随笔档案

log10(n!) = 0.5 * log10(2*pi*n) + n * log10(n/e).

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define e 2.71828182
#define pi acos(-1.0)
int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        double n;
        scanf("%lf",&n);
        double t = 0.5 * log10(2.0*pi*n) + n*log10(n*1.0/e);
        printf("%d\n",(int)t + 1);   
    }
    return 0;   
}