Problem:
Score : 100 points

Problem Statement

Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins.
Find the probability of having more heads than tails.

Constraints

N is an odd number.

1 \leq N \leq 2999

p_i is a real number and has two decimal places.

0 < p_i < 1

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 \ldots p_N

Output

Print the probability of having more heads than tails.
The output is considered correct when the absolute error is not greater than 10^{-9}.

Sample Input 1

3
0.30 0.60 0.80

Sample Output 1

0.612

The probability of each case where we have more heads than tails is as follows:

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Head) is 0.3 × 0.6 × 0.8 = 0.144;

The probability of having (Coin 1, Coin 2, Coin 3) = (Tail, Head, Head) is 0.7 × 0.6 × 0.8 = 0.336;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Tail, Head) is 0.3 × 0.4 × 0.8 = 0.096;

The probability of having (Coin 1, Coin 2, Coin 3) = (Head, Head, Tail) is 0.3 × 0.6 × 0.2 = 0.036.

Thus, the probability of having more heads than tails is 0.144 + 0.336 + 0.096 + 0.036 = 0.612.

Sample Input 2

1
0.50

Sample Output 2

0.5

Outputs such as 0.500, 0.500000001 and 0.499999999 are also considered correct.

Sample Input 3

5
0.42 0.01 0.42 0.99 0.42

Sample Output 3

0.3821815872

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
pair<ll,ll> memo[3005][3005];
pair<ll,ll>prob[3005];
pair<ll,ll>solve(int idx,int tail,int n)
{
    if(2*tail>=n) return make_pair(0,1);
    if(idx==n) return make_pair(1,1);
    if(memo[idx][tail].first!=-1) return memo[idx][tail];
    
    pair<ll,ll>he=solve(idx+1,tail,n),ta=solve(idx+1,tail+1,n);
    he.first*=prob[idx].first;
    he.second*=prob[idx].second;
    ta.first*=(prob[idx].second-prob[idx].first);
    ta.second*=prob[idx].second;
    ll gg=__gcd(he.first,he.second);
    he.first/=gg;
    he.second/=gg;
    gg=__gcd(ta.first,ta.second);
    ta.first/=gg;
    ta.second/=gg;
    pair<ll,ll>ret;
    gg=__gcd(he.second,ta.second);
    ret.second=he.second*ta.second/gg;
    ret.first=( (ret.second*he.first)/he.second )+ ( (ret.second*ta.first)/ta.second);
    gg=__gcd(ret.first,ret.second);
    ret.first/=gg;
    ret.second/=gg;
    return memo[idx][tail]=ret;
    
}
int main()
{
    for(int i=0;i<=3000;i++)
    {
        for(int j=0;j<=3000;j++)
        {
            memo[i][j]=make_pair(-1,-1);
        }
    }
    int n;
    cin>>n;
    double num;
    for(int i=0;i<n;i++)
    {
        cin>>num;
        prob[i].first=num*100;
        prob[i].second=100;
        ll gg=__gcd(prob[i].first,prob[i].second);
        prob[i].first/=gg;
        prob[i].second/=gg;
        //cout<<prob[i].first<<"/"<<prob[i].second<<"--\n";
    }
    
    pair<ll,ll>ans=solve(0,0,n);
    printf("%lf\n",ans.first*1.0/ans.second);
}