The naïve polynomial expectation solution

By the expected linearity of

can be added together

So x^3 x^2 K separate consideration

X^2 x^3 See Bzoj 4318:osu! Expected DP

Let's talk a little bit about K.

Each state can be viewed as a 01-string

For each k change must be at 0 end or 1 end

So you can offset it every time.

But of course there are exceptions.

When the first and last one is the same, it will not be able to offset

So just add that to the situation.

 #include <bits/stdc++.h> using namespace std; typedef long long ll; inline int read () {int x=0
    , F=1;char Ch=getchar (); while (ch< ' 0 ' | |
    Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}
    while (ch<= ' 9 ' &&ch>= ' 0 ') {x= (x<<3) + (x<<1) +ch-' 0 '; Ch=getchar ();}
return x*f;
} int N; 
Double K,p[1000010],l1[1000010],l2[1000010],n1[1000010],f[1000010][2],ans;
    int main () {register int i,j;
    scanf ("%d%lf", &n,&k);
        for (int i=1;i<=n;i++) {scanf ("%lf", &p[i]);
        l1[i]= (l1[i-1]+1) *p[i];
        n1[i]= (n1[i-1]+1) * (1-p[i]);
        l2[i]= (l2[i-1]+2*l1[i-1]+1) *p[i];
        f[i][1]=f[i-1][1]+ (3*l2[i-1]+3*l1[i-1]+1) *p[i];
    f[i][2]=f[i-1][2]-(2*n1[i-1]+1) * (1-p[i]);
    } Ans+=f[n][1]+f[n][2];
    ans=ans+p[1]*p[n]*k-(1-p[1]) * (1-p[n]) *k;
    printf ("%lf\n", ans);
return 0; }