 0 ; i <= tmp.length() - 3 ; i += 3){
String substring = tmp.substring(i, i+3);
octal.append(map.get(substring));
}
int index = FindZero(octal);
/*
* 如果index为-1,则8进制数前面没有0,
* 反之则有0,取无0的子字符串。
*/
if(index == -1)
return octal.toString();
else
return octal.substring(index+1).toString();
}
//统计8进制数的字符串的最前面连续0的个数,返回的最后一个0的下标
public static int FindZero(StringBuffer octal){
int index = -1;
for ( int i = 0 ; i < octal.length() ; i++){
if ( octal.charAt(i) == '0'){
index = i;
continue;
}else{
break;
}
}
return index;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
map = new HashMap<String, String>();
for ( int i = 0 ; i < Octal.length ; i++){
char[] ch = new char[1];
ch[0] = (char) (i+'0');
String str = new String(ch);
map.put(Octal[i], str);
}
int num = in.nextInt();
for (int i = 0 ; i < num ; i++){
String hex = in.next();
String binary = HexToBinary(hex);
String octal = BinaryToOctal(binary);
System.out.println(octal);
}
in.close();
}
}