[BZOJ1426]收集邮票(概率期望dp)
题面
https://darkbzoj.tk/problem/1426
题解
记\(Pr(i,j)\)为买了i次，正好买到j种票的概率。（“正好”的含义是：第i次刚好解锁一种新的票）
则所求答案为\(\sum_{x}{\frac{x(x+1)}{2}}Pr(x,n)\)。
将其拆开，设\(\sum_x{x^2}Pr(x,i)=f_2[i],\sum_x{xPr(x,i)}=f_1[i]\)。
\[f_1[i+1]=\sum\limits_{x}xPr(x,i+1)
\]
\[=\sum\limits_{x}x{\sum\limits_{j=0}^{x-1}}Pr(j,i){\frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
\]
\[={\sum\limits_{j=0}^{+\infty}}Pr(j,i)\sum\limits_{x=j+1}^{+\infty}x{\frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
\]
\[={\sum\limits_{j=0}^{+\infty}}Pr(j,i)\sum\limits_{d=1}^{+\infty}(d+j){\frac{i^{d-1}*(n-i)}{n^{d}}}
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}\sum\limits_{d=1}^{+\infty}(d+j)(\frac{i}{n})^{d}
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}(j\sum\limits_{d=1}^{+\infty}(\frac{i}{n})^{d}+\sum\limits_{d=1}^{+\infty}d(\frac{i}{n})^d)
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}(j*{\frac{i}{n-i}}+{\frac{ni}{(n-i)^2}})
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i)(j+{\frac{n}{(n-i)}})
\]
由于\(\sum_{j=0}^{+\infty}Pr(j,i)j=f_1[i]\)以及\(\sum_{j=0}^{+\infty}Pr(j,i)=1\)，得到
\[f_1[i+1]=f_1[i]+{\frac{n}{n-i}}
\]
f2也类似：
\[f_2[i+1]=\sum\limits_{x}x^2Pr(x,i+1)
\]
\[=\sum\limits_{x}x^2{\sum\limits_{j=0}^{x-1}}Pr(j,i){\frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
\]
\[={\sum\limits_{j=0}^{+\infty}}Pr(j,i)\sum\limits_{x=j+1}^{+\infty}x^2{\frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
\]
\[={\sum\limits_{j=0}^{+\infty}}Pr(j,i)\sum\limits_{d=1}^{+\infty}(d+j)^2{\frac{i^{d-1}*(n-i)}{n^{d}}}
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}\sum\limits_{d=1}^{+\infty}(d^2+2dj+j^2)(\frac{i}{n})^{d}
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}(j^2\sum\limits_{d=1}^{+\infty}(\frac{i}{n})^{d}+ 2j\sum\limits_{d=1}^{+\infty}{d(\frac{i}{n})^d}+\sum\limits_{d=1}^{+\infty}d^2(\frac{i}{n})^d)
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}(j^2{\frac{i}{n-i}}+ 2j{\frac{ni}{(n-i)^2}}+\sum\limits_{d=1}^{+\infty}d^2(\frac{i}{n})^d)
\]
其中\(\sum_{d=1}^{+\infty}d^2(\frac{i}{n})^d\)可以用扰动法求解：
\[S=\sum\limits_{d=1}^{+\infty}d^2(\frac{i}{n})^d
\]
\[=\sum\limits_{d=0}^{+\infty}(d+1)^2(\frac{i}{n})^{d+1}
\]
\[=\frac{i}{n}\sum\limits_{d=0}^{+\infty}(d^2+2d+1)(\frac{i}{n})^d
\]
\[=\frac{i}{n}(\sum\limits_{d=0}^{+\infty}d^2(\frac{i}{n})^d+2\sum\limits_{d=0}^{+\infty}d(\frac{i}{n})^d+\sum\limits_{d=0}^{+\infty}(\frac{i}{n})^d)
\]
\[=\frac{i}{n}(S+2\times\frac{ni}{(n-i)^2}+\frac{n}{n-i})
\]
所以算出\(S=\frac{ni^2+n^2i}{(n-i)^3}\)。代入原式，可得
\[f_2[i+1]=\sum\limits_{j=0}^{+\infty}Pr(j,i){\times}{\frac{n-i}{i}}(j^2{\frac{i}{n-i}}+ 2j{\frac{ni}{(n-i)^2}}+\frac{ni^2+n^2i}{(n-i)^3})
\]
\[=\sum\limits_{j=0}^{+\infty}Pr(j,i)(j^2+\frac{2jn}{n-i}+\frac{ni+n^2}{(n-i)^2})
\]
\[=f_2[i]+\frac{2n}{n-i}f_1[i]+\frac{ni+n^2}{(n-i)^2}
\]
所以有了下面两个式子：
\[\begin{cases} f_1[i+1]=f_1[i]+{\frac{n}{n-i}} \\ f_2[i+1]=f_2[i]+\frac{2n}{n-i}f_1[i]+\frac{ni+n^2}{(n-i)^2} \end{cases}
\]
就可以展开递推求解\(f_1,f_2\)。最后的答案就是\(\frac{f_1[n]+f_2[n]}{2}\)。
代码
此题笔头工作很多，代码几乎没有
#include<bits/stdc++.h>
using namespace std;
#define rg register
#define In inline
const int N = 1e4;
In double sqr(double x){
return x * x;
}
int n;
double f1[N+5],f2[N+5];
int main(){
cin >> n;
for(rg int i = 0;i < n;i++){
f1[i+1] = f1[i] + 1.0 * n / (n - i);
f2[i+1] = f2[i] + f1[i] * 2 * n / (n - i) + (n * i + n * n) / sqr(n - i);
}
printf("%.2f\n",0.5 * (f1[n]+f2[n]));
return 0;
}
