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]=sumlimits_{x}xPr(x,i+1)
]
[=sumlimits_{x}x{sumlimits_{j=0}^{x-1}}Pr(j,i){frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
]
[={sumlimits_{j=0}^{+infty}}Pr(j,i)sumlimits_{x=j+1}^{+infty}x{frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
]
[={sumlimits_{j=0}^{+infty}}Pr(j,i)sumlimits_{d=1}^{+infty}(d+j){frac{i^{d-1}*(n-i)}{n^{d}}}
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}sumlimits_{d=1}^{+infty}(d+j)(frac{i}{n})^{d}
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}(jsumlimits_{d=1}^{+infty}(frac{i}{n})^{d}+sumlimits_{d=1}^{+infty}d(frac{i}{n})^d)
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}(j*{frac{i}{n-i}}+{frac{ni}{(n-i)^2}})
]
[=sumlimits_{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]=sumlimits_{x}x^2Pr(x,i+1)
]
[=sumlimits_{x}x^2{sumlimits_{j=0}^{x-1}}Pr(j,i){frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
]
[={sumlimits_{j=0}^{+infty}}Pr(j,i)sumlimits_{x=j+1}^{+infty}x^2{frac{i^{x-j-1}*(n-i)}{n^{x-j}}}
]
[={sumlimits_{j=0}^{+infty}}Pr(j,i)sumlimits_{d=1}^{+infty}(d+j)^2{frac{i^{d-1}*(n-i)}{n^{d}}}
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}sumlimits_{d=1}^{+infty}(d^2+2dj+j^2)(frac{i}{n})^{d}
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}(j^2sumlimits_{d=1}^{+infty}(frac{i}{n})^{d}+ 2jsumlimits_{d=1}^{+infty}{d(frac{i}{n})^d}+sumlimits_{d=1}^{+infty}d^2(frac{i}{n})^d)
]
[=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}(j^2{frac{i}{n-i}}+ 2j{frac{ni}{(n-i)^2}}+sumlimits_{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=sumlimits_{d=1}^{+infty}d^2(frac{i}{n})^d
]
[=sumlimits_{d=0}^{+infty}(d+1)^2(frac{i}{n})^{d+1}
]
[=frac{i}{n}sumlimits_{d=0}^{+infty}(d^2+2d+1)(frac{i}{n})^d
]
[=frac{i}{n}(sumlimits_{d=0}^{+infty}d^2(frac{i}{n})^d+2sumlimits_{d=0}^{+infty}d(frac{i}{n})^d+sumlimits_{d=0}^{+infty}(frac{i}{n})^d)
]
[=frac{i}{n}(S+2 imesfrac{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]=sumlimits_{j=0}^{+infty}Pr(j,i){ imes}{frac{n-i}{i}}(j^2{frac{i}{n-i}}+ 2j{frac{ni}{(n-i)^2}}+frac{ni^2+n^2i}{(n-i)^3})
]
[=sumlimits_{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：
[egin{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
",0.5 * (f1[n]+f2[n]));
return 0;
}