hdu1018Big number(N！的位数-斯特林公式)_b - big number hdu

本文介绍了一种计算大整数阶乘位数的方法，利用斯特林公式解决了传统方法无法处理大整数的问题，并提供了C语言实现的示例代码。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31557 Accepted Submission(s): 14695

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

题目大意：给你个数N,求N的阶乘的位数？

解题思路：首先已知，一个数N的位数我们可用 log10(N) + 1求出；显然N!的位数在N非常大时是不能用一般方法解决的。

这里就要用到斯特林公式：

N！= sqrt (2*3.1415926*n)*(n/e)^n 即下面的公式:

那么就可以利用这个公式计算出N！的位数为：

num(N!) = log10 (2*sqrt(acos(-1.0)*N) + N*log10(N) - N*log(exp(1.0)) +1 ;

其中：log10() 、acos(x) 、exp(x).均为c语言的库函数，

log10(double x) 求以10为底X的对数。

acos(double x) 求余弦的反函数的值。

exp(double x) 求指数e^x的值。

那么代码就是：

#include<stdio.h>
#include<math.h>
int main()
{
	double n;
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
	   int num=0;
	   scanf("%lf",&n);
	   num=log10(sqrt(2*acos(-1.0)*n)) + n*log10(n) - n*log10(exp(1.0));
	   printf("%d\n",num+1);
	} 
return 0;
}