hdu1018 Big Number stirling公式

Stirling公式：n!与sqrt(2πn) * n^n * e^(-n)的值十分接近

所以log10(n!) = log(n!) / log(10) = ( n*log(n) - n + 0.5*log(2*π*n))/log(n);

代码：

#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <map>
#include <set>
#define sss(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a) memset(a,0,sizeof(a))
#define ss(a,b) scanf("%d%d",&a,&b)
#define s(a) scanf("%d",&a)
#define p(a) printf("%d\n", a)
#define INF 0x3f3f3f3f
#define w(a) while(a)
#define PI acos(-1.0)
#define LL long long
#define eps 10E-9
#define N 100000
#define mod 100000000
using namespace std;
void mys(int& res)
{
int flag=0;
char ch;
while(!(((ch=getchar())>='0'&&ch<='9')||ch=='-'))
if(ch==EOF)  res=INF;
if(ch=='-')  flag=1;
else if(ch>='0'&&ch<='9')  res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')  res=res*10+ch-'0';
res=flag?-res:res;
}
void myp(int a)
{
if(a>9)
myp(a/10);
putchar(a%10+'0');
}
/*************************THE END OF TEMPLATE************************/
double stirling(int n){
return ceil((n*log(double(n))-n+0.5*log(2.0*n*PI))/log(10.0));
}
int main()
{
int cas,n;
scanf("%d",&cas);
while(cas--)
{
scanf("%d",&n);
if(n<=1)printf("1\n");
else printf("%.0lf\n",stirling(n));
}
return 0;
}