POJ 1423：Big Number 求N的阶乘的长度斯特林公式

Big Number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27027		Accepted: 8626

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

Sample Output

7

19

题意是要求N!有多少位。

因为有斯特林公式，所以求n！的位数即log10(n)=log10(sqrt(2*acos(-1.0)*n))+n*log10(n/exp(1.0));

代码：

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

#include <cstring>

#pragma warning(disable:4996)

using namespace std;

int main()

{

	int test;

	long long n;

	cin >> test;

	while (test--)

	{

		cin >> n;

		double re = log10(sqrt(2 * acos(-1.0)*n)) + n*log10(n / exp(1.0));

		cout << (int)re + 1 << endl;

	}

	return 0;

}

巴特西

POJ 1423：Big Number 求N的阶乘的长度斯特林公式

最新文章

热门文章

巴特西

POJ 1423：Big Number 求N的阶乘的长度 斯特林公式

最新文章

热门文章

POJ 1423：Big Number 求N的阶乘的长度斯特林公式