//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&×!=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&×)
{
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;
}
posted on 2010-11-24 13:35
cometrue 阅读(361)
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