使用java 和 php 不限位数 进行进制转换 （自定义BigNumber）

转自：https://blog.youkuaiyun.com/xinshijimanon/article/details/46681191

1. /* 
2.  * 0~9用0~9表示,10~35用A~Z表示,36~61用a~z表示 
3.  */  
4. @SuppressWarnings("serial")  
5. public class BigNumber extends Exception{     
6.     private String MyNumber;//大数      
7.     private int SystemNumber;//进制数  
8.     public BigNumber(){  
9.         this("0",10);  
10.     }  
11.     public BigNumber(String MyNumber){  
12.         this(MyNumber,10);  
13.     }  
14.     public BigNumber(String MyNumber,int SystemNumber){  
15.         this.MyNumber=MyNumber;  
16.         this.SystemNumber=SystemNumber;  
17.         this.Islegal();   
18.     }  
19.     public String getMyNumber() {  
20.         return this.MyNumber;  
21.     }  
22.     public void setMyNumber(String myNumber) {  
23.         this.MyNumber = myNumber;  
24.     }  
25.     public int getSystemNumber() {  
26.         return this.SystemNumber;  
27.     }  
28.     public void setSystemNumber(int systemNumber) {  
29.         this.SystemNumber = systemNumber;  
30.     }  
31.     public void Islegal(){//判断该字符串是否合法  
32.         if(this.MyNumber==null||this.MyNumber.length()==0){//大数为空或大数字符串长度为0  
33.             throw (new  NumberFormatException("错误!大数为空或大数字符串长度为0 "+this.MyNumber));  
34.         }  
35.         if(this.SystemNumber<=1||this.SystemNumber>=63){//进制数不合法  
36.             throw (new  NumberFormatException("错误!进制数不合法 "+this.SystemNumber));  
37.         }  
38.         if(this.MyNumber.equals("+")||this.MyNumber.equals("-")||this.MyNumber.equals(".")){  
39.             throw (new  NumberFormatException("错误!大数不合法 "+this.MyNumber));  
40.         }  
41.         if(this.MyNumber.equals("+.")||this.MyNumber.equals("-.")){  
42.             throw (new  NumberFormatException("错误!大数不合法 "+this.MyNumber));  
43.         }  
44.         for(int i=0,a,k=-1;i<this.MyNumber.length();i++){  
45.             a=Chartoint(this.MyNumber.charAt(i));  
46.             if(i!=0&&(a==-1||a==-2)){  
47.                 throw (new  NumberFormatException("错误!符号位只能在大数首部 "+this.MyNumber.charAt(i)+"("+i+")"));  
48.             }  
49.             if(a>=this.SystemNumber||a==-4){  
50.                 throw (new  NumberFormatException("错误!该字符不在该进制合法字符中 "+this.MyNumber.charAt(i)+"("+i+")"));  
51.             }  
52.             if(a==-3){  
53.                 if(k==-1){  
54.                     k=i;  
55.                 }else{  
56.                     throw (new  NumberFormatException("错误!出现了第二个小数点 "+this.MyNumber.charAt(i)+"("+i+")"));  
57.                 }  
58.             }  
59.         }  
60.     }  
61.     private static String Add_Positive(String s1,String s2,int n){//自定义两个正数加法运算       
62.         String Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2);//分别获取整数部分及小数部分  
63.         int Inlength=Math.max(Ins1.length(), Ins2.length()),Dolength=Math.max(Dos1.length(), Dos2.length());  
64.         Ins1=AddToLeft(Ins1,Inlength+1)+AddToRight(Dos1,Dolength);  
65.         Ins2=AddToLeft(Ins2,Inlength+1)+AddToRight(Dos2,Dolength);  
66.         String s="";  
67.         for(int i=Inlength+Dolength,m,y=0;i>=0;i--){  
68.             m=Chartoint(Ins1.charAt(i))+Chartoint(Ins2.charAt(i))+y;  
69.             s=Inttochar(m%n)+s;  
70.             y=m/n;  
71.         }  
72.         int pointplace=s.length()-Dolength;  
73.         s=Format(s.substring(0,pointplace)+"."+s.substring(pointplace),true);     
74.         return s;  
75.     }         
76.     private static String Sub_Positive(String s1,String s2,int n){//自定义两个正数减法运算;  
77.         if(ComparetoString(s1,s2)<0){//如果s1<s2则返回-(s2-s1)  
78.             return "-"+Sub_Positive(s2,s1,n);  
79.               
80.         }                     
81.         String Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2);//分别获取整数部分及小数部分  
82.         int Inlength=Math.max(Ins1.length(), Ins2.length()),Dolength=Math.max(Dos1.length(), Dos2.length());  
83.         String R=getComplement(new BigNumber(AddToLeft(Ins2,Inlength)+"."+AddToRight(Dos2,Dolength),n)).MyNumber;  
84.         R=Add_Positive(s1,R,n);  
85.         s2="0."+AddToLeft("1",Dolength);  
86.         R=Add_Positive(R,s2,n);  
87.         return Format(R.substring(1),true);       
88.     }  
89.     private static String Mult_Positive(String s1,String s2,int n){//自定义两个正数乘法运算;   
90.         String Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2);//分别获取整数部分及小数部分  
91.         Ins1+=Dos1;  
92.         Ins2+=Dos2;       
93.         int[]A=new int[Ins1.length()+Ins2.length()-1];  
94.         for(int i=Ins1.length()-1;i>=0;i--){  
95.             for(int j=Ins2.length()-1;j>=0;j--){  
96.                 A[i+j]+=Chartoint(Ins1.charAt(i))*Chartoint(Ins2.charAt(j));  
97.             }  
98.         }  
99.         String s="";  
100.         for(int i=A.length-1;i>=1;i--){  
101.             s=Inttochar(A[i]%n)+s;  
102.             A[i-1]+=A[i]/n;  
103.         }  
104.         s=Inttochar(A[0]/n)+""+Inttochar(A[0]%n)+s;//特别要注意+""  
105.         int pointlength=Dos1.length()+Dos2.length();//获取小数点后的位数  
106.         return Format(s.substring(0,s.length()-pointlength)+"."+s.substring(s.length()-pointlength),true);        
107.     }     
108.     private static String  Division_Positive(String s1, String s2,int n,int f){//自定义两个正数相除(f为小数点后的位数,能被除尽，且小数点后的位数少于f,则直接返回结果);  
109.         String Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2); //分别获取整数部分及小数部分       
110.         int Dolength=Math.max(Dos1.length(), Dos2.length());  
111.         Ins1+=AddToRight(Dos1,Dolength);  
112.         Ins2+=AddToRight(Dos2,Dolength);  
113.         String[]B=new String[n];B[0]="0";  
114.         for(int i=1;i<n;i++){  
115.             B[i]=Add_Positive(B[i-1],Ins2,n);//B[i]=Mult_Positive(Ins2,Inttochar(i)+"", n);  
116.         }  
117.         int i=Math.min(Ins1.length(), Ins2.length())-2,j;  
118.         String Ds0="",Ds1=Ins1.substring(0,i);//初始化整数部分余数部分           
119.         while(true){  
120.             if(i!=Ins1.length()){  
121.                 if(i<Ins1.length()){  
122.                     Ds1+=Ins1.charAt(i);  
123.                 }else{  
124.                     Ds1+="0";  
125.                     if(Ds1.equals("00")||i-Ins1.length()-1==f){  
126.                         break;  
127.                     }  
128.                 }  
129.                 //下面开始求Ds1/Ins2的整数部分及余数,结果分别赋值为Ds0,Ds1;采用二分查找获取整数部分  
130.                 j=Binary(B,Ds1);  
131.                 Ds0+=Inttochar(j);  
132.                 Ds1=Sub_Positive(Ds1,B[j],n);                 
133.             }else{  
134.                 Ds0+=".";  
135.             }  
136.             i++;  
137.         }  
138.         return Format(Ds0,true);      
139.     }  
140.     private static int Binary(String[]A,String key){//二分查找key(数组A已经按从小到大的顺序排好序)  
141.         int low=0,height=A.length-1,middle,Compare;  
142.         while(low<=height){  
143.             middle=(low+height)/2;  
144.             Compare=ComparetoString(key,A[middle]);  
145.             if(Compare==0){  
146.                 return middle;  
147.             }else if(Compare>0){  
148.                 low=middle+1;  
149.             }else{  
150.                 height=middle-1;  
151.             }  
152.         }  
153.         return height;  
154.     }  
155.     public BigNumber Add(BigNumber b){//自定义加法运算  
156.         if(this.SystemNumber!=b.SystemNumber){  
157.             throw (new  NumberFormatException("错误!这两个大数进制数不一致 ("+this.SystemNumber+","+b.SystemNumber+")"));  
158.         }  
159.         String s;  
160.         if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){  
161.             s=Add_Positive(this.MyNumber,b.MyNumber,this.SystemNumber);  
162.         }else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){  
163.             s=Sub_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber);             
164.         }else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){             
165.             s=Sub_Positive(b.MyNumber,Opposite(this.MyNumber),this.SystemNumber);  
166.         }else{  
167.             s=Add_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber);  
168.             s=(s.equals("0")?"":"-")+s;  
169.         }  
170.         return new BigNumber(s,this.SystemNumber);  
171.     }  
172.     public BigNumber Sub(BigNumber b){//自定义减法运算  
173.         return this.Add(new BigNumber(Opposite(b.MyNumber),b.SystemNumber));  
174.     }  
175.     public BigNumber Mult(BigNumber b){//自定义乘法运算  
176.         if(this.SystemNumber!=b.SystemNumber){  
177.             throw (new  NumberFormatException("错误!这两个大数进制数不一致 ("+this.SystemNumber+","+b.SystemNumber+")"));  
178.         }  
179.         String s;  
180.         if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){  
181.             s=Mult_Positive(this.MyNumber,b.MyNumber,this.SystemNumber);  
182.         }else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){  
183.             s=Mult_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber);     
184.             s=(s.equals("0")?"":"-")+s;  
185.         }else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){             
186.             s=Mult_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumber);  
187.             s=(s.equals("0")?"":"-")+s;  
188.         }else{  
189.             s=Mult_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber);              
190.         }  
191.         return new BigNumber(s,this.SystemNumber);        
192.     }  
193.     public BigNumber Division(BigNumber b,int f){//自定义除法运算(f为小数点后的位数,能被除尽,且小数点后的位数少于f,则直接返回精确结果);  
194.         if(b.toString().equals("-0")||b.toString().equals("0")){  
195.             throw (new  NumberFormatException("错误!除数不能为零:"+b));  
196.         }  
197.         if(this.SystemNumber!=b.SystemNumber){  
198.             throw (new  NumberFormatException("错误!这两个大数进制数不一致! ("+this.SystemNumber+","+b.SystemNumber+")"));  
199.         }  
200.         if(f<0){  
201.             throw (new  NumberFormatException("错误!小数点后的位数应为非负值!("+f+")"));  
202.         }  
203.         String s;  
204.         if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){  
205.             s=Division_Positive(this.MyNumber,b.MyNumber,this.SystemNumber,f);  
206.         }else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){  
207.             s=Division_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber,f);     
208.             s=(s.equals("0")?"":"-")+s;  
209.         }else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){             
210.             s=Division_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumber,f);  
211.             s=(s.equals("0")?"":"-")+s;  
212.         }else{  
213.             s=Division_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber,f);            
214.         }  
215.         return new BigNumber(s,this.SystemNumber);   
216.     }  
217.     public BigNumber Division(BigNumber b){//自定义除法运算,默认为保留50位小数(若能被除尽,且小数点后的位数少于50,则直接返回精确结果);  
218.         return Division(b,50);  
219.     }  
220.     private static int ComparetoString(String s1,String s2){//判断两个正数的大小s1>s2返回1,s1=s2返回0,s1<s2返回-1  
221.         String Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2);  
222.         int Dolength=Math.max(Dos1.length(), Dos2.length());  
223.         Ins1+=AddToRight(Dos1,Dolength);  
224.         Ins2+=AddToRight(Dos2,Dolength);  
225.         if(Ins1.length()>Ins2.length()){  
226.             return 1;  
227.         }else if(Ins1.length()<Ins2.length()){  
228.             return -1;  
229.         }  
230.         for(int i=0;i<Ins1.length();i++){  
231.             if(Ins1.charAt(i)>Ins2.charAt(i)){  
232.                 return 1;  
233.             }else if(Ins1.charAt(i)<Ins2.charAt(i)){  
234.                 return -1;  
235.             }  
236.         }  
237.         return 0;  
238.     }  
239.     public int Compareto(BigNumber b){  
240.         if(this.SystemNumber!=b.SystemNumber){  
241.             throw (new  NumberFormatException("错误!这两个大数进制数不一致 ,暂时无法比较("+this.SystemNumber+","+b.SystemNumber+")"));  
242.         }  
243.         String s1=Format(this.MyNumber,false),s2=Format(b.MyNumber,false);  
244.         if(IsPositive(s1)&&IsPositive(s2)){  
245.             return ComparetoString(s1,s2);//int a=s1.CompareTo(s2); return a!=0?(a>0?1:-1):0;  
246.         }else if(IsPositive(s1)&&!IsPositive(s2)){  
247.             return 1;  
248.         }else if(!IsPositive(s1)&&IsPositive(s2)){  
249.             return -1;  
250.         }else{  
251.             return -ComparetoString(s1.substring(1),s2.substring(1));  
252.         }  
253.     }  
254.     private static String ConverFromTen(long n,int R){//将十进制数n转换为R进制数  
255.         if(R==10){  
256.             return n+"";  
257.         }  
258.         String s="";  
259.         while(true){  
260.             s=Inttochar((int)(n%R))+s;  
261.             n/=R;  
262.             if(n==0){  
263.                 return s;  
264.             }  
265.         }         
266.     }   
267.     /* 
268.      * 将R1进制正整数s转化为R2进制数  
269.      * (由于Conversion频繁调用该函数,为提高效率,引入两个辅助数组Hex1,Hex2): 
270.      * Hex1[i]是十进制数i转换为R2进制数后的数,0<=i<=R1;Hex2[i]=Hex1[R1]^i,0<=i<=s.length() 
271.      * Hex2[i]是十进制数R1转换为R2进制数后的i次方 
272.      */  
273.     private static String Conver(String s,int R2,String[]Hex1,String[]Hex2){  
274.         String Ins="0";//初始化新进制对应的大数  
275.         for(int i=0;i<s.length();i++){             
276.             Ins=Add_Positive(Ins,Mult_Positive(Hex1[Chartoint(s.charAt(s.length()-1-i))],Hex2[i],R2),R2);                              
277.         }  
278.         return Ins;  
279.     }  
280.     private static String Conversion(String MyNumber,int R1,int R2,int N,int f){//将R1进制大数MyNumber转换为R2进制的大数(f为小数点后的位数,能被除尽,且小数点后的位数少于f,则直接返回结果),N为将MyNumber分组后每组字符串的长度;  
281.         if(R2==R1){  
282.             int a=getDecimal(MyNumber).length()-f;  
283.             if(a>0){  
284.                 return MyNumber.substring(0,MyNumber.length()-a);  
285.             }  
286.             return MyNumber;  
287.         }  
288.         if(MyNumber.charAt(0)=='+'||MyNumber.charAt(0)=='-'){  
289.             char c=MyNumber.charAt(0);  
290.             return c+Conversion(MyNumber.substring(1),R1,R2,N,f);  
291.         }  
292.         String[]Hex1=new String[R1+1];  
293.         for(int i=0;i<=R1;i++){//将0~R1的所有数字转换为R2进制  
294.             Hex1[i]=ConverFromTen(i,R2);  
295.         }  
296.         String[]Hex2=new String[N+1];Hex2[0]="1";  
297.         for(int i=1;i<=N;i++){//求十进制数R1转换为R2进制数后的i次方  
298.             Hex2[i]=Mult_Positive(Hex2[i-1],Hex1[R1],R2);  
299.         }             
300.         String Ins1=getInteger(MyNumber),Dos1=getDecimal(MyNumber);//获取整数部分及小数部分          
301.         /* 
302.          * 经过计算当转化后的R2进制数只需保留f位小数时,原R1进制数MyNumber只需保留a位 
303.          * 当MyNumber小数位很多，而f较小时在一定程度上能提高效率 
304.          */  
305.         int a=(int)Math.ceil(f*Math.log(R2)/Math.log(R1))+1;  
306.         if(a<=Dos1.length()){  
307.             Dos1=Dos1.substring(0,a);  
308.         }  
309.           
310.         Ins1=AddToLeft(Ins1,Ins1.length()+(N-Ins1.length()%N)%N);//扩充Ins1  
311.         Dos1=AddToRight(Dos1,Dos1.length()+(N-Dos1.length()%N)%N);//扩充Dos1  
312.         String Ins="0",Dos="0",x="1";//new BigNumber(Conver1(R1,R2),R2).Power(N).MyNumber;初始化新进制对应的大数的整数部分及小数部分 ,设置初值及权                       
313.         int Count=Math.max(Ins1.length()/N,Dos1.length()/N);//循环次数     
314.         for(int i=0,i1=Ins1.length()/N,i2=Dos1.length()/N;i<Count;i++){                
315.             if(i<i1){  
316.                 Ins=Add_Positive(Ins,Mult_Positive(Conver(Ins1.substring(Ins1.length()-(i+1)*N,Ins1.length()-i*N),R2,Hex1,Hex2),x,R2),R2);  
317.             }  
318.             if(i<i2){                                                      
319.                 Dos=Add_Positive(Dos,Mult_Positive(Conver(Dos1.substring(Dos1.length()-(i+1)*N,Dos1.length()-i*N),R2,Hex1,Hex2),x,R2),R2);  
320.                 if(i==i2-1){  
321.                     Dos=Division_Positive(Dos,Mult_Positive(x,Hex2[N],R2),R2,f);  
322.                 }  
323.             }     
324.             x=Mult_Positive(x,Hex2[N],R2);  
325.         }  
326.         return Ins+Dos.substring(1);//或Add_Positive(Ins,Dos,R2);                      
327.     }  
328.     public BigNumber HexConversion(int R,int N,int f){  
329.         if(R<=1||R>=63){//进制数不合法  
330.             throw (new  NumberFormatException("错误!进制数不合法 "+this.SystemNumber));  
331.         }  
332.         if(N<=0){  
333.             throw (new  NumberFormatException("错误!每小组的字符串长度不能小于1!("+N+")"));  
334.         }  
335.         if(f<0){  
336.             throw (new  NumberFormatException("错误!小数点后的位数应为非负值!("+f+")"));  
337.         }  
338.         return new BigNumber(Conversion(this.MyNumber,this.SystemNumber,R,N,f),R);  
339.     }    
340.     public BigNumber HexConversion(int R,int f){  
341.         int N=(int)Math.pow(Math.max(getInteger(this.MyNumber).length(),getDecimal(this.MyNumber).length()),0.6)+1;  
342.         return HexConversion(R,N,f);  
343.     }   
344.     public BigNumber HexConversion(int R){  
345.         return HexConversion(R,50);  
346.     }        
347.       
348.     /*public BigNumber Power(int n,int f){ 
349.         if(n<0){ 
350.             return new BigNumber("1",this.SystemNumber).Division(this.Power(-n,f),f); 
351.         } 
352.         String s=ConverFromTen(n,2); 
353.         BigNumber r=new BigNumber("1",this.SystemNumber),k=this; 
354.         for(int i=0;i<s.length();i++){                
355.             if(s.charAt(s.length()-1-i)=='1'){ 
356.                 r=r.Mult(k); 
357.             } 
358.             k=k.Mult(k); 
359.         }        
360.         return r;                
361.     } 
362.     public BigNumber Power(int n){ 
363.         return Power(n,50); 
364.     }*/  
365.       
366.     private static BigNumber getComplement(BigNumber b){//获得 b的每位取反后的大数  
367.         String s="";  
368.         char c=b.MyNumber.charAt(0);  
369.         if(c=='+'||c=='-'||c=='.'){  
370.             s+=c;  
371.         }else{  
372.             s+=Inttochar(b.SystemNumber-1-Chartoint(c));  
373.         }  
374.         for(int i=1;i<b.MyNumber.length();i++){  
375.             c=b.MyNumber.charAt(i);  
376.             if(c=='.'){  
377.                 s+=c;  
378.             }else{  
379.                 s+=Inttochar(b.SystemNumber-1-Chartoint(c));  
380.             }  
381.         }  
382.         return new BigNumber(s,b.SystemNumber);  
383.     }  
384.     private static String AddToLeft(String s,int length){//若s的长度大于或等于length,则直接返回s,否则字符串s左边补零,直到s长度达到length  
385.         if(s.length()>=length){  
386.             return s;  
387.         }  
388.         while(s.length()<length){  
389.             s="0000000000"+s;  
390.         }  
391.         return s.substring(s.length()-length,s.length());  
392.     }  
393.     private static String AddToRight(String s,int length){//若s的长度大于或等于length,则直接返回s,否则字符串s右边补零,直到s长度达到length  
394.         if(s.length()>=length){  
395.             return s;  
396.         }  
397.         while(s.length()<length){  
398.             s=s+"0000000000";  
399.         }  
400.         return s.substring(0,length);  
401.     }  
402.     private static String Format(String MyNumber,boolean DealZero){//对于合法的大数,格式化大数为a.b(或-a.b)的形式,若b=0且DealZero为真则返回a(或-a)的形式  
403.         new BigNumber(MyNumber,62);//若该大数不合法则抛出异常  
404.         String s;  
405.         if(MyNumber.charAt(0)=='+'||MyNumber.charAt(0)=='-'){  
406.             s=MyNumber.substring(1);  
407.         }else{  
408.             s=MyNumber;  
409.         }  
410.         if(getPointPlace(s)==0){  
411.             s="0"+s;  
412.         }else if(getPointPlace(s)==s.length()-1){  
413.             s=s+"0";  
414.         }else if(getPointPlace(s)==s.length()){  
415.             s=s+".0";  
416.         }  
417.         int Ins=0,Dos=s.length(),pointplace=getPointPlace(s);  
418.         for(int i=0;i<pointplace-1;i++){  
419.             if(s.charAt(i)=='0'){  
420.                 Ins++;  
421.             }else{  
422.                 break;  
423.             }  
424.         }         
425.         for(int i=s.length()-1;i>pointplace+1;i--){  
426.             if(s.charAt(i)=='0'){  
427.                 Dos--;  
428.             }else{  
429.                 break;  
430.             }  
431.         }     
432.         s=(MyNumber.charAt(0)=='-'?"-":"")+s.substring(Ins,Dos);  
433.         if(DealZero&&s.substring(s.length()-2).equals(".0")){  
434.             return s.substring(0,s.length()-2);  
435.         }else{  
436.             return s;  
437.         }  
438.     }     
439.     private static boolean IsPositive(String MyNumber){//判断该大数是否为正数(负数返回false,0和正数返回true)  
440.         if(MyNumber.charAt(0)=='-'){  
441.             return false;  
442.         }  
443.         return true;  
444.     }  
445.     private static int  getPointPlace(String MyNumber){//返回该大数小数点位置  
446.         for(int i=0;i<MyNumber.length();i++){  
447.             if(MyNumber.charAt(i)=='.'){  
448.                 return i;  
449.             }  
450.         }  
451.         return MyNumber.length();  
452.     }     
453.     private static String Abs(String MyNumber){//返回该大数的绝对值  
454.         MyNumber=Format(MyNumber,true);  
455.         return MyNumber.substring(MyNumber.charAt(0)=='-'?1:0);  
456.     }  
457.     private static String Opposite(String MyNumber){//返回该大数的相反数       
458.         return MyNumber.charAt(0)=='-'?Abs(MyNumber):"-"+Abs(MyNumber);  
459.     }  
460.     private static String getInteger(String MyNumber){//返回该大数的整数部分       
461.         MyNumber=Format(MyNumber,false);  
462.         return MyNumber.substring(0,getPointPlace(MyNumber));  
463.     }  
464.     private static String getDecimal(String MyNumber){//返回该大数的小数部分   
465.         MyNumber=Format(MyNumber,false);  
466.         String s=getPointPlace(MyNumber)==MyNumber.length()?"0":MyNumber.substring(getPointPlace(MyNumber)+1);  
467.         return (MyNumber.charAt(0)=='-'?"-":"")+s;    
468.     }  
469.     private static int Chartoint(char c){//字符化为数字  
470.         if(c>='0'&&c<='9'){  
471.             return c-'0';  
472.         }else if(c>='A'&&c<='Z'){  
473.             return c-'A'+10;  
474.         }else if(c>='a'&&c<='z'){  
475.             return c-'a'+36;  
476.         }else if(c=='+'){//正号  
477.             return -1;  
478.         }else if(c=='-'){//负号  
479.             return -2;  
480.         }else if(c=='.'){//小数点  
481.             return -3;  
482.         }else{//其他  
483.             return -4;  
484.         }  
485.     }  
486.     private static char Inttochar(int n){//数字化为字符  
487.         if(n>=0&&n<=9){//0~9用0~9表示  
488.             return (char)(n+48);  
489.         }else if(n>=10&&n<=35){//10~35用A~Z表示  
490.             return (char)(n+55);  
491.         }else if(n>=36&&n<=61){//36~61用a~z表示  
492.             return (char)(n+61);  
493.         }else{//小于0或大于61的数用'~'表示  
494.             return '~';  
495.         }  
496.     }  
497.     public String toString(){  
498.         return Format(this.MyNumber,true);  
499.     }     
500. }  

顺便写了个测试类测试进制转换的效果与效率

测试类:

[java]  view plain  copy

1. public class TestBigNumber {  
2.     public static void main(String[]args){  
3.         String MyNumber="1313624242ADAFSFGDRG424242ADAF679707464342424242ADAFS24242ADAFSFGDRG424242ADAFFGDRG142356679707464342424242ADAFSFGDRG67866679707464342424242ADAFSFGDRG96068563556667970746V6679707464342424423536858686242ADAFSFGD69979RG67970746434242427979742ADAFS.FGDRG646676679707464342424242ADAFSFGDRG9707464342424242ADAFSFGDRG342424242ADAFSFGDRG67970746434266797074643426679707464342424242ADAFSFGDRG424242ADAFSFGDRG424242ADAFSFGDRG24242342426679707464342424242ADAFSFGDRG534364547455686679707464342424242ADAFSFGDRG6679707464342424242ADAFSFGDRGDADA6679707464342424242ADAFSFGDRGAFSFS6679707464342424242ADAFSFGDRG";  
4.         int R1=39,R2=62,R3=2;  
5.         int f=MyNumber.length()-MyNumber.indexOf('.')-1;  
6.         BigNumber b=new BigNumber(MyNumber,R1);  
7.         BigNumber b1,b2,b3,b4;  
8.         long c1=System.currentTimeMillis();  
9.         b1=b.HexConversion(R2, f);  
10.         System.out.println("将"+b.getSystemNumber()+"进制数\n"+b.getMyNumber()+"转化为"+R2+"进制数并保留"+f+"位小数后去除无效位的结果为:\n"+b1);  
11.         long c2=System.currentTimeMillis();  
12.         b2=b.HexConversion(R3, f);  
13.         System.out.println("将"+b.getSystemNumber()+"进制数\n"+b.getMyNumber()+"转化为"+R3+"进制数并保留"+f+"位小数后去除无效位的结果为:\n"+b2);  
14.         long c3=System.currentTimeMillis();  
15.         b3=b1.HexConversion(R1, f);  
16.         System.out.println("将"+b1.getSystemNumber()+"进制数\n"+b1.getMyNumber()+"转化为"+R1+"进制数并保留"+f+"位小数后去除无效位的结果为:\n"+b3);  
17.         long c4=System.currentTimeMillis();  
18.         b4=b2.HexConversion(R1, f);  
19.         System.out.println("将"+b2.getSystemNumber()+"进制数\n"+b2.getMyNumber()+"转化为"+R1+"进制数并保留"+f+"位小数后去除无效位的结果为:\n"+b4);  
20.         long c5=System.currentTimeMillis();  
21.         System.out.println("四次转换分别耗时为"+(c2-c1)+"ms,"+(c3-c2)+"ms,"+(c4-c3)+"ms,"+(c5-c4)+"ms");  
22.     }     
23. }  

测试结果:

[java]  view plain  copy

1. 将39进制数  
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转化为62进制数并保留348位小数后去除无效位的结果为:  
3. CSicxuSbDy6b15mLRWpdQ03beazrJG5zcEBpMTZ7oxowiNnCN2V2M9xDYHbAF9g6yYEPhEPyQi1Jki1eamKZv9ceksCKkZzNjHDFv7wnmlCJSPVJtkFovcB65NUUYnHyMRYOMrt7wpQg1cy9UFqoVJOg6uicQZDDxdsxW7OdFtG60YEkAx3uuL6izr87hebmroc4ZHum8izmVRxylA1mARwFa.OVmjfCSTpwpEPSYDMTLpzKxV1LWaoEOdZV8Q5Ie2y8XeUDJ5M5Yh4QksANCxWcYlDW1v3XpvZ5AXC7XLlR5uyHK63kwPWpwJhLh6nm0ypGg1iwIWo1N7DP6JQEz3cP5kJx90WtGH6BOEmMgsRPC4cOnUYeS6Ckx8KmdNtMl1P2qN6cHsxZFYOgPg3MRHuNVMhuApscNlpkWlM11Yj5X8nAuLW140Pj4PEn5jgktOGO2y20LPgqM77GGz6d0PyF0H3EL7OwAVLVv1ARt4jlKbomwr1rRyqvV975kZtJVatHkNsr8PCLHGz63p59NIQRkXOd6cyROVnwCpRtmBi3dXU4Vlo0Z7  
4. 将39进制数  
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转化为2进制数并保留348位小数后去除无效位的结果为:  
6. 110101101010110010111000101001101101000100001111010000100111110011011011000111010010000001100101000011100001001100010000111001010100011011101110010111011001011101001110011110100011101010100110111001010011011110011001011111001111000010010000010101110100010010000011111111011100010001110111111111111101110000110010001001110100100001010111100110001101110001001011110000101011000010100110100101111000110111111100110010101111111101011100110100100100101111011010001111001111111010101100100010110100001100101010111101010110100101000100110100011110001011000011110011001010001110000001001101110011011001011100110101011011101110000011010100001011100101111001000110110111010010100101100101011110101010001101100010000000000111100110010111011110010101011011001111100110000100010000011001011110101111001001001011111110110110110110101011100101100011111011011010100001000100100101010100111100111110110001010001000011001011000111111101100111011011110110001101101010000100101000110010100000110011111100001001101000000010101000000010110000001000101111001111111100010100000111110010010100000110011100100111001010001101010111010110100100011011000011100001001100110010010100110001010001011110101111100111110110000001111101110011100110101101101101011110010111100001001000001110100100000001011111100010101110110110.011001010011011010110001001001110110111110110101111110111110010011011101011111000001100101000110110001001011101011100010000101010010110101111011111101111100010100010100010101100001100100111111011110001101111111010111111010011001000110101011101000100111100010111100010110110110100001000001110011110010110110110101110111000100100111111100110010010011  
7. 将62进制数  
8. CSicxuSbDy6b15mLRWpdQ03beazrJG5zcEBpMTZ7oxowiNnCN2V2M9xDYHbAF9g6yYEPhEPyQi1Jki1eamKZv9ceksCKkZzNjHDFv7wnmlCJSPVJtkFovcB65NUUYnHyMRYOMrt7wpQg1cy9UFqoVJOg6uicQZDDxdsxW7OdFtG60YEkAx3uuL6izr87hebmroc4ZHum8izmVRxylA1mARwFa.OVmjfCSTpwpEPSYDMTLpzKxV1LWaoEOdZV8Q5Ie2y8XeUDJ5M5Yh4QksANCxWcYlDW1v3XpvZ5AXC7XLlR5uyHK63kwPWpwJhLh6nm0ypGg1iwIWo1N7DP6JQEz3cP5kJx90WtGH6BOEmMgsRPC4cOnUYeS6Ckx8KmdNtMl1P2qN6cHsxZFYOgPg3MRHuNVMhuApscNlpkWlM11Yj5X8nAuLW140Pj4PEn5jgktOGO2y20LPgqM77GGz6d0PyF0H3EL7OwAVLVv1ARt4jlKbomwr1rRyqvV975kZtJVatHkNsr8PCLHGz63p59NIQRkXOd6cyROVnwCpRtmBi3dXU4Vlo0Z7转化为39进制数并保留348位小数后去除无效位的结果为:  
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  
10. 将2进制数  
11. 110101101010110010111000101001101101000100001111010000100111110011011011000111010010000001100101000011100001001100010000111001010100011011101110010111011001011101001110011110100011101010100110111001010011011110011001011111001111000010010000010101110100010010000011111111011100010001110111111111111101110000110010001001110100100001010111100110001101110001001011110000101011000010100110100101111000110111111100110010101111111101011100110100100100101111011010001111001111111010101100100010110100001100101010111101010110100101000100110100011110001011000011110011001010001110000001001101110011011001011100110101011011101110000011010100001011100101111001000110110111010010100101100101011110101010001101100010000000000111100110010111011110010101011011001111100110000100010000011001011110101111001001001011111110110110110110101011100101100011111011011010100001000100100101010100111100111110110001010001000011001011000111111101100111011011110110001101101010000100101000110010100000110011111100001001101000000010101000000010110000001000101111001111111100010100000111110010010100000110011100100111001010001101010111010110100100011011000011100001001100110010010100110001010001011110101111100111110110000001111101110011100110101101101101011110010111100001001000001110100100000001011111100010101110110110.011001010011011010110001001001110110111110110101111110111110010011011101011111000001100101000110110001001011101011100010000101010010110101111011111101111100010100010100010101100001100100111111011110001101111111010111111010011001000110101011101000100111100010111100010110110110100001000001110011110010110110110101110111000100100111111100110010010011转化为39进制数并保留348位小数后去除无效位的结果为:  
12. 1313624242ADAFSFGDRG424242ADAF679707464342424242ADAFS24242ADAFSFGDRG424242ADAFFGDRG142356679707464342424242ADAFSFGDRG67866679707464342424242ADAFSFGDRG96068563556667970746V6679707464342424423536858686242ADAFSFGD69979RG67970746434242427979742ADAFS.FGDRG646676679707464342424242ADAFSFGDRG9707464342424242ADAFSFGDRG2UXNCMJNcZJJTAYBcNcPCOYPEHSET1EJ9c650HV24PNO1CDYUMXBEVT8P294CL2Lc81MQIRODQD6YUH3HVa7ADOUHW696PF2c0KU8XJJa6WYYASbGYRTUCCVO0BD8AGZEMSW2MWTCI3NbHFZW8GIC0FTRYNV8TCVAQF9N6AZHMDB5N16IWGUPHU812HN7AMT97LIUFX9DEZ5B841I6KMS3JGNMYVF79FXQHK8Y8DJKUUV3AaJVEWS80ZSaC0IKcVCXAaB5I3WBC2KAIGG0AIbR2Z2Ob  
13. 四次转换分别耗时为468ms,169ms,195ms,75ms  

使用php：

<?php
//超大整数（10进制）转换为二进制数
$n='12121111111141417908308410289576890247569807456709384756890743906790486790845';
//$n='15';//1111
//$n='257';//100000001
$r='';//结果
while ($n){
  //$n整除2，商$m、余数$k
  $k=0;
  $m='';
  do{
    $k=$k*10 + substr($n,0,1);
    if ($m!='' || $k>1) $m.=floor($k/2);
    $k=$k % 2;
    $n=substr($n,1);
    //$r=$k . $r;
  }while($n!='');
  //echo "r=$r;m=$m\n";//break;
  //下一轮除法
  $n=$m;
  $r=$k . $r;
}
echo $r;
?>