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。。。

 1 /**************************************************************
 2     Problem: 3000
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:28 ms
 7     Memory:816 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cmath>
12
13 using namespace std;
14 typedef long double Lf;
15 typedef long long ll;
16 const Lf pi = acos(-1.0);
17 const Lf e = exp(1);
18 const Lf eps = 1e-10;
19
20 int n, k;
21 Lf ans;
22
23 Lf log(Lf x, Lf y) {
24     return log(x) / log(y);
25 }
26
27 int main() {
28     int i;
29     while (scanf("%d%d", &n, &k) != EOF) {
30         if (n <= 10000) {
31             for (ans = 0.0, i = 1; i <= n; ++i) ans += log(i);
32             ans /= log(k);
33             printf("%.0Lf\n", ceil(ans + eps));
34         } else
35         printf("%lld\n", (ll) (0.5 * log(2 * pi * n, k) + n * log(n, k) - n * log(e, k)) + 1);
36     }
37 }