【HackerRank Week of Code 31】Colliding Circles

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。

[egin{aligned} E(n-1)=&E(n)+frac1{nchoose2}sum_{0leq i<jleq n}2r_ir_j\ =&E(n)+frac1{nchoose2}left[left(sum r_i ight)^2-sum r_i^2 ight]\=&E(n)+frac1{nchoose2}left[left(sum r_i ight)^2-E(n) ight]\ 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
", E[n] * acos(-1));
	
	return 0;
}