概率论 - BZOJ - 4001 TJOI2015

TJOI2015 #

Problem's Link
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Mean: 

求节点数为n的有根树期望的叶子结点数.（n≤10^9）

analyse:

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$ans = \frac{f_n}{h_n} = \frac{n h_{n-1}}{h_n} = \frac{n(n+1)}{2(2n-1)}$

Time complexity: O(N)

 

view code

#include <bits/stdc++.h>
using namespace std;
typedef long double lf;
int main() {
   lf n;
   scanf("%Lf", &n);
   printf("%.9Lf\n", n*(n+1)/2/(n*2-1));
   return 0;
}