51nod 1130 N的阶乘的长度 V2（斯特林近似）

使用 S t i r l i n g Stirling Stirling 公式

n ! = 2 π n ( n e ) n n!=sqrt{2pi n}(dfrac{n}{e})^n n!=2πn ​(en​)n

⇒ l e n ( n ! ) = log ⁡ 10 ( n ! ) + 1 Rightarrow len(n!)=log_{10}(n!)+1 ⇒len(n!)=log10​(n!)+1

= 1 2 log ⁡ 10 ( 2 π n ) + n log ⁡ 10 ( n e ) + 1 =dfrac{1}{2}log_{10}(2pi n)+nlog_{10}(dfrac{n}{e})+1 =21​log10​(2πn)+nlog10​(en​)+1

时间复杂度： O ( 1 ) O(1) O(1)

#include
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
#define PLL pair
#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		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y){
	if(x
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("%fn",exp(1));
	double s = 0.5*log10(2*pi*n)+n*log10(n/exp(1)) +1;
	printf("%lldn",(ll)s);
	}
	return 0;
}