[笔记] 斯特林公式

先看一道题

题目大意:求一个\(N\)的阶乘恰好大于\(x^x\)

用斯特林公式算出位数,然后二分查找,一定注意精度

Detail

\[\displaystyle{\lim\limits_{n \rightarrow \infty}\frac{n!}{\sqrt{2 \pi n}\left(\frac{e}{n}\right)^n}=1}\]

\[\displaystyle{\frac{\lg{2 \pi n}}{2}+n\lg{\frac{e}{n}}}\]

Code

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;const double Pi = acos(-1.0);const double e = exp(1.0);typedef long long ll;long long x;int main(){    cin >> x;    ll l = 1, r = 100000000000000L, mid;//字面值太大要加个L保险    while(l < r){        mid = l + ((r - l) >> 1);        if(double(double(log10(sqrt(2 * Pi * mid))) + double(mid * double(log10(mid / e)))) < x*double(log10(x)))            l = mid + 1;        else             r = mid;    }    printf("%lld\n", l);    return 0;}