题目大意：共$2n$个价格$p_i$。两人轮流取。你每次取最大的，对方每次随机取。问你取的期望和是多少。
题解：从小到大排序，$\sum\limits_{i=0}^{2n-1} \frac{i*p_i}{2n-1}$
卡点：无
C++ Code：
#include<cstdio>
#include<algorithm>
using namespace std;
long long n,s[200005];
long long ans;
int main(){
scanf("%lld",&n);
for (int i=0;i<n*2;i++)scanf("%lld",&s[i]);
sort(s,s+n*2);
for (long long i=0;i<n*2;i++)ans+=i*s[i];
printf("%.10lf",ans/(double)(2*n-1));
return 0;
}
Atcoder Code Festival 2017 Team Relay J - Indifferent
原文：https://www.cnblogs.com/Memory-of-winter/p/9269853.html