「HAOI2015」按位或

「HAOI2015」按位或

解题思路 :

这类期望题一眼 \(\text{Min-Max}\) 容斥，只需要稍微推一下如何求 \(E(minS)\) 即可。
\[ E(minS) = \frac{1}{\sum_{T \cap S\neq \emptyset} p_T} \\ = 1 - \frac{1}{\sum_{T \cap S = \emptyset}p_T} \\ = 1 - \frac{1}{\sum_{T \cap (U-S) = T}p_T} \\ = 1 - \frac{1}{\sum_{T \subseteq (U-S)}p_T} \]
对 \(p\) 做莫比乌斯变换得到：
\[ p'_S=\sum_{T \subseteq S} p_T \\ E(minS) = 1-\frac{1}{p'_{(U-S)}} \]
然后直接 \(\text{Min-Max}\) 容斥就做完了，总复杂度 \(O(n2^n)\)。

code

/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
    int ch = 0, f = 0; x = 0;
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
    for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    if(f) x = -x;
}
const int N = 2000005;
const double eps = 1e-6;
double p[N], ans;
int cnt[N], n;
int main(){
    read(n);
    for(int i = 0; i < (1 << n); i++) scanf("%lf", &p[i]);
    for(int i = 0; i < n; i++)
        for(int s = 0; s < (1 << n); s++)
            if((1 << i) & s) p[s] += p[s^(1<<i)], cnt[s]++;
    for(int i = 0; i < n; i++)
        if(1.0 - p[(1<<n)-(1<<i)-1] < eps) return puts("INF"), 0;
    for(int s = 0; s < (1 << n); s++){
        double res = 1.0 - p[(1<<n)-s-1];
        if(res > eps) ans += (1.0 / res) * (cnt[s] & 1 ? 1.0 : -1.0);
    }
    printf("%.10lf", ans);
    return 0;
}