来自lyd课件
发现s和last(s),next(s)成树结构，然后把式子化简成kx+b的形式，做树形dp即可
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,t,a[105];
double p,q;
struct qwe
{
double k,b;
qwe(double K=0,double B=0)
{
k=K,b=B;
}
};
qwe dfs(int s,int mx)
{
if(s>t)
return qwe(0,0);
double ne,b=0,k=0,r=1;
ne=1.0/(double)mx;
for(int i=1;i<=mx;i++)
{
qwe nw=dfs(s+a[i],i);
b=b+nw.b,k=k+nw.k;
}
q=s?p:0.0;
r=1.0-(1.0-q)*ne*k;
k=q/r;
b=((1.0-q)*ne*b+1)/r;
return qwe(k,b);
}
int main()
{
while(~scanf("%lf%d%d",&p,&t,&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
printf("%.3lf
",dfs(0,n).b);
}
return 0;
}