题目链接
题解
最值反演
也叫做(min-max)容斥，在计算期望时有奇效
[max{S} = sumlimits_{T in S} (-1)^{|T| + 1}min{T}
]
证明：
记(S = {a_i})，其中对于(i < j)有(a_i < a_j)
那么我们计算每一个(a_i)的贡献，有
[egin{aligned}
sumlimits_{T in S} (-1)^{|T| + 1}min{T} &= sumlimits_{i = 1}^{n}a_isumlimits_{j = 0}^{n - i}(-1)^{j}{n - i choose j}\
&= sumlimits_{i = 1}^{n}a_i[n - i = 0]\
&= sumlimits_{i = 1}^{n}a_i[n = i]\
&= a_n\
&= max{S}
end{aligned}
]
证毕
对于此题，我们想要求出所有卡片集合(S)中最晚出现的卡片的期望，就转化为计算(S)的所有子集中最早出现的期望
对于一个集合(T)，显然出现一张卡片的期望为
[frac{1}{sumlimits_{i in T}p_i}
]
枚举子集计算即可
复杂度(O(2^n))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 25,maxm = 100005,INF = 1000000000;
double p[maxn],ans;
int n;
int main(){
while (~scanf("%d",&n)){
REP(i,n) scanf("%lf",&p[i]);
int maxv = (1 << n) - 1; ans = 0;
for (int s = 1; s <= maxv; s++){
int cnt = 0; double tmp = 0;
for (int e = s,j = 1; e; e >>= 1,j++)
if (e & 1) cnt++,tmp += p[j];
ans += (cnt & 1) ? 1 / tmp : -1 / tmp;
}
printf("%.8lf
",ans);
}
return 0;
}