機率論 - BZOJ - 4001 TJOI2015TJOI2015 Problem's Link

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Mean: 

求節點數為n的有根樹期望的葉子結點數.（n≤10^9）

analyse:

方案數就是卡特蘭數，$h_0=1, h_n = \sum_{i=0}^{n-1} h_i h_{n-1-i} \(。 設葉子數量和為\)f_n\(，則得到\)f_n = 2 \sum_{i=0}^{n-1} f_i h_{n-1-i}$

設\(H(x)\)表示\(h_n\)的母函數，\(F(x)\)表示\(f_n\)的母函數

容易得到:\[H(x) = x H^2(x) + 1\] \[F(x) = 2 x F(x) H(x) + x\]即:\[H(x) = \frac{1-\sqrt{1-4x}}{2x}\] \[F(x) = \frac{x}{1-\sqrt{1-4x}}\]發現\[(xH(x))' = \sum_{i=0}^{\infty} (i+1)h_i x^i = \frac{1}{\sqrt{1-4x}} = \frac{F(x)}{x}\]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\(f_i = i h_{i-1}\)

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

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;

}