hdu4336 Card Collector 【最值反演】

题目链接

hdu4336

题解

最值反演

也叫做\(min-max\)容斥，在计算期望时有奇效

\[max\{S\} = \sum\limits_{T \in S} (-1)^{|T| + 1}min\{T\} \]

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

\[\begin{aligned} \sum\limits_{T \in S} (-1)^{|T| + 1}min\{T\} &= \sum\limits_{i = 1}^{n}a_i\sum\limits_{j = 0}^{n - i}(-1)^{j}{n - i \choose j}\\ &= \sum\limits_{i = 1}^{n}a_i[n - i = 0]\\ &= \sum\limits_{i = 1}^{n}a_i[n = i]\\ &= a_n\\ &= max\{S\} \end{aligned} \]

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

\[\frac{1}{\sum\limits_{i \in T}p_i} \]

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\n",ans);
	}
	return 0;
}