https://www.hackerrank.com/contests/w31/challenges/colliding-circles/problem
设E(n)为序列长度为n时的期望值。
\[\begin{aligned}
E(n-1)=&E(n)+\frac1{n\choose2}\sum_{0\leq i<j\leq n}2r_ir_j\\
=&E(n)+\frac1{n\choose2}\left[\left(\sum r_i\right)^2-\sum r_i^2\right]\\=&E(n)+\frac1{n\choose2}\left[\left(\sum r_i\right)^2-E(n)\right]\\
\end{aligned}
\]
E(n-1)=&E(n)+\frac1{n\choose2}\sum_{0\leq i<j\leq n}2r_ir_j\\
=&E(n)+\frac1{n\choose2}\left[\left(\sum r_i\right)^2-\sum r_i^2\right]\\=&E(n)+\frac1{n\choose2}\left[\left(\sum r_i\right)^2-E(n)\right]\\
\end{aligned}
\]
\(O(n)\)递推。
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 100003;
double a[N], sum = 0, E[N];
int n, k;
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; ++i) scanf("%lf", &a[i]), sum += a[i], E[n] += a[i] * a[i];
for (int i = 1; i <= k; ++i) {
E[n - 1] = (1.0 - (2.0 / n / (n - 1))) * E[n] + sum * sum * 2 / n / (n - 1);
--n;
}
printf("%.10lf\n", E[n] * acos(-1));
return 0;
}