任何长度超过long的整数的数据类型?

9rbhqvlz  于 2021-06-26  发布在  Java
关注(0)|答案(3)|浏览(557)

我想把二进制转换成整数,乘以17,然后再转换成二进制。这是我的密码:

Scanner scan = new Scanner(System.in);
String n = scan.nextLine();
long j = Long.parseLong(n, 2);
j = j * 17;
System.out.println(Long.toBinaryString(j));

我最初做了一个int,但是当我得到一个更大的测试用例时,我改变了它:

10001111110001000101000001000100111100110101100011000011011001111000100110110000110101110101100001001100010111000101000100010010011000000010010

它有一个numberformatexception,这是有意义的,因为long只能存储有限数量的数字,所以对于非常长的整数有数据类型吗?

pinkon5k

pinkon5k1#

使用 BigInteger 当计算超过 long ,例如。

String input = "10001111110001000101000001000100111100110101100011000011011001111000100110110000110101110101100001001100010111000101000100010010011000000010010";

// parse binary string
BigInteger num1 = new BigInteger(input, 2);

// multiply by 17
BigInteger num2 = num1.multiply(BigInteger.valueOf(17));

// format as binary string
String output = num2.toString(2);

System.out.println(output);

输出

100110001100000010010101010010010100001010001110010011111001111000000010010010111110010011001101110100010010001000010110001000111000011000100110010
de90aj5v

de90aj5v2#

例如:

import org.junit.Test;

import java.util.function.Function;

public class BinTest
{
    String binNo1 = "100011111100010001010000010001001111001101011000";
    String binNo2 = "10001111110001000101000001000100111100110101100011000011011001111000100110110000110101110101100001001100010111000101000100010010011000000010010";

    @Test
    public void testIt()
    {
        //System.out.println( bin17A( binNo1 ) );
        System.out.println( bin17S( binNo2 ) );
    }

    public static String bin17S( String bin )
    {
        // * 16
        String bin16 = bin + "0000";
        String bin01 = "0000" + bin;
        StringBuilder result = new StringBuilder();

        Function<Character, Integer> parser = c -> (c == '1') ? 1 : 0;

        int carry = 0;
        for ( int i = bin16.length() - 1; i >= 0; i-- )
        {
            int value = parser.apply( bin16.charAt( i ) )
                    + parser.apply( bin01.charAt( i ) )
                    + carry;

            carry = value / 2;

            result.insert(0, value % 2 );
        }
        while (carry > 0)
        {
            result.insert(0,carry % 2 );
            carry = carry / 2;
        }
        return result.toString();
    }

    public static String bin17A( String bin )
    {
        long j = Long.parseLong( bin, 2 );
        j = j * 17;
        return Long.toBinaryString( j );
    }

}
7z5jn7bk

7z5jn7bk3#

你试过biginteger还是bigdecimal。
https://www.baeldung.com/java-bigdecimal-biginteger
这两种类型特别适用于要求数字具有大范围或任意范围的情况,如某个值>或=到1x10^307和小于1x10^307

public void whenBigDecimalCreated_thenValueMatches() {
    BigDecimal bdFromString = new BigDecimal("0.1");
    BigDecimal bdFromCharArray = new BigDecimal(new char[] {'3','.','1','6','1','5'});
    BigDecimal bdlFromInt = new BigDecimal(42);
    BigDecimal bdFromLong = new BigDecimal(123412345678901L);
    BigInteger bigInteger = BigInteger.probablePrime(100, new Random());
    BigDecimal bdFromBigInteger = new BigDecimal(bigInteger);

    assertEquals("0.1",bdFromString.toString());
    assertEquals("3.1615",bdFromCharArray.toString());
    assertEquals("42",bdlFromInt.toString());
    assertEquals("123412345678901",bdFromLong.toString());
    assertEquals(bigInteger.toString(),bdFromBigInteger.toString());
}

那应该对你有帮助。

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