java自定义大数(进制可在2-62之间相互转换).doc

java自定义大数(进制可在262之间相互转化) 闲着没事做，写了一个自定义大数，并实现了两个大数的加减乘除以及比较，以及262进制任意数(包括但不限于正整数与负整数)的相互转化大数类: BigNumber.java/* * 09用09表示,1035用AZ表示,3661用az表示 */SuppressWarnings(serial)public class BigNumber extends Exceptionprivate String MyNumber;/大数private int SystemNumber;/进制数public BigNumber()this(0,10); public BigNumber(String MyNumber) this(MyNumber,10); public BigNumber(String MyNumber,int SystemNumber) this.MyNumber=MyNumber; this.SystemNumber=SystemNumber; this.Islegal(); public String getMyNumber() return this.MyNumber;public void setMyNumber(String myNumber) this.MyNumber = myNumber;public int getSystemNumber() return this.SystemNumber;public void setSystemNumber(int systemNumber) this.SystemNumber = systemNumber; public void Islegal()/判断该字符串是否合法 if(this.MyNumber=null|this.MyNumber.length()=0)/大数为空或大数字符串长度为0throw (new NumberFormatException(错误!大数为空或大数字符串长度为0 +this.MyNumber); if(this.SystemNumber=63)/进制数不合法throw (new NumberFormatException(错误!进制数不合法 +this.SystemNumber); if(this.MyNumber.equals(+)|this.MyNumber.equals(-)|this.MyNumber.equals(.) throw (new NumberFormatException(错误!大数不合法 +this.MyNumber); if(this.MyNumber.equals(+.)|this.MyNumber.equals(-.) throw (new NumberFormatException(错误!大数不合法 +this.MyNumber); for(int i=0,a,k=-1;i=this.SystemNumber|a=-4) throw (new NumberFormatException(错误!该字符不在该进制合法字符中 +this.MyNumber.charAt(i)+(+i+); if(a=-3) if(k=-1) k=i; else throw (new NumberFormatException(错误!出现了第二个小数点 +this.MyNumber.charAt(i)+(+i+); protected static String Add_Positive(String s1,String s2,int n)/自定义两个正数加法运算 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 int Inlength=Math.max(Ins1.length(), Ins2.length(),Dolength=Math.max(Dos1.length(), Dos2.length(); Ins1=AddToLeft(Ins1,Inlength+1)+AddToRight(Dos1,Dolength); Ins2=AddToLeft(Ins2,Inlength+1)+AddToRight(Dos2,Dolength); String s=; for(int i=Inlength+Dolength,m,y=0;i=0;i-) m=Chartoint(Ins1.charAt(i)+Chartoint(Ins2.charAt(i)+y; s=Inttochar(m%n)+s; y=m/n; int pointplace=s.length()-Dolength; s=Format(s.substring(0,pointplace)+.+s.substring(pointplace),true); return s; protected static String Sub_Positive(String s1,String s2,int n)/自定义两个正数减法运算; if(ComparetoString(s1,s2)0)/如果s1=0;i-) for(int j=Ins2.length()-1;j=0;j-) Ai+j+=Chartoint(Ins1.charAt(i)*Chartoint(Ins2.charAt(j); String s=; for(int i=A.length-1;i=1;i-) s=Inttochar(Ai%n)+s; Ai-1+=Ai/n; s=Inttochar(A0/n)+Inttochar(A0%n)+s;/特别要注意+ int pointlength=Dos1.length()+Dos2.length();/获取小数点后的位数 return Format(s.substring(0,s.length()-pointlength)+.+s.substring(s.length()-pointlength),true); protected 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); 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 int Dolength=Math.max(Dos1.length(), Dos2.length(); Ins1+=AddToRight(Dos1,Dolength); Ins2+=AddToRight(Dos2,Dolength); StringB=new Stringn;B0=0; for(int i=1;in;i+) Bi=Add_Positive(Bi-1,Ins2,n);/Bi=Mult_Positive(Ins2,Inttochar(i)+, n); int i=Math.min(Ins1.length(), Ins2.length()-2,j; String Ds0=,Ds1=Ins1.substring(0,i);/初始化整数部分余数部分 while(true) if(i!=Ins1.length() if(iIns1.length() Ds1+=Ins1.charAt(i); else Ds1+=0; if(Ds1.equals(00)|i-Ins1.length()-1=f) break; /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 j=Binary(B,Ds1); Ds0+=Inttochar(j);Ds1=Sub_Positive(Ds1,Bj,n); else Ds0+=.; i+; return Format(Ds0,true); protected static int Binary(StringA,String key)/二分查找key(数组A已经按从小到大的顺序排好序) int low=0,height=A.length-1,middle,Compare;while(low0)low=middle+1;elseheight=middle-1;return height; public BigNumber Add(BigNumber b)/自定义加法运算 if(this.SystemNumber!=b.SystemNumber) throw (new NumberFormatException(错误!这两个大数进制数不一致 (+this.SystemNumber+,+b.SystemNumber+); String s; if(IsPositive(this.MyNumber)&IsPositive(b.MyNumber) s=Add_Positive(this.MyNumber,b.MyNumber,this.SystemNumber); else if(IsPositive(this.MyNumber)&!IsPositive(b.MyNumber) s=Sub_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber); else if(!IsPositive(this.MyNumber)&IsPositive(b.MyNumber) s=Sub_Positive(b.MyNumber,Opposite(this.MyNumber),this.SystemNumber); else s=Add_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber); s=(s.equals(0)?:-)+s; return new BigNumber(s,this.SystemNumber); public BigNumber Sub(BigNumber b)/自定义减法运算 return this.Add(new BigNumber(Opposite(b.MyNumber),b.SystemNumber); public BigNumber Mult(BigNumber b)/自定义乘法运算 if(this.SystemNumber!=b.SystemNumber) throw (new NumberFormatException(错误!这两个大数进制数不一致 (+this.SystemNumber+,+b.SystemNumber+); String s; if(IsPositive(this.MyNumber)&IsPositive(b.MyNumber) s=Mult_Positive(this.MyNumber,b.MyNumber,this.SystemNumber); else if(IsPositive(this.MyNumber)&!IsPositive(b.MyNumber) s=Mult_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber); s=(s.equals(0)?:-)+s; else if(!IsPositive(this.MyNumber)&IsPositive(b.MyNumber) s=Mult_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumber); s=(s.equals(0)?:-)+s; else s=Mult_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber); return new BigNumber(s,this.SystemNumber); 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); if(b.toString().equals(-0)|b.toString().equals(0) throw (new NumberFormatException(错误!除数不能为零:+b); if(this.SystemNumber!=b.SystemNumber) throw (new NumberFormatException(错误!这两个大数进制数不一致! (+this.SystemNumber+,+b.SystemNumber+); if(fs2返回1,s1=s2返回0,s1Ins2.length() return 1; else if(Ins1.length()Ins2.length() return -1; for(int i=0;iIns2.charAt(i) return 1; else if(Ins1.charAt(i)0?1:-1):0; else if(IsPositive(s1)&!IsPositive(s2) return 1; else if(!IsPositive(s1)&IsPositive(s2) return -1; else return -ComparetoString(s1.substring(1),s2.substring(1); protected static String Conver1(long n,int R)/将十进制数n转换为R进制数 if(R=10) return n+; String s=; while(true) s=Inttochar(int)(n%R)+s; n/=R; if(n=0) return s; /* * 将R1进制正整数s转化为R2进制数 * (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): * Hex1i是十进制数i转换为R2进制数后的数,0=i=R1;Hex2i=Hex1R1i,0=i=s.length() * Hex2i是十进制数R1转换为R2进制数后的i次方 */ protected static String Conver(String s,int R1,int R2,StringHex1,StringHex2)String Ins=0;/初始化新进制对应的大数for(int i=0;i0) return MyNumber.substring(0,MyNumber.length()-a); return MyNumber; if(MyNumber.charAt(0)=+|MyNumber.charAt(0)=-) char c=MyNumber.charAt(0); return c+Conversion(MyNumber.substring(1),R1,R2,N,f); StringHex1=new StringR1+1;for(int i=0;i=R1;i+)/将0R1的所有数字转换为R2进制Hex1i=Conver1(i,R2);StringHex2=new StringN+1;Hex20=1;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Hex2i=Mult_Positive(Hex2i-1,Hex1R1,R2); String Ins1=getInteger(MyNumber),Dos1=getDecimal(MyNumber);/获取整数部分及小数部分/* * 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 * 当MyNumber小数位很多，而f较小时在一定程度上能提高效率 */int a=(int)Math.ceil(f*Math.log(R2)/Math.log(R1)+1;if(a=Dos1.length()Dos1=Dos1.substring(0,a);Ins1=AddToLeft(Ins1,Ins1.length()+(N-Ins1.length()%N)%N);/扩充Ins1Dos1=AddToRight(Dos1,Dos1.length()+(N-Dos1.length()%N)%N);/扩充Dos1String 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 int Count=Math.max(Ins1.length()/N,Dos1.length()/N);/循环次数 for(int i=0,i1=Ins1.length()/N,i2=Dos1.length()/N;iCount;i+) if(ii1)Ins=Add_Positive(Ins,Mult_Positive(Conver(Ins1.substring(Ins1.length()-(i+1)*N,Ins1.length()-i*N),R1,R2,Hex1,Hex2),x,R2),R2);if(ii2) Dos=Add_Positive(Dos,Mult_Positive(Conver(Dos1.substring(Dos1.length()-(i+1)*N,Dos1.length()-i*N),R1,R2,Hex1,Hex2),x,R2),R2);if(i=i2-1) Dos=Division_Positive(Dos,Mult_Positive(x,Hex2N,R2),R2,f); x=Mult_Positive(x,Hex2N,R2);return Ins+Dos.substring(1);/或Add_Positive(Ins,Dos,R2); public BigNumber HexConversion(int R,int N,int f) if(R=63)/进制数不合法throw (new NumberFormatException(错误!进制数不合法 +this.SystemNumber); if(N=0) throw (new NumberFormatException(错误!每小组的字符串长度不能小于1!(+N+); if(f0) throw (new NumberFormatException(错误!小数点后的位数应为非负值!(+f+); return new BigNumber(Conversion(this.MyNumber,this.SystemNumber,R,N,f),R); public BigNumber HexConversion(int R,int f) int N=(int)Math.pow(Math.max(getInteger(this.MyNumber).length(),getDecimal(this.MyNumber).length(),0.6)+1; return HexConversion(R,N,f); public BigNumber HexConversion(int R) return HexConversion(R,50); public BigNumber Power(int n,int f)/求an,n为十进制整数(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); if(n0) return new BigNumber(1,this.SystemNumber).Division(this.Power(-n,f),f); String s=Conver1(n,2);BigNumber r=new BigNumber(1,this.SystemNumber),k=this; for(int i=0;is.length();i+) if(s.charAt(s.length()-1-i)=1) r=r.Mult(k); k=k.Mult(k); return r; public BigNumber Power(int n)/求ab,b为十进制整数 return Power(n,50); protected static BigNumber getComplement(BigNumber b)/获得 b的每位取反后的大数 String s=; char c=b.MyNumber.charAt(0); if(c=+|c=-|c=.) s+=c; else s+=Inttochar(b.SystemNumber-1-Chartoint(c); for(int i=1;i=length) return s; while(s.length()=length) return s; while(s.length()length) s=s+0000000000; return s.substring(0,length); protected 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 new BigNumber(MyNumber,62);/若该大数不合法则抛出异常String s; if(MyNumber.charAt(0)=+|MyNumber.charAt(0)=-) s=MyNumber.substring(1); else s=MyNumber; if(getPointPlace(s)=0) s=0+s; else if(getPointPlace(s)=s.length()-1) s=s+0; else if(getPointPlace(s)=s.length() s=s+.0; int Ins=0,Dos=s.length(),pointplace=getPointPlace(s); for(int i=0;ipointplace+1;i-) if(s.charAt(i)=0) Dos-; else break; s=(MyNumber.charAt(0)=-?-:)+s.substring(Ins,Dos);if(DealZero&s.substring(s.length()-2).equals(.0)return s.substring(0,s.length()-2);elsereturn s; protected 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) if(MyNumber.charAt(0)=-) return false; return true; protected static int getPointPlace(String MyNumber)/返回该大数小数点位置 for(int i=0;i=0&c=A&c=a&c=0&n=10&n=36&n=61)/3661用az表示return (char)(n+61);else/小于0或大于61的数用表示return ;public String toString()return Format(this.MyNumber,true);顺便写了个测试类,限于篇幅这里只测试进制转换的效果与效率测试类: TestBigNumber.javapublic class TestBigNumber public static void main(Stringargs)String MyNumber=1313624242ADAFSFGDRG424242ADAF679707464342424242ADAFS24242ADAFSFGDRG424242ADAFFGD