[笔记] 斯特林公式

[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{limlimits_{n ightarrow infty}frac{n!}{sqrt{2 pi n}left(frac{e}{n} ight)^n}=1} ]

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

Code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
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
", l);
	return 0;
}
原文地址:https://www.cnblogs.com/LMSH7/p/9605391.html