BZOJ3000 Big Number

由Stirling公式：

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

故：$$egin{align} ans &= log_k n! + 1 \ &approx log_k [sqrt{2 pi n} (frac{n}{e})^n] + 1 \ &= frac{1}{2} log_k 2 pi n + n * (log_k n - log_k e) + 1\ end {align}$$

又$log_a b = frac{log a}{log b}$

而且要注意n比较小的时候近似值差别比较大。。。可以直接暴力。。。

/**************************************************************
    Problem: 3000
    User: rausen
    Language: C++
    Result: Accepted
    Time:28 ms
    Memory:816 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
 
using namespace std;
typedef long double Lf;
typedef long long ll;
const Lf pi = acos(-1.0);
const Lf e = exp(1);
const Lf eps = 1e-10;
 
int n, k;
Lf ans;
 
Lf log(Lf x, Lf y) {
    return log(x) / log(y);
}
 
int main() {
    int i;
    while (scanf("%d%d", &n, &k) != EOF) {
        if (n <= 10000) {
            for (ans = 0.0, i = 1; i <= n; ++i) ans += log(i);
            ans /= log(k);
            printf("%.0Lf
", ceil(ans + eps));
        } else
        printf("%lld
", (ll) (0.5 * log(2 * pi * n, k) + n * log(n, k) - n * log(e, k)) + 1);
    }
}