杭电1018-Big Number（大数）

最新推荐文章于 2020-07-04 15:32:33 发布

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#阶乘 #数学 #斯塔林 #1018

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本文介绍了一个计算特定整数阶乘位数的方法，使用斯塔林公式简化计算过程，适用于加密等应用场景中需要处理非常大的整数的情况。

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 31165 Accepted Submission(s): 14464

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 ≤ 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

还记得那个神奇的斯塔林公式吗？

直接就是公式了！

#include<cstdio>
#include<cstring>
#include<cmath>
#define pi acos(-1.0)
using namespace std;
int n;
int solve()
{
	return (int)((n*log(n)-n+(log(2*pi*n))/2)/log(10))+1;
}
int main()
{
	int N;
	scanf("%d",&N);
	while(N--)
	{
		scanf("%d",&n);
		int t=solve();
		printf("%d\n",t);
	}
	return 0;
}