hdu 1018 Big Number

题意：给你一个m，求m!的位数。

数学证明：

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

对公式两边取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!的位数为： b + 1 = log10(1)+ log10( 2)  + log10(3) + . . . . . .  + log10(m) - log10(a) + 1。

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

（去掉 log10(a) < 1，毕竟a不好求，且10>a>=1  ）。

测试例子：

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  位数( m! ) = b + 1；

Java AC 代码：

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);}}}

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31092    Accepted Submission(s): 14427

Problem Description

In many applications very large integers numbers are required. 

在许多的应用程序中，需要处理非常大的整数。

Some of these applications are using keys for secure transmission of data, encryption, etc. 

例如：这些应用程序使用安全密尺来传输加密数据，等等。

In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

在这道题中，你能得到一个整数，你必须去判断这个数的阶乘有多少位。

Input

Input consists of several lines of integer numbers. 

输入语句将包含多个。

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.

 输入的第一行是一个整数n，表示测试事件的个数，接着有n行输入，（1 ≤ n ≤ 10^7）。

Output

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

 把输入数的阶乘位数输出，在单独的一行显示。

Sample Input

21020

 

Sample Output

719

 

Source

Asia 2002, Dhaka (Bengal)

 

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