编码为Java的阶乘23

Question

我需要最多写23个阶乘！我最多可以做20个阶乘！但之后我迷路了，因为人数太大了。 我无法使用BigInteger。

我必须存储最右边的数字为0，所以示例输出：

10！ = 3628800-> fac = 36288，num10 = 2

23！ = 2585..976640000-> fac = 2585..97664，num10 = 4

import java.util.Scanner;

public class Factorial10{
    public static void main(String[] args){
        long fac;       // long: factorial is very large
        long pre_fac;       // to check overflow
        int i, n;
        int num10;

        Scanner sc = new Scanner(System.in);

        System.out.print("n? ");
        n = sc.nextInt();

        // Start from fac = 0! = 1
        for(i= 1, fac= 1L; i<n; i++){
            pre_fac = fac;
            fac *= i;

            // Checking if it is multiple of 10, 
            // and adjust the variables fac and num10

            //while((fac % 10) == 0){
            //  num10++;
            //  fac /= 10;
            //}

            // check if overflowed
            if(pre_fac != fac /i){
                System.out.println("Overflowed at " + i + "! = " + fac);
                fac = pre_fac;      // roll back to the previous, unoverflowed
                break;
            }
        }

        System.out.println((i-1) + "! = " + fac + "(fac = , num10 = )");
    }
}
```

Answer 1

当n大时，您可以说：

n！ = sqrt（2 * pi n）（n / e）^ n

https://fr.wikipedia.org/wiki/Factorielle

Answer 2

选择float，而不是long

    double fac;      
    double pre_fac;