HDU 1018 Big Number 斯特林数近似n!

题目链接：点击打开链接

斯特林数：点击打开链接

题意是计算n! 的位数

即ans = log10(n!) = log10(sqrt(2πn)) + n*log10(n/e)

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e8;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) { putchar('-');x = -x; }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double Pi = acos(-1.);
const double e = 2.718281828459045;
double n;
int main(){
int T; rd(T);while(T--){   cin>>n;
		if(n == 1){puts("1");continue;}
		double ans = log10(sqrt(2.*Pi*n));
		ans += n * log10(n/e);
		cout<<fixed<<setprecision(0)<<ceil(ans)<<endl;
	}
    return 0;
}