HODJ Big Number（java 阶乘）

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24822    Accepted Submission(s): 11266

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

 

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

 

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

 

Sample Input

21020

 

Sample Output

719
超时代码：以为用大数可以解决，结果发现求千万数量级的阶乘内存也无法受得了
import java.math.BigInteger;import java.util.Scanner;public class Main {            public static void main(String[] args) {                Scanner in = new Scanner(System.in);        int n = in.nextInt();        while(n-->0){            String str = in.next();            BigInteger number = new BigInteger(str);            BigInteger sum = BigInteger.ONE;            while(number.intValue()>0){                                sum = sum.multiply(number);                number = number.subtract(BigInteger.ONE);                            }                        System.out.println(sum.toString().length());                    }            }}

AC代码1：利用数学中对数公式可以求解出，lg(N!)=lg(1)+lg(2)+…+lg(N)
import java.util.Scanner;public class Main {        public static void main(String[] args) {                Scanner in = new Scanner(System.in);        int n = in.nextInt();        while(n-->0){            double d = in.nextDouble();            double sum = 0;            for(int i=1; i<=d; i++){                    sum += Math.log10(i);            }                System.out.println((int)sum+1);                    }            }}

AC代码2：采用斯特林公式计算能够在时间上快速解决
链接：http://baike.baidu.com/view/4113061.htm?fr=aladdin
import java.util.Scanner;public class Main {        public static void main(String[] args) {                Scanner in = new Scanner(System.in);        int n = in.nextInt();        while(n-->0){            int number = in.nextInt();            double sum1 = Math.log10(Math.sqrt(2*Math.PI*number));            double sum2 = number*Math.log10(number/Math.E);            System.out.println((int)(sum1+sum2+1));                    }            }}