poj1423（斯特林公式）

最新推荐文章于 2019-10-13 15:43:43 发布

胖头鱼爱算法

最新推荐文章于 2019-10-13 15:43:43 发布

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分类专栏： # 算法竞赛-数论文章标签：斯特林公式

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本文介绍了一个使用斯特林公式来计算任意大整数阶乘位数的方法，并提供了一段C++代码实现，该算法可以高效地解决在安全传输、加密等领域中遇到的大整数阶乘位数计算问题。

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Big Number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28472		Accepted: 9055

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

斯特林公式

#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
ll steling(ll n)
{
	return ll(log10(sqrt(4*acos(0.0)*n))+n*log10(n/exp(1.0)))+1;
//	return ll(log10(sqrt(4*acos(0.0)*n))+n*log10(n/exp(1.0)))+1;
	
//	res=(long)( (log10(sqrt(4.0*acos(0.0)*n)) + n*(log10(n)-log10(exp(1.0)))) + 1 );
}
int main()
{
	int n;
	cin>>n;
	while(n--){
		int m;
		cin>>m;
		if(m==1)
			cout<<1<<endl;
		else
		cout<<steling(m)<<endl;
	}
	return 0;
}