BZOJ3000 Big Number

由Stirling公式：

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

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

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();
 const Lf eps = 1e-;

 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 <= ) {
             for (ans = 0.0, i = ; i <= n; ++i) ans += log(i);
             ans /= log(k);
             printf("%.0Lf\n", ceil(ans + eps));
         } else
         printf("%lld\n", (ll) (0.5 * log( * pi * n, k) + n * log(n, k) - n * log(e, k)) + );
     }
 }