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;
}