题意
求节点数为(n)的有根树期望的叶子结点数。((n le 10^9))
分析
神题就打表找规律..
题解
方案数就是卡特兰数,$h_0=1, h_n = sum_{i=0}^{n-1} h_i h_{n-1-i} (。
设叶子数量和为)f_n(,则得到)f_n = 2 sum_{i=0}^{n-1} f_i h_{n-1-i}(
设)H(x)(表示)h_n(的母函数,)F(x)(表示)f_n(的母函数
容易得到:)(H(x) = x H^2(x) + 1)$ $$F(x) = 2 x F(x) H(x) + x$$即:$$H(x) = frac{1-sqrt{1-4x}}{2x}$$ $$F(x) = frac{x}{1-sqrt{1-4x}}$$发现$$(xH(x))' = sum_{i=0}^{infty} (i+1)h_i x^i = frac{1}{sqrt{1-4x}} = frac{F(x)}{x}$$即$$F(x) = sum_{i=0}^{infty} (i+1)h_i x^{i+1} = sum_{i=1}^{infty} i h_{i-1} x^i = sum_{i=0}^{infty} f_i x^i$$即(f_i = i h_{i-1})
所以(ans = frac{f_n}{h_n} = frac{n h_{n-1}}{h_n} = frac{n(n+1)}{2(2n-1)})
#include <bits/stdc++.h>
using namespace std;
typedef long double lf;
int main() {
lf n;
scanf("%Lf", &n);
printf("%.9Lf
", n*(n+1)/2/(n*2-1));
return 0;
}