hdu 1018

一：log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)

二：n! = sqrt(2*π*n) * ((n/e)^n) * (1 +1/(12*n) + 1/(288*n*n) + O(1/n^3))
π = acos(-1)=3.1415927;
e = exp(1)=2.718281828459;
两边对10取对数
忽略log10(1 + 1/(12*n) +1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0
得到公式
log10(n!) = log10(sqrt(2 * pi * n)) + n* log10(n / e)。

一：

#include<stdio.h>#include<math.h>int main(){    int i,n,t;    double temp=0;    scanf("%d",&t);    while(t--)    {        temp=0;        scanf("%d",&n);        for(i=1;i<=n;i++)            temp+=log10(i);        printf("%d\n",(int)temp+1);    }    return 0;}

二：

#include"stdio.h"#include"math.h"int main(){    int n;    double temp;    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        temp=0.5*log10(2*3.1415927*n)+n*log10(n/2.718281828459);        printf("%d\n",(int)temp+1);    }    return 0;}