[牛客OI测试赛2]F假的数学游戏(斯特灵公式)

题意

输入一个整数X，求一个整数N，使得N!恰好大于$X^X$。

Sol

考试的时候只会$O(n)$求$N!$的前缀和啊。

不过最后的结论挺好玩的

$n! \approx \sqrt{2 \pi n} (\frac{n}{e})^n$

然后就可以$O(1)$算啦

/*

*/

#include<iostream>

#include<cstdio>

#include<cstring>

#include<set>

#include<algorithm>

#include<map>

#include<cmath>

#define Pair pair<int, int>

#define fi first

#define se second

#define MP(x, y) make_pair(x, y)

#define LL long long

const LL MAXN = 1e8 + , mod = , inv = ;

using namespace std;

inline LL read() {

    char c = getchar(); LL x = , f = ;

    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}

    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();

    return x * f;

}

const double pi = acos(-), e = exp(1.0);

LL N, X;

double up;

bool check(double n) {

    return 0.5 * log( * pi * n) + n * log(n / e) >= up;

}

int main() {

    X = read();

    /*if(X == 7) {printf("10"); return 0;}

    if(X == 77) {printf("94"); return 0;}

    if(X == 777) {printf("892"); return 0;}

    if(X == 7777) {printf("8640"); return 0;}

    if(X == 77777) {printf("84657"); return 0;}

    if(X == 777777) {printf("834966"); return 0;}

    if(X == 7777777) {printf("8267019"); return 0;}

    if(X == 77777777) {printf("82052137"); return 0;}

    if(X == 777777777) {printf("815725636"); return 0;}

    if(X == 7777777777ll) {printf("8092563686"); return 0;}*/

    up = X * log(X);

//    for(LL i = 1; i <= 1e8; i++) lg[i] = log(i), s[i] = s[i - 1] + lg[i];

    //cout << lg[10];

    int times = ;

    double l = , r = 1e13, ans;

    while(times--) {

        LL mid = (l + r) / ;

        if(check(mid)) ans = mid, r = mid;

        else l = mid;

    }

    cout << (long long)ans;

    return ;

}

/*

2

4 6

4 6

*/