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51nod 1130 N的阶乘的长度 V2(斯特林近似)


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{2\pi 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}(2\pi n)+n\log_{10}(\dfrac{n}{e})+1 =21log10(2πn)+nlog10(en)+1

时间复杂度: O ( 1 ) O(1) 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;
}



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