在Java中计算大于int和long的数字的阶乘? [英] Calculating factorials with numbers bigger than ints and longs in java?

问题描述

开始在这里和Google搜索几天，然后问我的编程朋友. 不幸的是，我仍然不知道如何更改我的代码...

Been searching here and google for a couple of days as well as asking my programming friends. Unfortunately, i still don't understand how to change my code...

我的程序计算给定数字的阶乘.然后提供一个数字，该数字表示阶乘答案包括的位数.然后将这些数字的值相加，得出一个总数.

My program calculates the factorial of a given number. It then provides a number which represents how many digits the factorials answer includes. Then it sums the values of those digits together to give a total.

我的程序可用于1之间的任何数字！和31！...如果您输入超过31！ (例如50！或100！)，它不起作用，只会返回减号而没有总数.

My program works for any number between 1! and 31!... If you put in anything over 31! (for example 50! or 100!) it doesn't work and just throws back minus numbers and no total.

我希望你们能为我指出正确的方向或给我一些建议. 我知道使用BigIntegers也许是一种解决方案，但是我个人并不了解它们，因此请来到这里.

I was hoping you guys could point me in the right direction or give me some advice. I understand using BigIntegers maybe a solution however i don't understand them personally, hence coming here.

任何帮助将不胜感激.谢谢.

Any help would be much appreciated. Thanks.

    package java20;

    /**
    * Program to calculate the factorial of a given number.
    * Once implemented, it will calculate how many digits the answer includes.
    * It will then sum these digits together to provide a total.
     * @author shardy
     * date: 30/09/2012
     */

    //import java.math.BigInteger;
    public class Java20 {

    /**
    * @param args the command line arguments
    */
    public static void main(String[] args) {

    //Using given number stored in factorialNo, calculates factorial
    //currently only works for numbers between 1! and 31! :(
        int fact= 1;
        int factorialNo = 10;

        for (int i = 1; i <= factorialNo; i++)
            {
               fact=fact*i;
            }

        System.out.println("The factorial of " + factorialNo + 
                " (or " + factorialNo + "!) is: " + fact);

        //Using answer stored in fact, calculates how many digits the answer has
        final int answerNo = fact;
        final int digits = 1 + (int)Math.floor(Math.log10(answerNo));

        System.out.println("The number of digits in the factorials "
                + "answer is: " + digits);        

        //Using remainders, calculates each digits value and sums them together
        int number = fact;
        int reminder;
        int sum = 0;

        while(number>=1)
            {
             reminder=number%10; 
             sum=sum+reminder;
             number=number/10;
            }

        System.out.println("The total sum of all the " + digits 
                + " idividual digits from the answer of the factorial of " 
                + factorialNo + " is: " + sum);

      }
    }

推荐答案

您可以在Java中使用BigInteger，它可以根据需要使用任意数量的数字

You can use BigInteger in java, it has as much numbers as you want

    BigInteger fact= BigInteger.ONE;
    int factorialNo = 10;

    for (int i = 2; i <= factorialNo; i++){
      fact = fact.multiply(new BigInteger(String.valueOf(i)));
    }

    System.out.println("The factorial of " + factorialNo +
                                " (or " + factorialNo + "!) is: " + fact);
   final int digits = fact.toString().length();

   BigInteger number = new BigInteger(fact.toString());
   BigInteger reminder;
   BigInteger sum = BigInteger.ZERO;
   BigInteger ten = new BigInteger(String.valueOf(10));

   while(number.compareTo(BigInteger.ONE)>=0)
     {
     reminder=number.mod(ten);
     sum=sum.add(reminder);
     number=number.divide(ten);
     }

     System.out.println("The total sum of all the " + digits
                     + " idividual digits from the answer of the factorial of "
                     + factorialNo + " is: " + sum

编辑:对代码进行了改进，使其与作者的代码兼容

EDIT: the code is improved to be compatible with author's code

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