HDOJ題目地址：傳送門

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34843    Accepted Submission(s): 16536

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 2 10 20
Sample Output 7 19

題意：給你一個m，求m!的位數。

數學證明：

假設m! = a * 10 ^ b ，求m! 的位數（10>a>=1,b∈R)。

對公式兩邊取log10()；

得到： log10(a * 10 ^ b) = log10(m!)；

化解公式：log10(a) + log10(10 ^ b) = log10(1 * 2 *3 * ....... * m)；

-------->>  log10(a) + b = log10(1)+ log10( 2)  + log10(3) + . . . . . .  + log10(m)；

所以m!的位數為： b + 1 = log10(1)+ log10( 2)  + log10(3) + . . . . . .  + log10(m) - log10(a) + 1。

因此可以近似的寫成 b + 1 <= log10(1)+ log10( 2)  + log10(3) + . . . . . .  + log10(m)  + 1 

（去掉 log10(a) < 1，畢竟a不好求，且10>a>=1  ）。

測試例子：

1! = 1 = 1.0 * 10 ^ 0------------------------------------------------------------------- 1位 ->log10(1) = 0.0

2! = 2 = 2.0 * 10 ^ 0------------------------------------------------------------------- 1位 ->log10(2) = 0.3010299956639812

3! = 6 = 6.0 * 10 ^ 0------------------------------------------------------------------- 1位 ->log10(3) = 0.47712125471966244

4! = 24 = 2.4 * 10 ^ 1----------------------------------------------------------------- 2位 ->log10(4) = 0.6020599913279624

5! = 120 =  1.2 * 10 ^ 2--------------------------------------------------------------- 3位 ->log10(5) = 0.6989700043360189

6! = 720 = 7.2 * 10 ^ 2---------------------------------------------------------------- 3位 ->log10(6) = 0.7781512503836436

7! = 5040 =  5.04 * 10 ^ 3------------------------------------------------------------ 4位 ->log10(7) = 0.8450980400142568

8! = 40320 = 4.032 * 10 ^ 4--------------------------------------------------------- 5位 ->log10(8) = 0.9030899869919435

9! = 362880 = 3.6288 * 10 ^ 5；--------------------------------------------------- 6位 ->log10(9) = 0.9542425094393249

10！= 3628800 = 3.6288 * 10 ^ 6；---------------------------------------------- 7位 ->log10(10) = 1.0

log10(3.6288 * 10 ^ 6) = log10(3.6288) + log10(10*6) = 0.5597630328767937 + 6；

所以  位數( m! ) = b + 1；

import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
   public static void main(String[] args) {
	   Scanner sc = new Scanner(System.in);  
       int n = sc.nextInt();  
       int m = 0, answer;  
       while (n-- > 0) {  
           m = sc.nextInt();  
           double sum = 0.0;  
           for (int i = 1; i <= m; i++) {  
               sum += Math.log10(i * 1.0);  
           }  
           answer = (int) sum + 1;  
           System.out.println(answer);  
	   }
   }
}