Question

我在尝试添加两个＆＃34;大整数＆＃34;时遇到了问题。字节数组一起以相反的顺序存储以帮助数学。这是我的构造函数。

public class Intzilla {
public byte[] digits;
//private byte negative = -1;
private byte zero = 0;
private byte positive = 1;
private byte posNegZero = 0;
private boolean negative;
private String inputString;

public Intzilla() {
    this("0");
}

public Intzilla(String s) {
    String tempString = s;
    inputString = s;
    tempString = tempString.trim();
    if(tempString.substring(0,1).equals("-")){
        negative = true;
        tempString = tempString.substring(1);
    } else if(tempString.substring(0,1).equals("+")){
        negative = false;
        tempString = tempString.substring(1);
    }else {
        negative = false;
    }

    while((tempString.substring(0,1).equals("0")) && (tempString.length() > 1)){
        tempString = tempString.substring(1);
    }

    digits = new byte[tempString.length()];

    for(int i = 0; i < tempString.length(); i++){
        String currentChar = tempString.substring(i, i+1);
        byte tempDigits = Byte.parseByte(currentChar);
        digits[(digits.length - 1) -i] = tempDigits;
    }
}

到目前为止，这是我尝试过的plus方法。获得＆＃34;可能有损转换从转换为字节＆＃34;。

 public Intzilla plus(Intzilla addend) {
    byte carry = 0;
    byte mod = 10;
    Intzilla result = new Intzilla();

    for(int i = 0; i <= addend.digits.length-1; i++) {
        result.digits[i] = (byte)(this.digits[i] + addend.digits[i] + carry)% mod;
        carry = (byte)(this.digits[i] + addend.digits[i] + carry) / 10;
    }
    return result;
}

Answer 1

此格式有效

    byte [] newbytes = new byte [20];
    byte [] oldbytes = new byte [20];

    for (int x = 0; x < oldbytes.length; x++) {
        newbytes [x] = (byte) ((newbytes[x] + oldbytes [x]) / 100);
    }

将两个“Bigint”字节数组一起添加

1 个答案: