任何避免包含“9.223372036854776E18”的结果的方法

发布于 2024-11-26 13:58:04 字数 624 浏览 2 评论 0原文

我正在编写一个程序，将大量数字转换为字符串，然后将该字符串的字符相加。它工作正常，我唯一的问题是，Java 没有将其作为普通数字，而是将我的数字转换为标准形式，这使得解析字符串变得困难。对此有什么解决办法吗？

public static void main(String ags[]) {
    long nq = (long) Math.pow(2l, 1000l);
    long result = 0;
    String tempQuestion = Double.toString(nq);
    System.out.println(tempQuestion);
    String question = tempQuestion.substring(0, tempQuestion.length() - 2);
    for (int count = 0; count < question.length(); count++) {
        String stringResult = question.substring(count, count + 1);
        result += Double.parseDouble(stringResult);
    }
    System.out.println(result);

原文

I'm making a program that turns a large number into a string, and then adds the characters of that string up. It works fine, my only problem is that instead of having it as a normal number, Java converts my number into standard form, which makes it hard to parse the string. Are there any solutions to this?

public static void main(String ags[]) {
    long nq = (long) Math.pow(2l, 1000l);
    long result = 0;
    String tempQuestion = Double.toString(nq);
    System.out.println(tempQuestion);
    String question = tempQuestion.substring(0, tempQuestion.length() - 2);
    for (int count = 0; count < question.length(); count++) {
        String stringResult = question.substring(count, count + 1);
        result += Double.parseDouble(stringResult);
    }
    System.out.println(result);

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在你怀里撒娇 2024-12-03 13:58:05

BigInteger 易于使用，并且您不会面临精度问题的风险。（在这个特定的实例中，我认为不存在精度问题，因为 Math.pow(2, 1001) % 100000 返回正确的最后 5 位数字，但对于更大的数字，最终您将丢失信息.) 以下是如何使用 BigInteger：

groovy:000> b = new BigInteger(2L)
===> 2
groovy:000> b = b.pow(1001)
===> 214301721437253464189685009812000362112280962341106721488750077674070210224
98722449863967576313917162551893458351062936503742905713846280871969155149397149
60786913554964846197084214921012474228375590836430609294996716388253479753511833
1087892154125829142392955373084335320859663305248773674411336138752
groovy:000> ((b + "").toList().collect {new Integer(it)}).inject(0) {sum, n -> sum + n}
===> 1319

在 Java 中也是如此：

public class Example
{
  public static void main(String[] args)
  {
    int sum = 0;
    for (char ch : new java.math.BigInteger("2").pow(1001).toString().toCharArray()) {
      sum += Character.digit(ch, 10);
    }
    System.out.println(sum);
  }
}

BigInteger is easy to use and you don't risk precision problems with it. (In this particular instance I don't think there is a precision problem, because Math.pow(2, 1001) % 100000 returns the correct last 5 digits, but for bigger numbers eventually you will lose information.) Here's how you can use BigInteger:

groovy:000> b = new BigInteger(2L)
===> 2
groovy:000> b = b.pow(1001)
===> 214301721437253464189685009812000362112280962341106721488750077674070210224
98722449863967576313917162551893458351062936503742905713846280871969155149397149
60786913554964846197084214921012474228375590836430609294996716388253479753511833
1087892154125829142392955373084335320859663305248773674411336138752
groovy:000> ((b + "").toList().collect {new Integer(it)}).inject(0) {sum, n -> sum + n}
===> 1319

Here's the same thing in Java:

public class Example
{
  public static void main(String[] args)
  {
    int sum = 0;
    for (char ch : new java.math.BigInteger("2").pow(1001).toString().toCharArray()) {
      sum += Character.digit(ch, 10);
    }
    System.out.println(sum);
  }
}

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