java自定义大数(进制可在2~62之间)
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<p></p><p>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</p><p>BigNumber类:</p>
/* * 0~9用0~9表示,10~35用A~Z表示,36~61用a~z表示 */@SuppressWarnings("serial")public class BigNumber extends Exception{private 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<=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.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.MyNumber.length();i++){ a=Chartoint(this.MyNumber.charAt(i)); if(i!=0&&(a==-1||a==-2)){ throw (new NumberFormatException("错误!符号位只能在大数首部 "+this.MyNumber.charAt(i)+"("+i+")")); } if(a>=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+")")); } } } } private 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; } private 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<s2则返回-(s2-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="0."+AddToLeft("1",Dolength); R=Add_Positive(R,s2,n); return Format(R.substring(1),true); } private 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--){ for(int j=Ins2.length()-1;j>=0;j--){ A[i+j]+=Chartoint(Ins1.charAt(i))*Chartoint(Ins2.charAt(j)); } } String s=""; for(int i=A.length-1;i>=1;i--){ s=Inttochar(A[i]%n)+s; A[i-1]+=A[i]/n; } 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); } 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); 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]="0"; for(int i=1;i<n;i++){ B[i]=Add_Positive(B[i-1],Ins2,n);//B[i]=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(i<Ins1.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,B[j],n); }else{ Ds0+="."; } i++; } return Format(Ds0,true); } 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) int low=0,height=A.length-1,middle,Compare;while(low<=height){middle=(low+height)/2;Compare=ComparetoString(key,A[middle]);if(Compare==0){return middle;}else if(Compare>0){low=middle+1;}else{height=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(f<0){ throw (new NumberFormatException("错误!小数点后的位数应为非负值!("+f+")")); } String s; 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)){ s=Division_Positive(this.MyNumber,Opposite(b.MyNumber),this.SystemNumber,f); s=(s.equals("0")?"":"-")+s; }else if(!IsPositive(this.MyNumber)&&IsPositive(b.MyNumber)){ s=Division_Positive(Opposite(this.MyNumber),b.MyNumber,this.SystemNumber,f); s=(s.equals("0")?"":"-")+s; }else{ s=Division_Positive(Opposite(this.MyNumber),Opposite(b.MyNumber),this.SystemNumber,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); } 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 String 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()){ return 1; }else if(Ins1.length()<Ins2.length()){ return -1; } for(int i=0;i<Ins1.length();i++){ if(Ins1.charAt(i)>Ins2.charAt(i)){ return 1; }else if(Ins1.charAt(i)<Ins2.charAt(i)){ return -1; } } return 0; } public int Compareto(BigNumber b){ if(this.SystemNumber!=b.SystemNumber){ 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+")")); } String s1=Format(this.MyNumber,false),s2=Format(b.MyNumber,false); if(IsPositive(s1)&&IsPositive(s2)){ return ComparetoString(s1,s2);//int a=s1.CompareTo(s2); return a!=0?(a>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)); } } private static String ConverFromTen(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): * 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 */ private static String Conver(String s,int R2,String[]Hex1,String[]Hex2){String Ins="0";//初始化新进制对应的大数for(int i=0;i<s.length();i++){ Ins=Add_Positive(Ins,Mult_Positive(Hex1[Chartoint(s.charAt(s.length()-1-i))],Hex2[i],R2),R2); }return Ins; } 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; if(R2==R1){ int a=getDecimal(MyNumber).length()-f; if(a>0){ 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); } 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Hex1[i]=ConverFromTen(i,R2);}String[]Hex2=new String[N+1];Hex2[0]="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Hex2[i]=Mult_Positive(Hex2[i-1],Hex1[R1],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;i<Count;i++){ if(i<i1){Ins=Add_Positive(Ins,Mult_Positive(Conver(Ins1.substring(Ins1.length()-(i+1)*N,Ins1.length()-i*N),R2,Hex1,Hex2),x,R2),R2);}if(i<i2){ Dos=Add_Positive(Dos,Mult_Positive(Conver(Dos1.substring(Dos1.length()-(i+1)*N,Dos1.length()-i*N),R2,Hex1,Hex2),x,R2),R2);if(i==i2-1){ Dos=Division_Positive(Dos,Mult_Positive(x,Hex2[N],R2),R2,f); }} x=Mult_Positive(x,Hex2[N],R2);}return Ins+Dos.substring(1);//或Add_Positive(Ins,Dos,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){ throw (new NumberFormatException("错误!每小组的字符串长度不能小于1!("+N+")")); } if(f<0){ 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){ if(n<0){ return new BigNumber("1",this.SystemNumber).Division(this.Power(-n,f),f); } String s=ConverFromTen(n,2);BigNumber r=new BigNumber("1",this.SystemNumber),k=this; for(int i=0;i<s.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){ return Power(n,50); }*/ private 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<b.MyNumber.length();i++){ c=b.MyNumber.charAt(i); if(c=='.'){ s+=c; }else{ s+=Inttochar(b.SystemNumber-1-Chartoint(c)); } } return new BigNumber(s,b.SystemNumber); } 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 if(s.length()>=length){ return s; } while(s.length()<length){ s="0000000000"+s; } return s.substring(s.length()-length,s.length()); } 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 if(s.length()>=length){ return s; } while(s.length()<length){ s=s+"0000000000"; } return s.substring(0,length); } 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 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;i<pointplace-1;i++){ if(s.charAt(i)=='0'){ Ins++; }else{ break; } } for(int i=s.length()-1;i>pointplace+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);}else{return s;}} 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) if(MyNumber.charAt(0)=='-'){ return false; } return true; } private static int getPointPlace(String MyNumber){//返回该大数小数点位置 for(int i=0;i<MyNumber.length();i++){ if(MyNumber.charAt(i)=='.'){ return i; } } return MyNumber.length(); } private 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); } private static String Opposite(String MyNumber){//返回该大数的相反数 return MyNumber.charAt(0)=='-'?Abs(MyNumber):"-"+Abs(MyNumber); } private 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)); } private 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()?"0":MyNumber.substring(getPointPlace(MyNumber)+1); return (MyNumber.charAt(0)=='-'?"-":"")+s;} private static int Chartoint(char c){//字符化为数字if(c>='0'&&c<='9'){return c-'0';}else if(c>='A'&&c<='Z'){return c-'A'+10;}else if(c>='a'&&c<='z'){return c-'a'+36;}else if(c=='+'){//正号return -1;}else if(c=='-'){//负号return -2;}else if(c=='.'){//小数点return -3;}else{//其他return -4;}} private 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);}else if(n>=10&&n<=35){//10~35用A~Z表示return (char)(n+55);}else if(n>=36&&n<=61){//36~61用a~z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);}}顺便写了个测试类测试进制转换的效果与效率
public class TestBigNumber {public static void main(String[]args){String MyNumber="1313624242ADAFSFGDRG424242ADAF679707464342424242ADAFS24242ADAFSFGDRG424242ADAFFGDRG142356679707464342424242ADAFSFGDRG67866679707464342424242ADAFSFGDRG96068563556667970746V6679707464342424423536858686242ADAFSFGD69979RG67970746434242427979742ADAFS.FGDRG646676679707464342424242ADAFSFGDRG9707464342424242ADAFSFGDRG342424242ADAFSFGDRG67970746434266797074643426679707464342424242ADAFSFGDRG424242ADAFSFGDRG424242ADAFSFGDRG24242342426679707464342424242ADAFSFGDRG534364547455686679707464342424242ADAFSFGDRG6679707464342424242ADAFSFGDRGDADA6679707464342424242ADAFSFGDRGAFSFS6679707464342424242ADAFSFGDRG"; int R1=39,R2=62,R3=2; int f=MyNumber.length()-MyNumber.indexOf('.')-1; BigNumber b=new BigNumber(MyNumber,R1); BigNumber b1,b2,b3,b4;long c1=System.currentTimeMillis();b1=b.HexConversion(R2, f);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);long c2=System.currentTimeMillis();b2=b.HexConversion(R3, f);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);long c3=System.currentTimeMillis();b3=b1.HexConversion(R1, f);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);long c4=System.currentTimeMillis();b4=b2.HexConversion(R1, f);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);long c5=System.currentTimeMillis();System.out.println("四次转换分别耗时为"+(c2-c1)+"ms,"+(c3-c2)+"ms,"+(c4-c3)+"ms,"+(c5-c4)+"ms");}}测试结果:
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:CSicxuSbDy6b15mLRWpdQ03beazrJG5zcEBpMTZ7oxowiNnCN2V2M9xDYHbAF9g6yYEPhEPyQi1Jki1eamKZv9ceksCKkZzNjHDFv7wnmlCJSPVJtkFovcB65NUUYnHyMRYOMrt7wpQg1cy9UFqoVJOg6uicQZDDxdsxW7OdFtG60YEkAx3uuL6izr87hebmroc4ZHum8izmVRxylA1mARwFa.OVmjfCSTpwpEPSYDMTLpzKxV1LWaoEOdZV8Q5Ie2y8XeUDJ5M5Yh4QksANCxWcYlDW1v3XpvZ5AXC7XLlR5uyHK63kwPWpwJhLh6nm0ypGg1iwIWo1N7DP6JQEz3cP5kJx90WtGH6BOEmMgsRPC4cOnUYeS6Ckx8KmdNtMl1P2qN6cHsxZFYOgPg3MRHuNVMhuApscNlpkWlM11Yj5X8nAuLW140Pj4PEn5jgktOGO2y20LPgqM77GGz6d0PyF0H3EL7OwAVLVv1ARt4jlKbomwr1rRyqvV975kZtJVatHkNsr8PCLHGz63p59NIQRkXOd6cyROVnwCpRtmBi3dXU4Vlo0Z7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:110101101010110010111000101001101101000100001111010000100111110011011011000111010010000001100101000011100001001100010000111001010100011011101110010111011001011101001110011110100011101010100110111001010011011110011001011111001111000010010000010101110100010010000011111111011100010001110111111111111101110000110010001001110100100001010111100110001101110001001011110000101011000010100110100101111000110111111100110010101111111101011100110100100100101111011010001111001111111010101100100010110100001100101010111101010110100101000100110100011110001011000011110011001010001110000001001101110011011001011100110101011011101110000011010100001011100101111001000110110111010010100101100101011110101010001101100010000000000111100110010111011110010101011011001111100110000100010000011001011110101111001001001011111110110110110110101011100101100011111011011010100001000100100101010100111100111110110001010001000011001011000111111101100111011011110110001101101010000100101000110010100000110011111100001001101000000010101000000010110000001000101111001111111100010100000111110010010100000110011100100111001010001101010111010110100100011011000011100001001100110010010100110001010001011110101111100111110110000001111101110011100110101101101101011110010111100001001000001110100100000001011111100010101110110110.011001010011011010110001001001110110111110110101111110111110010011011101011111000001100101000110110001001011101011100010000101010010110101111011111101111100010100010100010101100001100100111111011110001101111111010111111010011001000110101011101000100111100010111100010110110110100001000001110011110010110110110101110111000100100111111100110010010011将62进制数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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将2进制数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:1313624242ADAFSFGDRG424242ADAF679707464342424242ADAFS24242ADAFSFGDRG424242ADAFFGDRG142356679707464342424242ADAFSFGDRG67866679707464342424242ADAFSFGDRG96068563556667970746V6679707464342424423536858686242ADAFSFGD69979RG67970746434242427979742ADAFS.FGDRG646676679707464342424242ADAFSFGDRG9707464342424242ADAFSFGDRG2UXNCMJNcZJJTAYBcNcPCOYPEHSET1EJ9c650HV24PNO1CDYUMXBEVT8P294CL2Lc81MQIRODQD6YUH3HVa7ADOUHW696PF2c0KU8XJJa6WYYASbGYRTUCCVO0BD8AGZEMSW2MWTCI3NbHFZW8GIC0FTRYNV8TCVAQF9N6AZHMDB5N16IWGUPHU812HN7AMT97LIUFX9DEZ5B841I6KMS3JGNMYVF79FXQHK8Y8DJKUUV3AaJVEWS80ZSaC0IKcVCXAaB5I3WBC2KAIGG0AIbR2Z2Ob四次转换分别耗时为468ms,169ms,195ms,75ms
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