POJ-1423 计算出n的阶乘的位数大数问题[Stirling公式]

Big Number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21661		Accepted: 6888

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 <= m <= 10^7 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

Source

Dhaka 2002

 

题意：计算出n的阶乘的位数

 

 1 #include <stdio.h>

 2 #include <math.h>

 3 

 4 int n;

 5 const double e = 2.7182818284590452354, pi = 3.141592653589793239;

 6 

 7 double f( int a )

 8 {

 9     return log10( sqrt( 2 * pi * a ) ) + a * log10( a / e );

10 }

11 

12 int main()

13 {

14     int cas, ans;

15     double i, s;

16     

17     scanf( "%d", &cas );

18     

19     while( cas-- )

20     {

21         scanf( "%d", &n );

22         if( n < 100000 )

23         {

24             for( s=0, i=1; i<=n; i++ )

25                 s += log10( i );

26         }

27         else s = f( n );

28         ans = (int)s;

29         if( ans <= s )

30             ans++;

31         

32         printf( "%d/n", ans );

33     }    

34     return 0;

35 }

 

 

注：

Stirling公式介绍：

lim(n→∞) (n/e)^n*√(2πn) / n! = 1

也就是说当n很大的时候,n!与√(2πn) * (n/e) ^ n的值十分接近

这就是Stirling公式.

 

公式的证明

令a(n)=n! / [ n^(n+1/2) * e^(-n) ]

 

　　则a(n) / a(n+1) = (n+1)^(n+3/2) / [ n^(n+1/2) * (n+1) * e ]

 

　　=(n+1)^(n+1/2) / [ n^(n+1/2) * e]

 

　　=(1+1/n)^n * (1+1/n)^1/2 *1/e

 

　　当n→∞时,(1+1/n)^n→e,(1+1/n)^1/2→1

 

　　即lim(n→∞) a(n)/a(n+1)=1

 

　　所以lim(n→∞)a(n) 存在

 

　　设A=lim(n→∞)a(n)

 

　　A=lim(n→∞)n! / [ n^(n+1/2) * e^(-n) ]

 

　　利用 Wallis公式,π/2 = lim(n→∞)[ (2n)!! / (2n-1)!! ]^2 / (2n+1)

 

　　π/2 = lim(n→∞)[ (2n)!! / (2n-1)!! ]^2 / (2n+1)

 

　　=lim(n→∞)[ (2n)!! * (2n)!! / (2n)! ]^2 / (2n+1)

 

　　=lim(n→∞) 2^(4n) [ (n!)^2 / (2n)! ]^2 / (2n+1)

 

　　=lim(n→∞) 2^(4n) [ (A * n^(n+1/2) * e^(-n) )^2 / (A * (2n)^(2n+1/2) * e^(-2n) )]^2 / (2n+1)

 

　　=lim(n→∞) 2^(4n) [ 2^(-2n-1/2) * A * √n ]^2 / (2n+1)

 

　　=lim(n→∞) 2^(4n) * A^2 * 2^(-4n-1) * n/(2n+1)

 

　　=A^2 / 4

 

　　所以A=√(2π)

    

     lim(n→∞)n! / [ n^(n+1/2) * e^(-n) ] = √(2π)

    

     即lim(n→∞) √(2πn) * n^n * e^(-n) / n! = 1

 

公式的意义：

　Stirling公式的意义在于:当n足够大之后n!计算起来十分困难,虽然有很多关于n!的不等式,但并不能很好的对阶乘结果进行估计,尤其是n很大之后,误差将会非常大.但利用Stirling公式可以将阶乘转化成幂函数,使得阶乘的结果得以更好的估计.而且n越大,估计得就越准确.