FACVSPOW

http://www.spoj.com/problems/FACVSPOW/

求解n! > a^n最小的整数n

对于有n!和a^n的东西，一般是取ln

然后就是求解

(ln(1) + ln(2) + .... + ln(n)) / n > ln(a)的最小整数n

发现左边的函数单调，所以可以预处理出来，右边最大值是ln(1e6)

所以预处理5e6个。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 5e6 + 20;
double f[maxn];
void init() {
    for (int i = 1; i <= maxn - 20; ++i) {
        f[i] = (f[i - 1] * (i - 1) + log(i * 1.0)) / (i * 1.0);
    }
//    for (int i = 1; i <= 100; ++i) {
//        cout << f[i] << endl;
//    }
}
int a;
double up;
const double eps = 1e-12;
bool bigger(double a, double b) {
    return a - b > eps;
}
void work() {
    scanf("%d", &a);
    up = log(a * 1.0);
    int be = 1, en = maxn - 20;
    while (be <= en) {
        int mid = (be + en) >> 1;
        if (bigger(f[mid], up)) {
            en = mid - 1;
        } else be = mid + 1;
    }
    printf("%d
", be);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    init();
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}

据说 

ln(n!) ≈ (n * ln(n)) - n + (1 / 2 * ln(2 * PI * n));