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: Accepted)

#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<(int)(n);i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

double dp[3001][3001];

int main(){
    int N; cin >> N;
    vector<double> p(N);
    for(int i=0;i<N;i++) cin >> p[i];

    // dp[i][j] := i枚中j枚が表を向く確率
    dp[0][0] = 1;
    for(int i=0;i<N;i++){
        for(int j=0;j<=i;j++){
            dp[i+1][j+1] += dp[i][j] * (p[i]); 
            dp[i+1][j]   += dp[i][j] * (1-p[i]);
        }
    }

    double ans = 0;
    for(int i=(N+1)/2;i<=N;i++){
        ans += dp[N][i];
    }
    cout << Double(ans) << endl;
}