- problem: 求任意两个不同进制非负整数的转换(2进制~16进制),所给整数在long所能表达的范围之内。 不同进制的表示符号为(0,1,...,9,a,b,...,f)或者(0,1,...,9,A,B,...,F)。
- Input: 输入只有一行,包含三个整数a,n,b。a表示其后的n 是a进制整数,b表示欲将a进制整数n转换成b进制整数。 a,b是十进制整数,2 =< a,b <= 16。
- Output 输出包含一行,该行有一个整数为转换后的b进制数。输出时字母符号全部用大写表示,即(0,1,...,9,A,B,...,F)。
- Sample Input 15 Aab3 7
- Sample Output 210306
我在这里用java原有数制转换工具实现。
package com.fantJ.ACM;import java.util.Scanner;/** * https://vjudge.net/contest/199742#problem/B * 2 =< a,b <= 16 * Created by Fant.J. * 2017/12/4 10:13 */public class B数制转换 { public static void main(String []args){ Scanner scanner = new Scanner(System.in); String data = scanner.nextLine(); String[] datas = data.split(" "); Integer transitionFrom = Integer.valueOf(datas[0]); String character = datas[1]; Integer transitionTo = Integer.valueOf(datas[2]); try { if (transitionFrom >= 2 && transitionTo <= 16) { String result = Integer.toString(Integer.valueOf(character, transitionFrom), transitionTo); Integer resultNum = Integer.valueOf(result); System.out.print(resultNum); } else { System.out.println("非法输入"); } }catch (Exception e){ throw new RuntimeException("非法输入"); } }}复制代码
来看看Integer.valueOf(String s, int radix)源码
public static Integer valueOf(String s, int radix) throws NumberFormatException { return Integer.valueOf(parseInt(s,radix)); }复制代码
parseInt(s,radix)源码
public static final int MIN_VALUE = 0x80000000;public static final int MAX_VALUE = 0x7fffffff;public static final int MIN_RADIX = 2;public static final int MAX_RADIX = 36;public static int parseInt(String s, int radix) throws NumberFormatException { /* * WARNING: This method may be invoked early during VM initialization * before IntegerCache is initialized. Care must be taken to not use * the valueOf method. */ if (s == null) { throw new NumberFormatException("null"); } if (radix < Character.MIN_RADIX) { throw new NumberFormatException("radix " + radix + " less than Character.MIN_RADIX"); } if (radix > Character.MAX_RADIX) { throw new NumberFormatException("radix " + radix + " greater than Character.MAX_RADIX"); } int result = 0; boolean negative = false; int i = 0, len = s.length(); int limit = -Integer.MAX_VALUE; int multmin; int digit; if (len > 0) { char firstChar = s.charAt(0); if (firstChar < '0') { // Possible leading "+" or "-" if (firstChar == '-') { negative = true; limit = Integer.MIN_VALUE; } else if (firstChar != '+') throw NumberFormatException.forInputString(s); if (len == 1) // Cannot have lone "+" or "-" throw NumberFormatException.forInputString(s); i++; } multmin = limit / radix; while (i < len) { // Accumulating negatively avoids surprises near MAX_VALUE digit = Character.digit(s.charAt(i++),radix); if (digit < 0) { throw NumberFormatException.forInputString(s); } if (result < multmin) { throw NumberFormatException.forInputString(s); } result *= radix; if (result < limit + digit) { throw NumberFormatException.forInputString(s); } result -= digit; } } else { throw NumberFormatException.forInputString(s); } return negative ? result : -result; }复制代码
然后再分析Character.digit方法 emmmm java用Unicode在处理。
int digit(int ch, int radix) { int value = -1; if (radix >= Character.MIN_RADIX && radix <= Character.MAX_RADIX) { int val = getProperties(ch); int kind = val & 0x1F; if (kind == Character.DECIMAL_DIGIT_NUMBER) { value = ch + ((val & 0x3E0) >> 5) & 0x1F; } else if ((val & 0xC00) == 0x00000C00) { // Java supradecimal digit value = (ch + ((val & 0x3E0) >> 5) & 0x1F) + 10; } } return (value < radix) ? value : -1; }复制代码
知识补充:
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2 to 8:以10010为例,要转成8进制,则从右向左看,每3个为一组,不足的补零,变成010 010,加上权值后为22,即为8进制数!
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8 to 2:与上面的相反,以27为例,要转为2进制,则每个位作为一组分开,变成2 7,通过权值变换后为010 111(为1的替换为权值,然后相加等于7,则4+2+1,即每个位都是1,故为111),最后得到的2进制数为010111,去掉左边的0,最终结果是10111。
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2 to 16: 以101110为例,要转成16进制,类似,从右向左看,每4个为一组,不足的补零,变成0010 1110,加上权值后为2E,有个规律,8进制的各个位<=7,16进制的各个位<=15,也就是说16进制中的数可以是1、2、3、4……9、A、B、C、D、E、F。
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16 to 2:以EF为例,每个为作为一组分开,变成E F,通过权值变换后为1110 1111,最后得到的2进制是 11101111
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8 to 16:以27为例,8进制和16进制之间的转换需要用2进制来作为过渡,先转成2进制为010 111,然后从右向左数,将现在的3个一组变为4个一组,不足的补零,变为0001 0111,然后权值变换后为1 7,也就是16进制数 17