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;
- }
来源: http://www.bubuko.com/infodetail-2672710.html