題解

聽說這是一道論文題orz

\(\sum_{k = 1}^{\infty} k(p^{k} - p^{k - 1})\)
答案是這個多項式的第\(2^N - 1\)項的系數
我們反演一下，卷積變點積
\(\hat{f_{S}} = \sum_{k = 1}^{\infty} k(\hat{p_{S}}^{k} - \hat{p_{S}}^{k - 1})\)

這是個等比數列啊，怎麽推呢= =

設答案為\(S\)
\(S = \infty \hat_{p}^{\infty} - \sum_{k = 0}^{\infty} \hat_{p}^{k}\)
\(\hat{p}S = \infty \hat{p}^{\infty} - \sum_{k = 1}^{\infty} \hat{p}^{k}\)

所以就有
\(\hat{f} = \left\{\begin{matrix} -\frac{1}{1 - \hat{p}} & \hat{p} < 1\\ 0 & \hat{p} = 1 \end{matrix}\right.\)

最後把F反演回去就行

代碼

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
 

#include <ctime>
#include <vector>
//#define ivorysi
#define MAXN 2000005
#define eps 1e-8
#define mo 974711
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef
 
 double db;
const int64 MOD = 998244353;
int N,L;
db P[MAXN],F[MAXN];
bool dcmp(db a,db b) {
    return fabs(a - b) < eps;
}
template <class T>
void FMT(T *a,T ty) {
    for(int i = 1 ; i < L ; i <<= 1) {
    for(int j = 0 ; j < L ; ++j) {
        if(j & i) {
        a[j] = a[j] + ty * a[j ^ i]; 
        }
    }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    scanf("%d",&N);
    L = 1 << N;
    for(int i = 0 ; i < L ; ++i) scanf("%lf",&P[i]);
    FMT(P,1.0);
    for(int i = 0 ; i < L ; ++i) {
    if(dcmp(1.0,P[i])) F[i] = 0;
    else F[i] = -1/(1 - P[i]);
    }
    FMT(F,-1.0);
    if(dcmp(F[L - 1],0)) puts("INF");
    else printf("%.6lf\n",F[L - 1]);
    return 0;
}

【LOJ】#2127. 「HAOI2015」按位或