蓝桥杯java 基础训练 十六进制转八进制
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
一、
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String[] sts = new String[n];
for(int i=0;i<n;i++){
sts[i] = sc.next();
}
sc.close();
for(int i=0;i<n;i++){
String strBinary = toBinary(sts[i]);
int len_strBin = strBinary.length();
if(len_strBin%3==1) strBinary = "00"+strBinary;
if(len_strBin%3==2) strBinary = "0"+strBinary;
String strOctal = toOctal(strBinary);
System.out.println(strOctal);
}
}
private static String toOctal(String strBinary) {
int len = strBinary.length();
int k;
StringBuffer stb = new StringBuffer();
if(strBinary.substring(0, 3).equals("000"))
k=3;
else
k=0;
for(int i=k;i<len-2;i+=3){
switch (strBinary.substring(i, i+3)) {
case "000":stb.append("0");break;
case "001":stb.append("1");break;
case "010":stb.append("2");break;
case "011":stb.append("3");break;
case "100":stb.append("4");break;
case "101":stb.append("5");break;
case "110":stb.append("6");break;
case "111":stb.append("7");break;
default:break;
}
}
return stb.toString();
}
private static String toBinary(String strHex) {
int len_str = strHex.length();
StringBuffer stb = new StringBuffer();
for(int i=0;i<len_str;i++){
switch (strHex.charAt(i)) {
case '0':stb.append("0000");break;
case '1':stb.append("0001");break;
case '2':stb.append("0010");break;
case '3':stb.append("0011");break;
case '4':stb.append("0100");break;
case '5':stb.append("0101");break;
case '6':stb.append("0110");break;
case '7':stb.append("0111");break;
case '8':stb.append("1000");break;
case '9':stb.append("1001");break;
case 'A':stb.append("1010");break;
case 'B':stb.append("1011");break;
case 'C':stb.append("1100");break;
case 'D':stb.append("1101");break;
case 'E':stb.append("1110");break;
case 'F':stb.append("1111");break;
default:break;
}
}
return stb.toString();
}
}
二、
import java.util.HashMap;
import java.util.Scanner;
public class Test {
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();
}
}
三、
import java.util.Scanner;
public class Test1 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String[] st = new String[n];
for (int i = 0; i < n; i++) {
st[i] = sc.next();
}
sc.close();
for (int i = 0; i < n; i++) {
String str1 = ttos(st[i]);
int len_str1 = str1.length();
if (len_str1 % 3 == 1)
str1 = "00" + str1;
else if (len_str1 % 3 == 2)
str1 = "0" + str1;
ttoe(str1);
System.out.println();
}
}
public static String ttos(String str) {
int len_str = str.length();
StringBuilder str2 = new StringBuilder();
for (int i = 0; i < len_str; ++i) {
switch (str.charAt(i)) {
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.toString();
}
public static void ttoe(String str2) {
int len = str2.length();
int a;
a = (str2.charAt(0) - '0') * 4 + (str2.charAt(1) - '0') * 2 + (str2.charAt(2) - '0');
if (a != 0)
System.out.print(a);
for (int i = 3; i <= len - 2; i += 3) {
a = (str2.charAt(i) - '0') * 4 + (str2.charAt(i + 1) - '0') * 2 + (str2.charAt(i + 2) - '0');
System.out.print(a);
}
}
}
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