Mengikut prinsip menukar perpuluhan kepada perduaan, simulasi membahagi dengan 2 dan mengambil baki

Sepatutnya tidak terlalu sukar untuk menulis algoritma sendiri, dan biarkan ia di latar belakang untuk diproses

private static String process(ArrayList<String> imp,String HexStr){
        String rem = "";
        Long remaind = (long) 0;
        ArrayList<String> newList = new ArrayList<String>();
        Iterator<String> itor =  imp.iterator();
        while(itor.hasNext()){
            String ele = itor.next();
            Long num = Long.parseLong(rem+ele);
            Long quot = num / 16;
            remaind = num%16;
            rem = remaind.toString();
            if(quot > 0){ 
                newList.add(padString(quot.toString(),16,"0",0));
            }
        };
        String hexRem = Long.toHexString(remaind);
        if(newList.isEmpty()){
            return hexRem;
        } else{
            return process(newList,hexRem)+hexRem;
        }
        
    }

    public static String D2Hex(String dNum){
        String tmp = dNum;
        ArrayList<String> splitNum = new ArrayList<String>();
        while(tmp.length() > 16){
            String high = tmp.substring(0,tmp.length() - 16);
            splitNum.add(high);
            String low = tmp.substring(tmp.length() - 16);
            tmp = low;
        }
        splitNum.add(tmp);
        return process(splitNum,"");
    }
    public static String D2B(String dNum){
        String hex = D2Hex(dNum);
        StringBuilder bin= new StringBuilder();
        for(int i =0; i < hex.length(); i ++){        
            String tmp = Integer.toBinaryString( Integer.parseInt(hex.substring(i,i+1),16));
            bin.append(padString(tmp,4,"0",0));
        }
        return bin.toString();
    }
    
    
    

Hasil nombor super besar tidak disahkan
0------>0--->0000
135,9234------>14bd82--->0001010010111101109,90010111101109,90999,9099 9999 ------>2386f26fc0ffff--->0010001110000110111100100110111111 0000001111111111111111
0,00-t 2386f26fc10 000--->00100011100001101111001001101111111000001000000000000000
9,9999,9999, 9999,9999 ------>16345785d89ffff--->0001011000110100010101111000010111011000100111111110001101000101011110000101110110001001111111111111110000100111111111111111000100111111111111111000100111111111111111999,9999,9999 9,9999,9999,99. 01111100000001111 111111111111111111111111111111

Tulis apa sahaja yang anda mahukan

function toBin(str)
{
  var arr = [];
  var remainder,i,str2,num,char;
  while(str.length>0) {
    str2 = "";remainder=0;
    for(i=0;i<str.length;i++) { // str2 = str组成的十进制数 / 2
      num = str.charCodeAt(i)-0x30; // num to String
      num = remainder*10 + num;
      char = Math.floor(num/2).toString();
      // 忽略最高为的0 ， 即最高为如果是 0 则不放入 str2
      if(!(char === "0" && str2 === "")) { str2 += char;}
      remainder = num%2;
    }
    str = str2;
    arr.push(remainder); // 保存余数
  }
  return arr.reverse().join('');
  
}
console.log(toBin("3")); // 11
console.log(toBin("9")); // 1001
console.log(toBin("10")); // 1010
console.log(toBin("120")); // 1111000

Saya nak tanya, super besar, berapa besarnya

Jika ia hanya untuk paparan, anda boleh mempertimbangkan untuk membiarkannya dipindahkan di latar belakang