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 + 10, mod = 100000007, inv = 50000004;
using namespace std;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
const double pi = acos(-1), e = exp(1.0);
LL N, X;
double up;
bool check(double n) {
    return 0.5 * log(2 * 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 = 51;
    double l = 1, r = 1e13, ans;
    while(times--) {
        LL mid = (l + r) / 2;
        if(check(mid)) ans = mid, r = mid;
        else l = mid;
    }
    cout << (long long)ans;
    return 0;
}
/*
2
4 6
4 6
*/