51nod 1130 n factorial length V2 (Stirling approximation)

51nod 1130 N The length of the factorial of V2（ Stirling approximation ）

Use $Stirling$ The formula

$n!=\sqrt{2\pi n}(\frac{n}{e}{)}^n$

$\Rightarrow len(n!)={\log }_{10}(n!)+1$

$=\frac{1}{2}{\log }_{10}(2\pi n)+n{\log }_{10}(\frac{n}{e})+1$

Time complexity ：$O(1)$

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e3+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = {
    402653189,805306457,1610612741,998244353};
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n){
    
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\n",a[n]); 
}
template <typename T>		//x=max(x,y) x=min(x,y)
void cmx(T &x,T y){
    
	if(x<y) x=y;
}
template <typename T>
void cmn(T &x,T y){
    
	if(x>y) x=y;
}
int n;
const db pi = acos(-1.0);
int main(){
    
	int T;scanf("%d",&T);
	while(T--){
    
	scanf("%d",&n);
	//printf("%f\n",exp(1));
	double s = 0.5*log10(2*pi*n)+n*log10(n/exp(1)) +1;
	printf("%lld\n",(ll)s);
	}
	return 0;
}