斯特林公式&1018 Big Number

斯特林公式是一条用来取n的阶乘的近似值的数学公式。一般来说，当n很大的时候，n阶乘的计算量十分大，所以斯特林公式十分好用，而且，即使在n很小的时候（除1以外），斯特林公式的取值已经十分准确。
求n!的位数
利用斯特林（Stirling）公式的进行求解。下面是推导得到的公式：
　　res=(long)( (log10(sqrt(4.0*acos(0.0)n)) + n(log10(n)-log10(exp(1.0)))) + 1 );

Big Number

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

#include<bits/stdc++.h>using namespace std;using LL=int64_t;int main(){    LL T;    cin>>T;    for(int i=0;i<T;i++) {        int n;        cin>>n;        if(n==1) cout<<1<<endl;        else {            double res=(long)( (log10(sqrt(4.0*acos(0.0)*n)) + n*(log10(n)-log10(exp(1.0)))) + 1 );            cout<<(int)res<<endl;        }    }    return 0;}