InputThe first line of each test case contains one integer N (1 <= N <= 20), indicating the number of different cards you need the collect. The second line contains N numbers p1, p2, ..., pN, (p1 + p2 + ... + pN <= 1), indicating the possibility of each card to appear in a bag of snacks.

Note there is at most one card in a bag of snacks. And it is possible that there is nothing in the bag.OutputOutput one number for each test case, indicating the expected number of bags to buy to collect all the N different cards.

You will get accepted if the difference between your answer and the standard answer is no more that 10^-4.Sample Input

1

0.1

2

0.1 0.4

Sample Output

10.000

10.500

#include<bits/stdc++.h>

using namespace std;

const int maxn=;

double p[maxn],ans; int N;

void dfs(int pos,double now,int opt)

{

    if(pos==N+) {

        if(opt>){

           if(opt&) ans+=1.0/now;

           else ans-=1.0/now;

        }

        return ;

    }

    dfs(pos+,now,opt);

    dfs(pos+,now+p[pos],opt+);

}

int main()

{

    while(~scanf("%d",&N)){

        for(int i=;i<=N;i++) scanf("%lf",&p[i]);

        ans=; dfs(,0.0,);

        printf("%.4lf\n",ans);

    }

    return ;

}