java自定义大数（进制可在2-62之间相互转换）

java自定义大数(进制可在2~62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2~62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 /*

* 0~9用0~9表示,10~35用A~Z表示,36~61用a~z表示 */

@SuppressWarnings(\)

public class BigNumber extends Exception{

private String MyNumber;//大数 private int SystemNumber;//进制数 public BigNumber(){ }

this(\,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() { }

public void setMyNumber(String myNumber) { }

public int getSystemNumber() { }

public void setSystemNumber(int systemNumber) { }

this.SystemNumber = systemNumber; return this.SystemNumber; this.MyNumber = myNumber; return this.MyNumber;

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0

throw (new NumberFormatException(\错误!大数为空或大数字符串长度为0

\+this.MyNumber)); }

if(this.SystemNumber<=1||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.

throw (new NumberFormatException(\错误!大数不合法 \+this.MyNumber));

equals(\)){ }

if(this.MyNumber.equals(\)||this.MyNumber.equals(\)){ }

for(int i=0,a,k=-1;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;

a=Chartoint(this.MyNumber.charAt(i)); if(i!=0&&(a==-1||a==-2)){ }

if(a>=this.SystemNumber||a==-4){ }

if(a==-3){ }

if(k==-1){ }

k=i;

throw (new NumberFormatException(\错误!出现了第二个小数点 }else{

throw (new NumberFormatException(\错误!该字符不在该进制合法字符中 throw (new NumberFormatException(\错误!符号位只能在大数首部 throw (new NumberFormatException(\错误!大数不合法 \+this.MyNumber));

\+this.MyNumber.charAt(i)+\+i+\));

\+this.MyNumber.charAt(i)+\+i+\));

\+this.MyNumber.charAt(i)+\+i+\));

}

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

return \+Sub_Positive(s2,s1,n);

} 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()); String R=getComplement(new

BigNumber(AddToLeft(Ins2,Inlength)+\+AddToRight(Dos2,Dolength),n)).MyNumber;

R=Add_Positive(s1,R,n);

s2=\+AddToLeft(\,Dolength); R=Add_Positive(R,s2,n);

return Format(R.substring(1),true); }

protected static String Mult_Positive(String s1,String s2,int n){//自定义两个正数乘法运算; 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 Ins1+=Dos1; Ins2+=Dos2;

int[]A=new int[Ins1.length()+Ins2.length()-1]; for(int i=Ins1.length()-1;i>=0;i--){ }

String s=\;

for(int i=A.length-1;i>=1;i--){ }

s=Inttochar(A[i]%n)+s; A[i-1]+=A[i]/n;

for(int j=Ins2.length()-1;j>=0;j--){ }

A[i+j]+=Chartoint(Ins1.charAt(i))*Chartoint(Ins2.charAt(j));

s=Inttochar(A[0]/n)+\+Inttochar(A[0]%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); String[]B=new String[n];B[0]=\; for(int i=1;i

B[i]=Add_Positive(B[i-1],Ins2,n);//B[i]=Mult_Positive(Ins2,Inttochar(i)+\ }

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){

}

return Format(Ds0,true);

if(i!=Ins1.length()){ } i++;

if(i

//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j=Binary(B,Ds1); Ds0+=Inttochar(j);

Ds1=Sub_Positive(Ds1,B[j],n);

Ds1+=Ins1.charAt(i); Ds1+=\;

if(Ds1.equals(\)||i-Ins1.length()-1==f){ }

break;

}else{

查找获取整数部分

}else{

Ds0+=\;

}

protected 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)

int low=0,height=A.length-1,middle,Compare;

while(low<=height){ }

return height;

middle=(low+height)/2;

Compare=ComparetoString(key,A[middle]); if(Compare==0){ }

return middle; low=middle+1; height=middle-1; }else if(Compare>0){ }else{

}

public BigNumber Add(BigNumber b){//自定义加法运算 if(this.SystemNumber!=b.SystemNumber){ }

String s;

if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){

s=Sub_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber);

s=Add_Positive(this.MyNumber,b.MyNumber,this.SystemNumber);

}else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){

throw (new NumberFormatException(\错误!这两个大数进制数不一致

(\+this.SystemNumber+\+b.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.System

s=(s.equals(\)?\:\)+s;

Number); }

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){ }

String s;

if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){

s=Mult_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber);

s=(s.equals(\)?\:\)+s;

s=Mult_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumber);

s=(s.equals(\)?\:\)+s;

s=Mult_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.Syste

mNumber); }

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(\)||b.toString().equals(\)){ }

if(this.SystemNumber!=b.SystemNumber){ } if(f<0){ }

String s;

if(IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){

s=Division_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumbe

s=(s.equals(\)?\:\)+s;

r,f);

}else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){

s=Division_Positive(this.MyNumber,b.MyNumber,this.SystemNumber,f);

}else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){

throw (new NumberFormatException(\错误!小数点后的位数应为非负

值!(\+f+\));

throw (new NumberFormatException(\错误!这两个大数进制数不一致!

(\+this.SystemNumber+\+b.SystemNumber+\));

throw (new NumberFormatException(\错误!除数不能为零:\+b)); s=Mult_Positive(this.MyNumber,b.MyNumber,this.SystemNumber);

}else if(IsPositive(this.MyNumber)&&!IsPositive(b.MyNumber)){

throw (new NumberFormatException(\错误!这两个大数进制数不一致

(\+this.SystemNumber+\+b.SystemNumber+\));

}else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){

}else{

s=Division_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumbe

s=(s.equals(\)?\:\)+s;

r,f);

s=Division_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.S

ystemNumber,f); }

return new BigNumber(s,this.SystemNumber); }

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); return Division(b,50); }

protected 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

Ins1=getInteger(s1),Dos1=getDecimal(s1),Ins2=getInteger(s2),Dos2=getDecimal(s2);

int Dolength=Math.max(Dos1.length(), Dos2.length()); Ins1+=AddToRight(Dos1,Dolength); Ins2+=AddToRight(Dos2,Dolength); if(Ins1.length()>Ins2.length()){ }

for(int i=0;i

return 0; }

public int Compareto(BigNumber b){ if(this.SystemNumber!=b.SystemNumber){ }

String s1=Format(this.MyNumber,false),s2=Format(b.MyNumber,false); if(IsPositive(s1)&&IsPositive(s2)){

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+\));

if(Ins1.charAt(i)>Ins2.charAt(i)){ }

return 1; return -1;

}else if(Ins1.charAt(i)

}else if(Ins1.length()

return ComparetoString(s1,s2);//int a=s1.CompareTo(s2); return a!=0?(a>0?1:-1):0;

}else if(IsPositive(s1)&&!IsPositive(s2)){ } }

protected static String Conver1(long n,int R){//将十进制数n转换为R进制数 if(R==10){ }

String s=\; while(true){ } /*

* 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): * 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() * Hex2[i]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,String[]Hex1,String[]Hex2){

Ins=Add_Positive(Ins,Mult_Positive(Hex1[Chartoint(s.charAt(s.length()-1

}

return Ins;

String Ins=\;//初始化新进制对应的大数 for(int i=0;i

s=Inttochar((int)(n%R))+s; n/=R; if(n==0){ }

return s; return n+\; return 1; return -1;

return -ComparetoString(s1.substring(1),s2.substring(1));

}else if(!IsPositive(s1)&&IsPositive(s2)){ }else{

}

-i))],Hex2[i],R2),R2);

}

protected 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; if(R2==R1){

}

int a=getDecimal(MyNumber).length()-f; if(a>0){ }

return MyNumber;

return MyNumber.substring(0,MyNumber.length()-a);

if(MyNumber.charAt(0)=='+'||MyNumber.charAt(0)=='-'){ }

String[]Hex1=new String[R1+1];

for(int i=0;i<=R1;i++){//将0~R1的所有数字转换为R2进制 }

String[]Hex2=new String[N+1];Hex2[0]=\;

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 }

Hex2[i]=Mult_Positive(Hex2[i-1],Hex1[R1],R2);

Hex1[i]=Conver1(i,R2); char c=MyNumber.charAt(0);

return c+Conversion(MyNumber.substring(1),R1,R2,N,f);

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 * 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()){ }

Ins1=AddToLeft(Ins1,Ins1.length()+(N-Ins1.length()%N)%N);//扩充Ins1 Dos1=AddToRight(Dos1,Dos1.length()+(N-Dos1.length()%N)%N);//扩充Dos1 String Ins=\,Dos=\,x=\;//new

Dos1=Dos1.substring(0,a);

保留a位

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;i

if(i

Ins=Add_Positive(Ins,Mult_Positive(Conver(Ins1.substring(Ins1.length()-

}

if(i

(i+1)*N,Ins1.length()-i*N),R1,R2,Hex1,Hex2),x,R2),R2);

Dos=Add_Positive(Dos,Mult_Positive(Conver(Dos1.substring(Dos1.length()-

}

return Ins+Dos.substring(1);//或Add_Positive(Ins,Dos,R2);

} }

x=Mult_Positive(x,Hex2[N],R2);

if(i==i2-1){

Dos=Division_Positive(Dos,Mult_Positive(x,Hex2[N],R2),R2,f);

(i+1)*N,Dos1.length()-i*N),R1,R2,Hex1,Hex2),x,R2),R2);

}

public BigNumber HexConversion(int R,int N,int f){ if(R<=1||R>=63){//进制数不合法

throw (new NumberFormatException(\错误!进制数不合法

\+this.SystemNumber)); }

if(N<=0){ } if(f<0){ }

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a^n,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(n<0){ }

return new

BigNumber(\,this.SystemNumber).Division(this.Power(-n,f),f);

throw (new NumberFormatException(\错误!小数点后的位数应为非负

值!(\+f+\));

throw (new NumberFormatException(\错误!每小组的字符串长度不能小于

1!(\+N+\));

String s=Conver1(n,2);

BigNumber r=new BigNumber(\,this.SystemNumber),k=this;

if(s.charAt(s.length()-1-i)=='1'){ }

k=k.Mult(k);

r=r.Mult(k);

for(int i=0;i

public BigNumber Power(int n){//求a^b,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=='.'){ }

for(int i=1;i

return new BigNumber(s,b.SystemNumber); }

protected 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 if(s.length()>=length){ }

while(s.length()

return s.substring(s.length()-length,s.length()); }

protected static String AddToRight(String s,int length){//若s的长度大于或

return s;

c=b.MyNumber.charAt(i); if(c=='.'){ }

s+=c;

s+=Inttochar(b.SystemNumber-1-Chartoint(c)); }else{ s+=c;

s+=Inttochar(b.SystemNumber-1-Chartoint(c));

}else{

return r;

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 if(s.length()>=length){ }

while(s.length()

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;

s=MyNumber.substring(1); s=MyNumber;

if(MyNumber.charAt(0)=='+'||MyNumber.charAt(0)=='-'){ }

if(getPointPlace(s)==0){ }

int Ins=0,Dos=s.length(),pointplace=getPointPlace(s); for(int i=0;i

if(s.charAt(i)=='0'){ }

Ins++; break; }else{ s=\+s; s=s+\; s=s+\;

}else if(getPointPlace(s)==s.length()-1){ }else if(getPointPlace(s)==s.length()){ }else{

return s;

}

for(int i=s.length()-1;i>pointplace+1;i--){

if(s.charAt(i)=='0'){ }

s=(MyNumber.charAt(0)=='-'?\:\)+s.substring(Ins,Dos); if(DealZero&&s.substring(s.length()-2).equals(\)){

return s.substring(0,s.length()-2); }else{

Dos--; break; }else{

}

}

return 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 true; }

protected static int getPointPlace(String MyNumber){//返回该大数小数点位置

for(int i=0;i

return MyNumber.length(); }

protected static String Abs(String MyNumber){//返回该大数的绝对值 MyNumber=Format(MyNumber,true);

return MyNumber.substring(MyNumber.charAt(0)=='-'?1:0); }

protected static String Opposite(String MyNumber){//返回该大数的相反数

return MyNumber.charAt(0)=='-'?Abs(MyNumber):\+Abs(MyNumber); }

protected static String getInteger(String MyNumber){//返回该大数的整数部分 MyNumber=Format(MyNumber,false);

return MyNumber.substring(0,getPointPlace(MyNumber)); }

protected static String getDecimal(String MyNumber){//返回该大数的小数部分 MyNumber=Format(MyNumber,false); String

s=getPointPlace(MyNumber)==MyNumber.length()?\:MyNumber.substring(getPointPlace(MyNumber)+1);

return (MyNumber.charAt(0)=='-'?\:\)+s;

}

if(c>='0'&&c<='9'){

return c-'0'; return c-'A'+10; }else if(c>='A'&&c<='Z'){ }else if(c>='a'&&c<='z'){

protected static int Chartoint(char c){//字符化为数字

if(MyNumber.charAt(i)=='.'){ }

return i; return false;

}

} }

}

return c-'a'+36; return -1; return -2; return -3; return -4;

}else if(c=='+'){//正号 }else if(c=='-'){//负号 }else if(c=='.'){//小数点 }else{//其他

protected static char Inttochar(int n){//数字化为字符

if(n>=0&&n<=9){//0~9用0~9表示 }

return (char)(n+48); return (char)(n+55); return (char)(n+61); return '~';

}else if(n>=10&&n<=35){//10~35用A~Z表示 }else if(n>=36&&n<=61){//36~61用a~z表示 }else{//小于0或大于61的数用'~'表示

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.java

public class TestBigNumber {

public static void main(String[]args){

String

MyNumber=\DRG424242ADAFFGDRG142356679707464342424242ADAFSFGDRG67866679707464342424242ADAFSFGDRG96068563556667970746V6679707464342424423536858686242ADAFSFGD69979RG67970746434242427979742ADAFS.FGDRG646676679707464342424242ADAFSFGDRG9707464342424242ADAFSFGDRG342424242ADAFSFGDRG67970746434266797074643426679707464342424242ADAFSFGDRG424242ADAFSFGDRG424242ADAFSFGDRG242423424266797074643424

24242ADAFSFGDRG534364547455686679707464342424242ADAFSFGDRG6679707464342424242ADAFSFGDRGDADA6679707464342424242ADAFSFGDRGAFSFS6679707464342424242ADAFSFGDRG\; }

int R1=39,R2=62,R3=2; int f=100;

BigNumber b=new BigNumber(MyNumber,R1); }

long c1=System.currentTimeMillis();

System.out.println(\将\+R1+\进制数\+MyNumber+\转化为\+R2+\进制数并保留long c2=System.currentTimeMillis();

System.out.println(\将\+R1+\进制数\+MyNumber+\转化为\+R3+\进制数并保留long c3=System.currentTimeMillis();

System.out.println(\两次转换分别耗时为 \+(c2-c1)+\+(c3-c2)+\);

\+f+\位小数后去除无效位的结果为:\+b.HexConversion(R2, f));

\+f+\位小数后去除无效位的结果为:\+b.HexConversion(R3, f));

测试结果:

将39进制数

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转化为62进制数并保留100位小数后去除无效位的结果

为:CSicxuSbDy6b15mLRWpdQ03beazrJG5zcEBpMTZ7oxowiNnCN2V2M9xDYHbAF9g6yYEPhEPyQi1Jki1eamKZv9ceksCKkZzNjHDFv7wnmlCJSPVJtkFovcB65NUUYnHyMRYOMrt7wpQg1cy9UFqoVJOg6uicQZDDxdsxW7OdFtG60YEkAx3uuL6izr87hebmroc4ZHum8izmVRxylA1mARwFa.OVmjfCSTpwpEPSYDMTLpzKxV1LWaoEOdZV8Q5Ie2y8XeUDJ5M5Yh4QksANCxWcYlDW1v3XpvZ5AXC7XLlR5uyHK63kwPWpwJhLh6 将39进制数

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

RGDADA6679707464342424242ADAFSFGDRGAFSFS6679707464342424242ADAFSFGDRG转化为2进制数并保留100位小数后去除无效位的结果

为:110101101010110010111000101001101101000100001111010000100111110011011011000111010010000001100101000011100001001100010000111001010100011011101110010111011001011101001110011110100011101010100110111001010011011110011001011111001111000010010000010101110100010010000011111111011100010001110111111111111101110000110010001001110100100001010111100110001101110001001011110000101011000010100110100101111000110111111100110010101111111101011100110100100100101111011010001111001111111010101100100010110100001100101010111101010110100101000100110100011110001011000011110011001010001110000001001101110011011001011100110101011011101110000011010100001011100101111001000110110111010010100101100101011110101010001101100010000000000111100110010111011110010101011011001111100110000100010000011001011110101111001001001011111110110110110110101011100101100011111011011010100001000100100101010100111100111110110001010001000011001011000111111101100111011011110110001101101010000100101000110010100000110011111100001001101000000010101000000010110000001000101111001111111100010100000111110010010100000110011100100111001010001101010111010110100100011011000011100001001100110010010100110001010001011110101111100111110110000001111101110011100110101101101101011110010111100001001000001110100100000001011111100010101110110110.01100101001101101011000100100111011011111011010111111011111001001101110101111100000110010100011011 两次转换分别耗时为 407ms,312ms

RGDADA6679707464342424242ADAFSFGDRGAFSFS6679707464342424242ADAFSFGDRG转化为2进制数并保留100位小数后去除无效位的结果

为:110101101010110010111000101001101101000100001111010000100111110011011011000111010010000001100101000011100001001100010000111001010100011011101110010111011001011101001110011110100011101010100110111001010011011110011001011111001111000010010000010101110100010010000011111111011100010001110111111111111101110000110010001001110100100001010111100110001101110001001011110000101011000010100110100101111000110111111100110010101111111101011100110100100100101111011010001111001111111010101100100010110100001100101010111101010110100101000100110100011110001011000011110011001010001110000001001101110011011001011100110101011011101110000011010100001011100101111001000110110111010010100101100101011110101010001101100010000000000111100110010111011110010101011011001111100110000100010000011001011110101111001001001011111110110110110110101011100101100011111011011010100001000100100101010100111100111110110001010001000011001011000111111101100111011011110110001101101010000100101000110010100000110011111100001001101000000010101000000010110000001000101111001111111100010100000111110010010100000110011100100111001010001101010111010110100100011011000011100001001100110010010100110001010001011110101111100111110110000001111101110011100110101101101101011110010111100001001000001110100100000001011111100010101110110110.01100101001101101011000100100111011011111011010111111011111001001101110101111100000110010100011011 两次转换分别耗时为 407ms,312ms

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