P - Big Number

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

2
10
20

Sample Output

7
19 

给你一个数n，要你求他的阶乘有几位数字组成；一看到这题就条件反射的觉得要用大数才能解决，结果折腾了老久，竟然不对。然后去看大神的代码，要不是最近天气热，没吃啥东西，差点一口老血喷将出来；尼玛我敲了好长的代码，他给我就简简单单的几行，就用了一个函数log10（）来自变量的位数，而这题给的是阶乘，比如说log10(3!)=log(1*2*3)=log(1)+log(2)+log(3);一个函数，一个for循环，丫的就解决的干干精简。这就是和大神的差距啊。不过也算赚到了。利用log10（）的特点求位数 。   

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int a[100][100000];
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int n;cin>>n;double sum=0;//在这里sum由于要进行相加求和，只能是double，而在求单独一个
        for(int i=1;i<=n;i++)//数的位数时可以用int来盛装
        {
            sum+=log10(i);
        }
        cout<<(int)sum+1<<endl;
    }
}
除此之外，这道题还可以用斯特林公式来求

//此题必须用公式才能AC,数据规模太大了
//Stirling公式:    n! = ((2*pi*n)^(1/2))*((n/e)^n); 前提是n > 3
//由此可以导出lg(n!) = (lg(2*pi)+lg(n))/2 + n*(lg(n)-lg(e)); 
#include<iostream>
#include<cmath>

using namespace std;

int main()
{
	int test,n;
	long long int s;
	const long double c1 = 0.798179868358; //lg(2*pi)
	const long double c2 = 0.434294481903; //lg(e)
	long double c3;
	cin >> test;
	while(--test+1)
	{
		cin >> n;
		c3 = log10((double)n);
		s = 1;
		if(n > 3)
		{
			s = (c3 + c1)/2 + n * (c3 - c2) + 1;
			cout << s << endl;
		}
		else
			cout << 1 << endl; 
	}
	return 0;
}

据说这个公式可以解决和阶乘相关的一切问题，我表示太强了