求N!的位数

//求N!的位数

//N!=1*2*3**N，两边取常用对数，即可算出log10(N!),向上取整即为N!的位数
//hdoj 984MS 344K

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	double sum=0.0;
	int n,i,times,res;
	if(cin>>times&&times!=0)
	{
		while(times)
		{
			cin>>n;
			for(i=2;i<=n;++i)
				sum+=log10(i);
			res=ceil(sum);
			cout<<res<<endl;
			sum=0.0;
			--times;
		}
	}
	return 0;
}

//String公式的方法，N!~sqrt(2*pi*N)*(N/e)^N
//hdoj 0MS 360K
#include <iostream>
#include <cmath>
using namespace std;
const double pi=3.1415926;
int main()
{
	int n,times;
	long double sum;
	if(cin>>times&&times)
	{
		while(times)
		{
			cin>>n;
			sum=(long double)0.5*log10(2*pi*n)+(long double)n*(log10(n)-log10(exp(1)));
			cout<<(long)ceil(sum)<<endl;
			--times;
		}
	}
	return 0;
}